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		<id>https://www.astro-init.top/index.php?title=2022-2023%E5%AD%A6%E5%B9%B4CNAO%E9%A2%84%E8%B5%9B%E9%80%89%E6%8B%A9%E9%A2%98&amp;diff=2755</id>
		<title>2022-2023学年CNAO预赛选择题</title>
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		<updated>2025-03-27T19:56:07Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==试题==&lt;br /&gt;
&lt;br /&gt;
01. 今晚的月相更接近( )。&lt;br /&gt;
&lt;br /&gt;
A. 朔 B. 望 C. 下弦月 D.残月&lt;br /&gt;
&lt;br /&gt;
02. 以下哪颗恒星不属于“冬季大六边形“?( )&lt;br /&gt;
&lt;br /&gt;
A. 五车二 B. 参宿四 C. 南河三 D. 毕宿五&lt;br /&gt;
&lt;br /&gt;
03.M31是( )。&lt;br /&gt;
&lt;br /&gt;
A. 星云 B. 星团 C. 星系 D. 恒星&lt;br /&gt;
&lt;br /&gt;
04. “笔尖上的行星”是( )。&lt;br /&gt;
&lt;br /&gt;
A. 地球 B. 火星 C. 土星 D. 海王星&lt;br /&gt;
&lt;br /&gt;
05. 最早使用望远镜进行天文观测的科学家是( )。&lt;br /&gt;
&lt;br /&gt;
A. 哥白尼 B. 牛顿 C. 伽利略 D. 开普勒&lt;br /&gt;
&lt;br /&gt;
06. 象限仪流星雨的辐射点与以下哪个星座最近?( )&lt;br /&gt;
&lt;br /&gt;
A. 玉夫座 B. 武仙座 C. 狮子座 D. 六分仪座&lt;br /&gt;
&lt;br /&gt;
07. 天文学是自然科学六大基础学科之一，其主要的三个分支学科包括:天体测量学、天体力学、( )。&lt;br /&gt;
&lt;br /&gt;
A. 天体化学 B. 天体物理学 C. 天体生物学 D. 天文地理学&lt;br /&gt;
&lt;br /&gt;
08. 地月距离大约是( )个天文单位。&lt;br /&gt;
&lt;br /&gt;
A. 0.00001 B. 0.0025 C. 0.05 D. 1&lt;br /&gt;
&lt;br /&gt;
09. 北极星(小熊座 α)的目视亮度在全天恒星中排( )。&lt;br /&gt;
&lt;br /&gt;
A. 第 1 名 B. 第 5 名 C. 第 20 名左右 D. 第 50 名左右&lt;br /&gt;
&lt;br /&gt;
10. 以下节气(中气)出现在农历“正月”里概率最大的是( )。 &lt;br /&gt;
&lt;br /&gt;
A. 小寒 B. 大寒 C. 立春 D. 雨水&lt;br /&gt;
&lt;br /&gt;
11. 以下哪台已有或在建望远镜的有效口径最大?( )&lt;br /&gt;
&lt;br /&gt;
A. E-ELT B. GTC C. 昴星团望远镜 D. HET&lt;br /&gt;
&lt;br /&gt;
12. 在地球上任意一个位置，以下情况一定正确的是( )。&lt;br /&gt;
&lt;br /&gt;
A. 一年里每次太阳上中天的方位角相同。&lt;br /&gt;
&lt;br /&gt;
B. 一年里每次日出时(日心位于地平线上)太阳的方位角相同。 &lt;br /&gt;
&lt;br /&gt;
C. 一年里每次日出(日面出地平至完全出地平)的用时相同。 &lt;br /&gt;
&lt;br /&gt;
D. 其他几个选项都不正确。&lt;br /&gt;
&lt;br /&gt;
13. 在位于四川省甘孜藏族自治州稻城县的无名山天文台(100.108E,29.107N)，观测者看 到一颗流星自北向南划过消失在地平线附近，在以下哪个天文台站有可能观测到同一颗流星?( )&lt;br /&gt;
&lt;br /&gt;
A. 西藏阿里(80.027E,32.326N) &lt;br /&gt;
&lt;br /&gt;
B. 新疆慕士塔格(74.897E,38.330N)&lt;br /&gt;
&lt;br /&gt;
C. 云南高美谷(100.029E,26.694N) &lt;br /&gt;
&lt;br /&gt;
D. 河北兴隆(117.577E,40.396N)&lt;br /&gt;
&lt;br /&gt;
14. 下次日食将发生在( )。&lt;br /&gt;
&lt;br /&gt;
A. 2023 年 4 月 20 日 B. 2023 年 10 月 27 日 C. 2023 年 11 月 23 日 D. 2024 年 4 月 9 日&lt;br /&gt;
&lt;br /&gt;
15. 昨晚日落后不久，不考虑天气和遮挡的影响，你所在地能看到最亮的行星是?( )&lt;br /&gt;
&lt;br /&gt;
A. 水星 B. 金星 C. 木星 D. 天王星&lt;br /&gt;
&lt;br /&gt;
16. 以下恒星中哪个看上去颜色更偏红?( )&lt;br /&gt;
&lt;br /&gt;
A. 参宿七 B. 织女星 C. 角宿一 D. 心宿二&lt;br /&gt;
&lt;br /&gt;
17. 以下哪个梅西叶天体不属于室女星系团?( )&lt;br /&gt;
&lt;br /&gt;
A. M60 B. M84 C. M87 D. M101&lt;br /&gt;
&lt;br /&gt;
18. 2022 年 7 月 12 日，詹姆斯韦布望远镜公开了首批科学图像，其中( )的科学图像中只包括光谱观测结果，而不包括成像观测结果。&lt;br /&gt;
&lt;br /&gt;
A. 船底座大星云“宇宙悬崖” &lt;br /&gt;
&lt;br /&gt;
B. WASP-96b 系外行星&lt;br /&gt;
&lt;br /&gt;
C. 南指环星云 &lt;br /&gt;
&lt;br /&gt;
D. SMACSJ0723 星系团&lt;br /&gt;
&lt;br /&gt;
19. 彗星 C/2022E3(ZTF)于 2023 年 1 月 12 日过近日点，当时位于后发座。该彗星轨道偏心率接近1，轨道倾角 109°。2223 年 4 月 1 日该彗星最可能位于以下哪个星座附近?( ) &lt;br /&gt;
&lt;br /&gt;
A. 天龙座 B. 天鹰座 C. 天鹤座. D. 天鹅座&lt;br /&gt;
&lt;br /&gt;
20. 以下哪一颗小天体不是由业余天文学家 Gennadiy Borisov 发现的?( ) &lt;br /&gt;
&lt;br /&gt;
A. 2023 BU B. C/2019 Q4 C. C/2017 K2 D. C/2013 V2&lt;br /&gt;
&lt;br /&gt;
21. 某次上弦月恰逢月亮位于白道对黄道的升交点，在某地月亮位于上中天时观测，月亮视 面被照亮部分占整个月面的比例为( )。&lt;br /&gt;
&lt;br /&gt;
A. 小于等于 40% B. 大于 40%小于 50% C. 等于 50% D. 大于 50%小于 60%&lt;br /&gt;
&lt;br /&gt;
22. 在以下太阳系的天然卫星中，轨道运行周期比其母行星的自转周期还短的是( )。&lt;br /&gt;
&lt;br /&gt;
A. 木卫一 B. 木卫四 C. 火卫一 D. 火卫二&lt;br /&gt;
&lt;br /&gt;
23. “荧惑守心”是中国史籍中的一种天象记载，其中的“荧惑”指的是( )。 &lt;br /&gt;
&lt;br /&gt;
A. 天狼星 B. 土星 C. 火星 D. 木星&lt;br /&gt;
&lt;br /&gt;
24. 在中国，有一种载人航天飞船叫“神舟”;有一种载人空间站叫“天宫”;有一种中继通信卫星叫“鹊桥”;有一种暗物质探测器叫“悟空”，下面哪个暂时还不是中国航天项目 的名称?( )&lt;br /&gt;
&lt;br /&gt;
A. 洛神 B. 嫦娥 C. 羲和 D. 祝融&lt;br /&gt;
&lt;br /&gt;
25. 2023 年是比利时耶稣会士南怀仁诞辰 600 周年，下列哪架天文仪器是南怀仁设计监制的?( )&lt;br /&gt;
&lt;br /&gt;
A. 玑衡抚辰仪 B. 纪限仪 C. 浑仪 D. 水运仪象台&lt;br /&gt;
&lt;br /&gt;
26. 在距离太阳 200 天文单位的地方释放一个石块，若仅考虑太阳引力作用，它需要多长时间落到太阳表面?( )&lt;br /&gt;
&lt;br /&gt;
A. 5 年 B. 50 年 C. 500 年 D. 5000 年&lt;br /&gt;
&lt;br /&gt;
27. 假设一个视星等为 18 等的星系的光全部均匀分布在视半径为 6 角秒的圆形区域中，其面 视星等是( )等/平方角秒。&lt;br /&gt;
&lt;br /&gt;
A. 13 B. 18 C. 23 D. 28&lt;br /&gt;
&lt;br /&gt;
28. 假设宇宙匀速膨胀，已知今天的哈勃常数 H0=70 km/s/Mpc，请问宇宙的年龄大约为? ( )&lt;br /&gt;
&lt;br /&gt;
A. 1 亿年 B. 10 亿年 C. 100 亿年 D. 1000 亿年&lt;br /&gt;
&lt;br /&gt;
29. 在刘慈欣的知名科幻小说《三体》中，我们的邻居比邻星上存在着比地球文明更高级的 三体文明。三体人由于生存环境恶劣，准备向地球迁移。假设三体舰队从比邻星出发，一直以 1/10 光速的速度向地球航行，走了比邻星到地球距离的 9/10 后，由于故障原地暂停。此时他们看到的太阳亮度( )。&lt;br /&gt;
&lt;br /&gt;
A. 变亮 1 个星等 B. 变亮 5 个星等 C. 变亮 10 个星等 D. 变亮 100 个星等&lt;br /&gt;
&lt;br /&gt;
30. 比邻星到我们的距离是 4.2 光年，在地球上看，它的周年视差是多少?( ) &lt;br /&gt;
&lt;br /&gt;
A. 1.58 角秒 B. 0.24 角秒 C. 0.77 角秒 D. 4.2 角秒&lt;br /&gt;
&lt;br /&gt;
31. 视向速度为 1000 km/s 的超高速恒星对应的红移/蓝移是多少?( ) &lt;br /&gt;
&lt;br /&gt;
A. 3 B. 0.3 C. 0.03 D. 0.003&lt;br /&gt;
&lt;br /&gt;
32. 多位邵逸夫天文学奖得主目前已获得诺贝尔物理学奖。以下哪项发现已获得了邵逸夫天文学奖，但尚未获得诺贝尔物理学奖?( )&lt;br /&gt;
&lt;br /&gt;
A. 发现重子声学振荡 B. 直接观测引力波 C. 银河系中心的超大质量黑洞 D. 首个环绕类太阳恒星的系外行星&lt;br /&gt;
&lt;br /&gt;
33. 建造中的中国空间站巡天望远镜(CSST)的工作波段范围是( )。&lt;br /&gt;
&lt;br /&gt;
A. 紫外–可见光 B. 光学–近红外 C. 光学–射电 D. 紫外–亚毫米波&lt;br /&gt;
&lt;br /&gt;
34. 射电望远镜可以组成干涉阵列，望远镜的最大间隔越长，角分辨率越好，其角分辨率的 计算方法与光学望远镜近似。如果在我国国土最东端和最西端分别修建射电望远镜并组网， 在 21 厘米的波长上进行观测，理论角分辨率是( )。&lt;br /&gt;
&lt;br /&gt;
A. 40纳角秒 B.2微角秒 C.8毫角秒 D.1-2角秒(受限于大气视宁度)&lt;br /&gt;
&lt;br /&gt;
35. “赫罗图”是天文学家最喜欢的图形之一，是研究恒星非常重要的工具。它是以恒星光 谱型(或表面温度)为横坐标，以绝对星等(光度)为纵坐标作图。我们从赫罗图中可以获 得很多恒星的信息，包括恒星的温度、光度、半径，分类和演化轨迹等。下图中，哪条线可以代表恒星半径在赫罗图上的分布规律(箭头方向代表越来越大)?( )&lt;br /&gt;
[[文件:2023年全国天文竞赛预赛第35题图.jpg|缩略图|居中|2023年全国天文竞赛预赛第35题图]]&lt;br /&gt;
&lt;br /&gt;
==答案==&lt;br /&gt;
&lt;br /&gt;
BBCDC BBBDD ADCAB DDBCC DCCAB CCCBC DADCC&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2022-2023%E5%AD%A6%E5%B9%B4CNAO%E9%A2%84%E8%B5%9B%E9%80%89%E6%8B%A9%E9%A2%98&amp;diff=2754</id>
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		<updated>2025-03-27T19:48:41Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==试题==&lt;br /&gt;
&lt;br /&gt;
01. 今晚的月相更接近( )。&lt;br /&gt;
&lt;br /&gt;
A. 朔 B. 望 C. 下弦月 D.残月&lt;br /&gt;
&lt;br /&gt;
02. 以下哪颗恒星不属于“冬季大六边形“?( )&lt;br /&gt;
&lt;br /&gt;
A. 五车二 B. 参宿四 C. 南河三 D. 毕宿五&lt;br /&gt;
&lt;br /&gt;
03.M31是( )。&lt;br /&gt;
&lt;br /&gt;
A. 星云 B. 星团 C. 星系 D. 恒星&lt;br /&gt;
&lt;br /&gt;
04. “笔尖上的行星”是( )。&lt;br /&gt;
&lt;br /&gt;
A. 地球 B. 火星 C. 土星 D. 海王星&lt;br /&gt;
&lt;br /&gt;
05. 最早使用望远镜进行天文观测的科学家是( )。&lt;br /&gt;
&lt;br /&gt;
A. 哥白尼 B. 牛顿 C. 伽利略 D. 开普勒&lt;br /&gt;
&lt;br /&gt;
06. 象限仪流星雨的辐射点与以下哪个星座最近?( )&lt;br /&gt;
&lt;br /&gt;
A. 玉夫座 B. 武仙座 C. 狮子座 D. 六分仪座&lt;br /&gt;
&lt;br /&gt;
07. 天文学是自然科学六大基础学科之一，其主要的三个分支学科包括:天体测量学、天体力学、( )。&lt;br /&gt;
&lt;br /&gt;
A. 天体化学 B. 天体物理学 C. 天体生物学 D. 天文地理学&lt;br /&gt;
&lt;br /&gt;
08. 地月距离大约是( )个天文单位。&lt;br /&gt;
&lt;br /&gt;
A. 0.00001 B. 0.0025 C. 0.05 D. 1&lt;br /&gt;
&lt;br /&gt;
09. 北极星(小熊座 α)的目视亮度在全天恒星中排( )。&lt;br /&gt;
&lt;br /&gt;
A. 第 1 名 B. 第 5 名 C. 第 20 名左右 D. 第 50 名左右&lt;br /&gt;
&lt;br /&gt;
10. 以下节气(中气)出现在农历“正月”里概率最大的是( )。 &lt;br /&gt;
&lt;br /&gt;
A. 小寒 B. 大寒 C. 立春 D. 雨水&lt;br /&gt;
&lt;br /&gt;
11. 以下哪台已有或在建望远镜的有效口径最大?( )&lt;br /&gt;
&lt;br /&gt;
A. E-ELT B. GTC C. 昴星团望远镜 D. HET&lt;br /&gt;
&lt;br /&gt;
12. 在地球上任意一个位置，以下情况一定正确的是( )。&lt;br /&gt;
&lt;br /&gt;
A. 一年里每次太阳上中天的方位角相同。&lt;br /&gt;
&lt;br /&gt;
B. 一年里每次日出时(日心位于地平线上)太阳的方位角相同。 &lt;br /&gt;
&lt;br /&gt;
C. 一年里每次日出(日面出地平至完全出地平)的用时相同。 &lt;br /&gt;
&lt;br /&gt;
D. 其他几个选项都不正确。&lt;br /&gt;
&lt;br /&gt;
13. 在位于四川省甘孜藏族自治州稻城县的无名山天文台(100.108E,29.107N)，观测者看 到一颗流星自北向南划过消失在地平线附近，在以下哪个天文台站有可能观测到同一颗流星?( )&lt;br /&gt;
&lt;br /&gt;
A. 西藏阿里(80.027E,32.326N) &lt;br /&gt;
&lt;br /&gt;
B. 新疆慕士塔格(74.897E,38.330N)&lt;br /&gt;
&lt;br /&gt;
C. 云南高美谷(100.029E,26.694N) &lt;br /&gt;
&lt;br /&gt;
D. 河北兴隆(117.577E,40.396N)&lt;br /&gt;
&lt;br /&gt;
14. 下次日食将发生在( )。&lt;br /&gt;
&lt;br /&gt;
A. 2023 年 4 月 20 日 B. 2023 年 10 月 27 日 C. 2023 年 11 月 23 日 D. 2024 年 4 月 9 日&lt;br /&gt;
&lt;br /&gt;
15. 昨晚日落后不久，不考虑天气和遮挡的影响，你所在地能看到最亮的行星是?( )&lt;br /&gt;
&lt;br /&gt;
A. 水星 B. 金星 C. 木星 D. 天王星&lt;br /&gt;
&lt;br /&gt;
16. 以下恒星中哪个看上去颜色更偏红?( )&lt;br /&gt;
&lt;br /&gt;
A. 参宿七 B. 织女星 C. 角宿一 D. 心宿二&lt;br /&gt;
&lt;br /&gt;
17. 以下哪个梅西叶天体不属于室女星系团?( )&lt;br /&gt;
&lt;br /&gt;
A. M60 B. M84 C. M87 D. M101&lt;br /&gt;
&lt;br /&gt;
18. 2022 年 7 月 12 日，詹姆斯韦布望远镜公开了首批科学图像，其中( )的科学图像中只包括光谱观测结果，而不包括成像观测结果。&lt;br /&gt;
&lt;br /&gt;
A. 船底座大星云“宇宙悬崖” &lt;br /&gt;
&lt;br /&gt;
B. WASP-96b 系外行星&lt;br /&gt;
&lt;br /&gt;
C. 南指环星云 &lt;br /&gt;
&lt;br /&gt;
D. SMACSJ0723 星系团&lt;br /&gt;
&lt;br /&gt;
19. 彗星 C/2022E3(ZTF)于 2023 年 1 月 12 日过近日点，当时位于后发座。该彗星轨道偏心率接近1，轨道倾角 109°。2223 年 4 月 1 日该彗星最可能位于以下哪个星座附近?( ) &lt;br /&gt;
&lt;br /&gt;
A. 天龙座 B. 天鹰座 C. 天鹤座. D. 天鹅座&lt;br /&gt;
&lt;br /&gt;
20. 以下哪一颗小天体不是由业余天文学家 Gennadiy Borisov 发现的?( ) &lt;br /&gt;
&lt;br /&gt;
A. 2023 BU B. C/2019 Q4 C. C/2017 K2 D. C/2013 V2&lt;br /&gt;
&lt;br /&gt;
21. 某次上弦月恰逢月亮位于白道对黄道的升交点，在某地月亮位于上中天时观测，月亮视 面被照亮部分占整个月面的比例为( )。&lt;br /&gt;
&lt;br /&gt;
A. 小于等于 40% B. 大于 40%小于 50% C. 等于 50% D. 大于 50%小于 60%&lt;br /&gt;
&lt;br /&gt;
22. 在以下太阳系的天然卫星中，轨道运行周期比其母行星的自转周期还短的是( )。&lt;br /&gt;
&lt;br /&gt;
A. 木卫一 B. 木卫四 C. 火卫一 D. 火卫二&lt;br /&gt;
&lt;br /&gt;
23. “荧惑守心”是中国史籍中的一种天象记载，其中的“荧惑”指的是( )。 &lt;br /&gt;
&lt;br /&gt;
A. 天狼星 B. 土星 C. 火星 D. 木星&lt;br /&gt;
&lt;br /&gt;
24. 在中国，有一种载人航天飞船叫“神舟”;有一种载人空间站叫“天宫”;有一种中继通信卫星叫“鹊桥”;有一种暗物质探测器叫“悟空”，下面哪个暂时还不是中国航天项目 的名称?( )&lt;br /&gt;
&lt;br /&gt;
A. 洛神 B. 嫦娥 C. 羲和 D. 祝融&lt;br /&gt;
&lt;br /&gt;
25. 2023 年是比利时耶稣会士南怀仁诞辰 600 周年，下列哪架天文仪器是南怀仁设计监制的?( )&lt;br /&gt;
&lt;br /&gt;
A. 玑衡抚辰仪 B. 纪限仪 C. 浑仪 D. 水运仪象台&lt;br /&gt;
&lt;br /&gt;
26. 在距离太阳 200 天文单位的地方释放一个石块，若仅考虑太阳引力作用，它需要多长时间落到太阳表面?( )&lt;br /&gt;
&lt;br /&gt;
A. 5 年 B. 50 年 C. 500 年 D. 5000 年&lt;br /&gt;
&lt;br /&gt;
27. 假设一个视星等为 18 等的星系的光全部均匀分布在视半径为 6 角秒的圆形区域中，其面 视星等是( )等/平方角秒。&lt;br /&gt;
&lt;br /&gt;
A. 13 B. 18 C. 23 D. 28&lt;br /&gt;
&lt;br /&gt;
28. 假设宇宙匀速膨胀，已知今天的哈勃常数 H0=70 km/s/Mpc，请问宇宙的年龄大约为? ( )&lt;br /&gt;
&lt;br /&gt;
A. 1 亿年 B. 10 亿年 C. 100 亿年 D. 1000 亿年&lt;br /&gt;
&lt;br /&gt;
29. 在刘慈欣的知名科幻小说《三体》中，我们的邻居比邻星上存在着比地球文明更高级的 三体文明。三体人由于生存环境恶劣，准备向地球迁移。假设三体舰队从比邻星出发，一直以 1/10 光速的速度向地球航行，走了比邻星到地球距离的 9/10 后，由于故障原地暂停。此时他们看到的太阳亮度( )。&lt;br /&gt;
&lt;br /&gt;
A. 变亮 1 个星等 B. 变亮 5 个星等 C. 变亮 10 个星等 D. 变亮 100 个星等&lt;br /&gt;
&lt;br /&gt;
30. 比邻星到我们的距离是 4.2 光年，在地球上看，它的周年视差是多少?( ) &lt;br /&gt;
&lt;br /&gt;
A. 1.58 角秒 B. 0.24 角秒 C. 0.77 角秒 D. 4.2 角秒&lt;br /&gt;
&lt;br /&gt;
31. 视向速度为 1000 km/s 的超高速恒星对应的红移/蓝移是多少?( ) &lt;br /&gt;
&lt;br /&gt;
A. 3 B. 0.3 C. 0.03 D. 0.003&lt;br /&gt;
&lt;br /&gt;
32. 多位邵逸夫天文学奖得主目前已获得诺贝尔物理学奖。以下哪项发现已获得了邵逸夫天文学奖，但尚未获得诺贝尔物理学奖?( )&lt;br /&gt;
&lt;br /&gt;
A. 发现重子声学振荡 B. 直接观测引力波 C. 银河系中心的超大质量黑洞 D. 首个环绕类太阳恒星的系外行星&lt;br /&gt;
&lt;br /&gt;
33. 建造中的中国空间站巡天望远镜(CSST)的工作波段范围是( )。&lt;br /&gt;
&lt;br /&gt;
A. 紫外–可见光 B. 光学–近红外 C. 光学–射电 D. 紫外–亚毫米波&lt;br /&gt;
&lt;br /&gt;
34. 射电望远镜可以组成干涉阵列，望远镜的最大间隔越长，角分辨率越好，其角分辨率的 计算方法与光学望远镜近似。如果在我国国土最东端和最西端分别修建射电望远镜并组网， 在 21 厘米的波长上进行观测，理论角分辨率是( )。&lt;br /&gt;
&lt;br /&gt;
A. 40纳角秒 B.2微角秒 C.8毫角秒 D.1-2角秒(受限于大气视宁度)&lt;br /&gt;
&lt;br /&gt;
35. “赫罗图”是天文学家最喜欢的图形之一，是研究恒星非常重要的工具。它是以恒星光 谱型(或表面温度)为横坐标，以绝对星等(光度)为纵坐标作图。我们从赫罗图中可以获 得很多恒星的信息，包括恒星的温度、光度、半径，分类和演化轨迹等。下图中，哪条线可以代表恒星半径在赫罗图上的分布规律(箭头方向代表越来越大)?( )&lt;br /&gt;
&lt;br /&gt;
[[文件:2023年全国天文竞赛预赛第35题图|缩略图|左|2023年全国天文竞赛预赛第35题图]]&lt;br /&gt;
&lt;br /&gt;
==答案==&lt;br /&gt;
&lt;br /&gt;
BBCDC BBBDD ADCAB DDBCC DCCAB CCCBC DADCC&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=%E6%96%87%E4%BB%B6:2023%E5%B9%B4%E5%85%A8%E5%9B%BD%E5%A4%A9%E6%96%87%E7%AB%9E%E8%B5%9B%E9%A2%84%E8%B5%9B%E7%AC%AC35%E9%A2%98%E5%9B%BE.jpg&amp;diff=2753</id>
		<title>文件:2023年全国天文竞赛预赛第35题图.jpg</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=%E6%96%87%E4%BB%B6:2023%E5%B9%B4%E5%85%A8%E5%9B%BD%E5%A4%A9%E6%96%87%E7%AB%9E%E8%B5%9B%E9%A2%84%E8%B5%9B%E7%AC%AC35%E9%A2%98%E5%9B%BE.jpg&amp;diff=2753"/>
		<updated>2025-03-27T19:45:12Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;2023年全国天文竞赛预赛第35题图&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=%E7%94%A8%E6%88%B7:%E6%9F%90%E4%B9%9D%E6%B5%81%E8%BE%A3%E9%B8%A1%E6%B0%91%E7%A7%91&amp;diff=2752</id>
		<title>用户:某九流辣鸡民科</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=%E7%94%A8%E6%88%B7:%E6%9F%90%E4%B9%9D%E6%B5%81%E8%BE%A3%E9%B8%A1%E6%B0%91%E7%A7%91&amp;diff=2752"/>
		<updated>2025-03-27T17:11:40Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;2018年CNAO铜牌，IOAA银牌&lt;br /&gt;
&lt;br /&gt;
2019-2023，山东大学物理学院&lt;br /&gt;
&lt;br /&gt;
2023-，中国科学院国家天文台，天体物理学专业硕士研究生在读&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2243</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2243"/>
		<updated>2022-04-04T03:26:09Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_i=n_i k_B T_i$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_e=n_e k_B T_e$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(&amp;lt;math&amp;gt;p_e V=N_e k_B T_e&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)On the Earth, &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt; where &amp;lt;math&amp;gt;P_{E} \approx 10^{5} \mathrm{~Pa}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;T_{E} \approx 300 \mathrm{~K}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, &amp;lt;math&amp;gt;n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
(a)压力是来自各个方向的。太阳黑子内外的等离子体的压强必须相等，使黑子边界处保持平衡。&lt;br /&gt;
&lt;br /&gt;
太阳黑子内部的动力学压强为&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_i=n_i k_B T_i$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
其中 n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; 是黑子内的数密度 &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
同样有,&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_e=n_e k_B T_e$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(&amp;lt;math&amp;gt;p_e V=N_e k_B T_e&amp;lt;/math&amp;gt; 也是正确的。但是最好写成压强和数密度的关系)&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
由假设知,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n。平衡要求:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
于是有&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)在地球上， &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt;，其中&amp;lt;math&amp;gt;P_{E} \approx 10^{5} \mathrm{~Pa}&amp;lt;/math&amp;gt; ， &amp;lt;math&amp;gt;T_{E} \approx 300 \mathrm{~K}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
于是有, &amp;lt;math&amp;gt;n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
这意味着地球表面附近大气中粒子的数密度至少是太阳大气中等离子数密度的100倍（当然质量密度会更高）。&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2242</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2242"/>
		<updated>2022-04-04T03:25:52Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 中文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_i=n_i k_B T_i$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_e=n_e k_B T_e$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(&amp;lt;math&amp;gt;p_e V=N_e k_B T_e&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)On the Earth, &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt; where &amp;lt;math&amp;gt;P_{E} \approx 10^{5} \mathrm{~Pa}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;T_{E} \approx 300 \mathrm{~K}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, &amp;lt;math&amp;gt;n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
(a)压力是来自各个方向的。太阳黑子内外的等离子体的压强必须相等，使黑子边界处保持平衡。&lt;br /&gt;
&lt;br /&gt;
太阳黑子内部的动力学压强为&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_i=n_i k_B T_i$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
其中 n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; 是黑子内的数密度 &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
同样有,&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_e=n_e k_B T_e$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(&amp;lt;math&amp;gt;p_e V=N_e k_B T_e&amp;lt;/math&amp;gt; 也是正确的。但是最好写成压强和数密度的关系)&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
由假设知,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n。平衡要求:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
于是有&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)在地球上， &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt;，其中&amp;lt;math&amp;gt;P_{E} \approx 10^{5} \mathrm{~Pa}&amp;lt;/math&amp;gt; ， &amp;lt;math&amp;gt;T_{E} \approx 300 \mathrm{~K}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
于是有, &amp;lt;math&amp;gt;n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
这意味着地球表面附近大气中粒子的数密度至少是太阳大气中等离子数密度的100倍（当然质量密度会更高）。&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2241</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2241"/>
		<updated>2022-04-04T03:17:45Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_i=n_i k_B T_i$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_e=n_e k_B T_e$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(&amp;lt;math&amp;gt;p_e V=N_e k_B T_e&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)On the Earth, &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt; where &amp;lt;math&amp;gt;P_{E} \approx 10^{5} \mathrm{~Pa}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;T_{E} \approx 300 \mathrm{~K}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, &amp;lt;math&amp;gt;n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2240</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2240"/>
		<updated>2022-04-04T03:17:30Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_i=n_i k_B T_i$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$p_e=n_e k_B T_e$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(&amp;lt;math&amp;gt;p_e V=N_e k_B T_e&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)On the Earth, &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt; where &amp;lt;math&amp;gt;P_{E} \approx 10^{5} \mathrm{~Pa}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;T_{E} \approx 300 \mathrm{~K}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, &amp;lt;math&amp;gt;n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}&amp;lt;math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2239</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2239"/>
		<updated>2022-04-04T03:16:02Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_i=n_i k_B T_i$$&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_e=n_e k_B T_e$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(&amp;lt;math&amp;gt;p_e V=N_e k_B T_e&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)On the Earth, &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt; where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2238</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2238"/>
		<updated>2022-04-04T03:15:09Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 中文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{i}=n_{i} k_{B} T_{i}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(&amp;lt;math&amp;gt;p_{e} V=N_{e} k_{B} T_{e}&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)On the Earth, &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt; where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2237</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2237"/>
		<updated>2022-04-04T03:14:58Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 中文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{i}=n_{i} k_{B} T_{i}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(&amp;lt;math&amp;gt;p_{e} V=N_{e} k_{B} T_{e}&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)On the Earth, &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt; where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
:&amp;lt;math&amp;gt;\int_0^\infty e^{-x^2}\,dx&amp;lt;/math&amp;gt;&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2236</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2236"/>
		<updated>2022-04-04T03:13:58Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{i}=n_{i} k_{B} T_{i}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(&amp;lt;math&amp;gt;p_{e} V=N_{e} k_{B} T_{e}&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(b)On the Earth, &amp;lt;math&amp;gt;p_{E}=n_{E} k_{B} T_{E}&amp;lt;/math&amp;gt; where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2235</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2235"/>
		<updated>2022-04-04T03:13:09Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{i}=n_{i} k_{B} T_{i}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(&amp;lt;math&amp;gt;p_{e} V=N_{e} k_{B} T_{e}&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(b)On the Earth, $p_{E}=n_{E} k_{B} T_{E}$ where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2234</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2234"/>
		<updated>2022-04-04T03:12:30Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{i}=n_{i} k_{B} T_{i}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(&amp;lt;math&amp;gt;p_{e} V=N_{e} k_{B} T_{e}&amp;lt;/math&amp;gt; is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(b)On the Earth, $p_{E}=n_{E} k_{B} T_{E}$ where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2233</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2233"/>
		<updated>2022-04-04T03:11:42Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;p_{i}=n_{i} k_{B} T_{i}&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
($$p_{e} V=N_{e} k_{B} T_{e}$$ is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(b)On the Earth, $p_{E}=n_{E} k_{B} T_{E}$ where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2232</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2232"/>
		<updated>2022-04-04T03:11:22Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;p_{i}=n_{i} k_{B} T_{i}$$&amp;lt;/math&amp;gt;&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
($$p_{e} V=N_{e} k_{B} T_{e}$$ is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(b)On the Earth, $p_{E}=n_{E} k_{B} T_{E}$ where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2231</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2231"/>
		<updated>2022-04-04T03:07:17Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{i}=n_{i} k_{B} T_{i}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
($$p_{e} V=N_{e} k_{B} T_{e}$$ is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(b)On the Earth, $p_{E}=n_{E} k_{B} T_{E}$ where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2230</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2230"/>
		<updated>2022-04-04T03:06:49Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{i}=n_{i} k_{B} T_{i}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt; is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
($p_{e} V=N_{e} k_{B} T_{e}$ is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
Then&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
(b)On the Earth, $p_{E}=n_{E} k_{B} T_{E}$ where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2229</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2229"/>
		<updated>2022-04-04T03:06:05Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
The kinetic pressure inside the sunspot is &amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{i}=n_{i} k_{B} T_{i}$$&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,&amp;lt;br/&amp;gt;&lt;br /&gt;
$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
($p_{e} V=N_{e} k_{B} T_{e}$ is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&amp;lt;br/&amp;gt;&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&lt;br /&gt;
Then&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&lt;br /&gt;
*(b)On the Earth, $p_{E}=n_{E} k_{B} T_{E}$ where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2228</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2228"/>
		<updated>2022-04-04T03:04:49Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
*(a)The pressure is from all directions. The plasma inside the sunspot must have the same total pressure as the plasma outside to maintain equilibrium on the border.&lt;br /&gt;
:The kinetic pressure inside the sunspot is $$p_{i}=n_{i} k_{B} T_{i}$$&lt;br /&gt;
where n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;is the number density inside the sunspot. &amp;lt;br/&amp;gt;&lt;br /&gt;
And also,$$p_{e}=n_{e} k_{B} T_{e}$$&amp;lt;br/&amp;gt;&lt;br /&gt;
($p_{e} V=N_{e} k_{B} T_{e}$is also correct. But it’s better to write the pressure in terms of number density)&amp;lt;br/&amp;gt;&lt;br /&gt;
From the assumption,n&amp;lt;sub&amp;gt;i&amp;lt;/sub&amp;gt;=n&amp;lt;sub&amp;gt;e&amp;lt;/sub&amp;gt;=n. The equilibrium requires:&lt;br /&gt;
$$n k_{B} T_{i}+\frac{B_{i}^{2}}{2 \mu_{0}}=n k_{B} T_{e}+\frac{B_{e}^{2}}{2 \mu_{0}}$$&lt;br /&gt;
Then&lt;br /&gt;
$$n=\frac{1}{2 \mu_{0} k_{B}} \frac{\left(B_{i}^{2}-B_{e}^{2}\right)}{\left(T_{e}-T_{i}\right)}=1.43 \times 10^{23} \mathrm{~m}^{-3}$$&lt;br /&gt;
*(b)On the Earth, $p_{E}=n_{E} k_{B} T_{E}$ where $P_{E} \approx 10^{5} \mathrm{~Pa}$ and $T_{E} \approx 300 \mathrm{~K}$&amp;lt;br/&amp;gt;&lt;br /&gt;
Thus, $n_{E}=\frac{p_{E}}{k_{B} T_{E}}=2.4 \times 10^{25} m^{-3}$&amp;lt;br/&amp;gt;&lt;br /&gt;
This means that the number density of the atmospheric particles at the surface of the earth is at least 100 times larger that the number density of particles in the solar photosphere (of course the mass density is even higher).&lt;br /&gt;
&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2227</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2227"/>
		<updated>2022-04-04T02:55:08Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
==中文解答==&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2226</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2226"/>
		<updated>2022-04-04T02:54:54Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
==中文解答&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2225</id>
		<title>2018年IOAA理论第7题-太阳黑子</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2018%E5%B9%B4IOAA%E7%90%86%E8%AE%BA%E7%AC%AC7%E9%A2%98-%E5%A4%AA%E9%98%B3%E9%BB%91%E5%AD%90&amp;diff=2225"/>
		<updated>2022-04-04T02:50:56Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{需要解答}}&lt;br /&gt;
==英文原题==&lt;br /&gt;
&lt;br /&gt;
'''(T7) Sunspot (25 points)''' &lt;br /&gt;
&lt;br /&gt;
Magnetic fields are important in the physics of stars and sunspots. As an approximation, we can model the photosphere of the Sun consisting of a plasma, which can be simply treated as a single component ideal gas, and a magnetic field ('''B'''), which has an associated magnetic pressure $$p_B=\frac{B^2}{2\mu_0}$$. It behaves like any other physical pressure except that it is carried by the magnetic field rather than by the kinetic energy of particles.&lt;br /&gt;
&lt;br /&gt;
Assume that the number density of particles in the photosphere is constant everywhere, but the magnetic field inside the sunspot (B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T) is much stronger than outside (B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T). From the blackbody spectrum, the temperature inside the sunspot is T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K, while the temperature outside is T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K (which is why the sunspot looks darker). For the sunspot to be stable, the inside must be in equilibrium with the outside.&lt;br /&gt;
&lt;br /&gt;
(a) Estimate the number density of plasma particles in the solar photosphere.&lt;br /&gt;
&lt;br /&gt;
(b) Compare your answer with an estimate of the number density of particles in the atmosphere at the surface of the Earth.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
&lt;br /&gt;
在恒星和黑子的物理过程中磁场很重要。作为近似，我们将太阳光球层简化为一种等离子体，可以按照理想气体一样对待，并且具有一定的磁压$$p_B=\frac{B^2}{2\mu_0}$$，与当地的磁场（'''B'''）有关。磁压的行为与其他的压强类似，只不过它来自磁场而不是粒子的动能。&lt;br /&gt;
&lt;br /&gt;
假设光球层中粒子的数密度是处处相等的，但是太阳黑子内部的磁场(B&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;=0.1T)远大于外部(B&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;=5×10&amp;lt;sup&amp;gt;-3&amp;lt;/sup&amp;gt;T)。从黑体谱来看，太阳黑子内部的磁场为T&amp;lt;sub&amp;gt;in&amp;lt;/sub&amp;gt;~4000K，而外部的温度为T&amp;lt;sub&amp;gt;out&amp;lt;/sub&amp;gt;~6000K（这也就是为什么黑子看起来更暗）。为了使太阳黑子保持稳定，内部与外部的压强需要保持平衡。&lt;br /&gt;
&lt;br /&gt;
(a)估算太阳光球层中粒子的数密度。&lt;br /&gt;
&lt;br /&gt;
(b)将计算的结果与地表大气的数密度作比较。&lt;br /&gt;
&lt;br /&gt;
[[分类:磁场]]&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2202</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2202"/>
		<updated>2022-03-20T07:23:45Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 中文解析（by 侯志鹏） */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
== 标准答案（英文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 标准答案（中文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    在1660Mhz频率处，流量密度为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    在观测者处，源所张立体角为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    流量密度与总光度有以下关系&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    在该频率下，总光度可以用瑞利-金斯公式近似得到：&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    其中T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;为亮温度。于是我们有&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 中文解析（by 侯志鹏） ==&lt;br /&gt;
本题目的是通过观测得到的源的流量密度与持续时间估计源的亮温度，即通过观测到的流量密度计算源的辐射强度，再由辐射强度计算亮温度。要知道源的辐射强度，还需知道源的大小。本题给出了脉冲的持续时间，因此可以根据持续时间估计源的大小，直接的办法是假定源的半径小于光在脉冲持续时间内经过的距离。&lt;br /&gt;
&lt;br /&gt;
设源的半径为 r，源的距离为 R，辐射强度为 $$B_{\nu}$$，接收到的流量密度为 $$S_{\nu}$$&lt;br /&gt;
&lt;br /&gt;
$$B_{\nu}$$ 与 $$S_{\nu}$$ 的关系为：&lt;br /&gt;
&lt;br /&gt;
$$S_{\nu}=B_{\nu}\cdot\Omega$$， $$\Omega=\pi \frac{r^{2}}{R^{2}}$$为源所张立体角&lt;br /&gt;
&lt;br /&gt;
在射电波段，普朗克公式可近似为瑞利-金斯公式：&lt;br /&gt;
&lt;br /&gt;
$$B_{\nu}=\frac{2\nu^{2}}{c^{2}}kT$$&lt;br /&gt;
&lt;br /&gt;
该公式中辐射强度与热力学温度为线性关系，因此在射电天文领域常用亮温度来估计展源的亮度, 从该公式计算出的温度称为亮温度。&lt;br /&gt;
&lt;br /&gt;
于是得到：&lt;br /&gt;
&lt;br /&gt;
$$T_{b}=\frac{S_{\nu}c^{2}}{2k\nu^{2}\Omega}$$&lt;br /&gt;
&lt;br /&gt;
代入数值得到最终结果。  &lt;br /&gt;
&lt;br /&gt;
本题易错点在于，部分同学可能先根据流量密度算出总光度，再由总光度计算出源表面的流量密度，并将源表面的流量密度代入瑞丽-金斯公式进行计算。这种做法混淆了流量密度与辐射强度的概念，前者的定义为单位面积的辐射功率，后者的定义为单位面积在单位方向上的辐射功率，两者关系为$$F_{\nu}=\pi B_{\nu}$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2201</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2201"/>
		<updated>2022-03-20T07:23:17Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 中文解析（by 侯志鹏） */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
== 标准答案（英文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 标准答案（中文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    在1660Mhz频率处，流量密度为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    在观测者处，源所张立体角为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    流量密度与总光度有以下关系&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    在该频率下，总光度可以用瑞利-金斯公式近似得到：&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    其中T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;为亮温度。于是我们有&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 中文解析（by 侯志鹏） ==&lt;br /&gt;
本题目的是通过观测得到的源的流量密度与持续时间估计源的亮温度，即通过观测到的流量密度计算源的辐射强度，再由辐射强度计算亮温度。要知道源的辐射强度，还需知道源的大小。本题给出了脉冲的持续时间，因此可以根据持续时间估计源的大小，直接的办法是假定源的半径小于光在脉冲持续时间内经过的距离。&lt;br /&gt;
&lt;br /&gt;
设源的半径为 r，源的距离为 R，辐射强度为 $$B_{\nu}$$，接收到的流量密度为 $$S_{\nu}$$&lt;br /&gt;
&lt;br /&gt;
$$B_{\nu}$$ 与 $$S_{\nu}$$ 的关系为：&lt;br /&gt;
&lt;br /&gt;
$$S_{\nu}=B_{\nu}\cdot\Omega$$， $$\Omega=\pi \frac{r^{2}}{R^{2}}$$为源所张立体角&lt;br /&gt;
&lt;br /&gt;
在射电波段，普朗克公式可近似为瑞利-金斯公式：&lt;br /&gt;
&lt;br /&gt;
$$B_{\nu}=\frac{2\nu^{2}}{c^{2}}kT&lt;br /&gt;
&lt;br /&gt;
该公式中辐射强度与热力学温度为线性关系，因此在射电天文领域常用亮温度来估计展源的亮度, 从该公式计算出的温度称为亮温度。&lt;br /&gt;
&lt;br /&gt;
于是得到：&lt;br /&gt;
&lt;br /&gt;
$$T_{b}=\frac{S_{\nu}c^{2}}{2k\nu^{2}\Omega}$$&lt;br /&gt;
&lt;br /&gt;
代入数值得到最终结果。  &lt;br /&gt;
&lt;br /&gt;
本题易错点在于，部分同学可能先根据流量密度算出总光度，再由总光度计算出源表面的流量密度，并将源表面的流量密度代入瑞丽-金斯公式进行计算。这种做法混淆了流量密度与辐射强度的概念，前者的定义为单位面积的辐射功率，后者的定义为单位面积在单位方向上的辐射功率，两者关系为$$F_{\nu}=\pi B_{\nu}$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2200</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2200"/>
		<updated>2022-03-19T14:53:48Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 中文解析（by 侯志鹏） */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
== 标准答案（英文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 标准答案（中文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    在1660Mhz频率处，流量密度为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    在观测者处，源所张立体角为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    流量密度与总光度有以下关系&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    在该频率下，总光度可以用瑞利-金斯公式近似得到：&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    其中T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;为亮温度。于是我们有&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 中文解析（by 侯志鹏） ==&lt;br /&gt;
本题目的是通过观测得到的源的流量密度与持续时间估计源的亮温度，即通过观测到的流量密度计算源的辐射强度，再由辐射强度计算亮温度。要知道源的辐射强度，还需知道源的大小。本题给出了脉冲的持续时间，因此可以根据持续时间估计源的大小，直接的办法是假定源的半径小于光在脉冲持续时间内经过的距离。&lt;br /&gt;
&lt;br /&gt;
设源的半径为 r，源的距离为 R，辐射强度为 $$B_{\nu}$$，接收到的流量密度为 $$S_{\nu}$$&lt;br /&gt;
&lt;br /&gt;
$$B_{\nu}$$ 与 $$S_{\nu}$$ 的关系为：&lt;br /&gt;
&lt;br /&gt;
$$S_{\nu}=B_{\nu}\cdot\Omega$$， $$\Omega=\pi \frac{r^{2}}{R^{2}}$$为源所张立体角&lt;br /&gt;
&lt;br /&gt;
在射电波段，普朗克公式可近似为瑞利-金斯公式：&lt;br /&gt;
&lt;br /&gt;
$$B_{\nu}=\frac{2\nu^{2}}{c^{2}}kT&lt;br /&gt;
&lt;br /&gt;
该公式中辐射强度与热力学温度为线性关系，因此在射电天文领域常用亮温度来估计展源的亮度, 从该公式计算出的温度称为亮温度。&lt;br /&gt;
&lt;br /&gt;
于是得到：&lt;br /&gt;
&lt;br /&gt;
$$T_{b}=\frac{S_{\nu}c^{2}}{2k\nu^{2}\Omega}&lt;br /&gt;
&lt;br /&gt;
代入数值得到最终结果。  &lt;br /&gt;
&lt;br /&gt;
本题易错点在于，部分同学可能先根据流量密度算出总光度，再由总光度计算出源表面的流量密度，并将源表面的流量密度代入瑞丽-金斯公式进行计算。这种做法混淆了流量密度与辐射强度的概念，前者的定义为单位面积的辐射功率，后者的定义为单位面积在单位方向上的辐射功率，两者关系为$$F_{\nu}=\pi B_{\nu}$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2199</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2199"/>
		<updated>2022-03-19T14:53:04Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
== 标准答案（英文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 标准答案（中文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    在1660Mhz频率处，流量密度为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    在观测者处，源所张立体角为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    流量密度与总光度有以下关系&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    在该频率下，总光度可以用瑞利-金斯公式近似得到：&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    其中T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;为亮温度。于是我们有&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 中文解析（by 侯志鹏） ==&lt;br /&gt;
本题目的是通过观测得到的源的流量密度与持续时间估计源的亮温度，即通过观测到的流量密度计算源的辐射强度，再由辐射强度计算亮温度。要知道源的辐射强度，还需知道源的大小。本题给出了脉冲的持续时间，因此可以根据持续时间估计源的大小，直接的办法是假定源的半径小于光在脉冲持续时间内经过的距离。&lt;br /&gt;
设源的半径为 r，源的距离为 R，辐射强度为 $$B_{\nu}$$，接收到的流量密度为 $$S_{\nu}$$&lt;br /&gt;
$$B_{\nu}$$ 与 $$S_{\nu}$$ 的关系为：&lt;br /&gt;
$$S_{\nu}=B_{\nu}\cdot\Omega$$， $$\Omega=\pi \frac{r^{2}}{R^{2}}$$为源所张立体角&lt;br /&gt;
在射电波段，普朗克公式可近似为瑞利-金斯公式：&lt;br /&gt;
$$B_{\nu}=\frac{2\nu^{2}}{c^{2}}kT&lt;br /&gt;
该公式中辐射强度与热力学温度为线性关系，因此在射电天文领域常用亮温度来估计展源的亮度, 从该公式计算出的温度称为亮温度。&lt;br /&gt;
于是得到：&lt;br /&gt;
$$T_{b}=\frac{S_{\nu}c^{2}}{2k\nu^{2}\Omega}&lt;br /&gt;
代入数值得到最终结果。  &lt;br /&gt;
本题易错点在于，部分同学可能先根据流量密度算出总光度，再由总光度计算出源表面的流量密度，并将源表面的流量密度代入瑞丽-金斯公式进行计算。这种做法混淆了流量密度与辐射强度的概念，前者的定义为单位面积的辐射功率，后者的定义为单位面积在单位方向上的辐射功率，两者关系为$$F_{\nu}=\pi B_{\nu}$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2198</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2198"/>
		<updated>2022-03-19T14:52:14Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 中文解析（by 侯志鹏） */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
== 标准答案（英文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 标准答案（中文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    在1660Mhz频率处，流量密度为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    在观测者处，源所张立体角为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    流量密度与总光度有以下关系&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    在该频率下，总光度可以用瑞利-金斯公式近似得到：&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    其中T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;为亮温度。于是我们有&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 中文解析（by 侯志鹏） ==&lt;br /&gt;
  本题目的是通过观测得到的源的流量密度与持续时间估计源的亮温度，即通过观测到的流量密度计算源的辐射强度，再由辐射强度计算亮温度。要知道源的辐射强度，还需知道源的大小。本题给出了脉冲的持续时间，因此可以根据持续时间估计源的大小，直接的办法是假定源的半径小于光在脉冲持续时间内经过的距离。&lt;br /&gt;
  设源的半径为 r，源的距离为 R，辐射强度为 $$B_{\nu}$$，接收到的流量密度为 $$S_{\nu}$$&lt;br /&gt;
  $$B_{\nu}$$ 与 $$S_{\nu}$$ 的关系为：&lt;br /&gt;
  $$S_{\nu}=B_{\nu}\cdot\Omega$$， $$\Omega=\pi \frac{r^{2}}{R^{2}}$$为源所张立体角&lt;br /&gt;
  在射电波段，普朗克公式可近似为瑞利-金斯公式：&lt;br /&gt;
  $$B_{\nu}=\frac{2\nu^{2}}{c^{2}}kT&lt;br /&gt;
  该公式中辐射强度与热力学温度为线性关系，因此在射电天文领域常用亮温度来估计展源的亮度, 从该公式计算出的温度称为亮温度。&lt;br /&gt;
  于是得到：&lt;br /&gt;
  $$T_{b}=\frac{S_{\nu}c^{2}}{2k\nu^{2}\Omega}&lt;br /&gt;
  代入数值得到最终结果。  &lt;br /&gt;
  本题易错点在于，部分同学可能先根据流量密度算出总光度，再由总光度计算出源表面的流量密度，并将源表面的流量密度代入瑞丽-金斯公式进行计算。这种做法混淆了流量密度与辐射强度的概念，前者的定义为单位面积的辐射功率，后者的定义为单位面积在单位方向上的辐射功率，两者关系为$$F_{\nu}=\pi B_{\nu}$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2197</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2197"/>
		<updated>2022-03-19T14:50:58Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 中文解析（by 侯志鹏） */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
== 标准答案（英文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 标准答案（中文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    在1660Mhz频率处，流量密度为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    在观测者处，源所张立体角为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    流量密度与总光度有以下关系&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    在该频率下，总光度可以用瑞利-金斯公式近似得到：&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    其中T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;为亮温度。于是我们有&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 中文解析（by 侯志鹏） ==&lt;br /&gt;
&lt;br /&gt;
    本题目的是通过观测得到的源的流量密度与持续时间估计源的亮温度，即通过观测到的流量密度计算源的辐射强度，再由辐射强度计算亮温度。要知道源的辐射强度，还需知道源的大小。本题给出了脉冲的持续时间，因此可以根据持续时间估计源的大小，直接的办法是假定源的半径小于光在脉冲持续时间内经过的距离。&lt;br /&gt;
&lt;br /&gt;
    设源的半径为 r，源的距离为 R，辐射强度为 $$B_{\nu}$$，接收到的流量密度为 $$S_{\nu}$$&lt;br /&gt;
    $$B_{\nu}$$ 与 $$S_{\nu}$$ 的关系为：&lt;br /&gt;
    $$S_{\nu}=B_{\nu}\cdot\Omega$$， $$\Omega=\pi \frac{r^{2}}{R^{2}}$$为源所张立体角&lt;br /&gt;
    在射电波段，普朗克公式可近似为瑞利-金斯公式：&lt;br /&gt;
    $$B_{\nu}=\frac{2\nu^{2}}{c^{2}}kT&lt;br /&gt;
    该公式中辐射强度与热力学温度为线性关系，因此在射电天文领域常用亮温度来估计展源的亮度, 从该公式计算出的温度称为亮温度。&lt;br /&gt;
&lt;br /&gt;
    于是得到：&lt;br /&gt;
    $$T_{b}=\frac{S_{\nu}c^{2}}{2k\nu^{2}\Omega}&lt;br /&gt;
    代入数值得到最终结果。&lt;br /&gt;
    &lt;br /&gt;
    本题易错点在于，部分同学可能先根据流量密度算出总光度，再由总光度计算出源表面的流量密度，并将源表面的流量密度代入瑞丽-金斯公式进行计算。这种做法混淆了流量密度与辐射强度的概念，前者的定义为单位面积的辐射功率，后者的定义为单位面积在单位方向上的辐射功率，两者关系为$$F_{\nu}=\pi B_{\nu}$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2195</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2195"/>
		<updated>2022-03-19T12:24:16Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 中文解答（by 侯志鹏） */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
== 标准答案（英文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 标准答案（中文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    在1660Mhz频率处，流量密度为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    在观测者处，源所张立体角为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    流量密度与总光度有以下关系&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    在该频率下，总光度可以用瑞利-金斯公式近似得到：&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    其中T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;为亮温度。于是我们有&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 中文解析（by 侯志鹏） ==&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2194</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2194"/>
		<updated>2022-03-19T11:10:23Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
== 标准答案（英文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 标准答案（中文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    在1660Mhz频率处，流量密度为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    在观测者处，源所张立体角为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    流量密度与总光度有以下关系&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    在该频率下，总光度可以用瑞利-金斯公式近似得到：&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    其中T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;为亮温度。于是我们有&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 中文解答（by 侯志鹏） ==&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2193</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2193"/>
		<updated>2022-03-19T11:06:17Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
== 标准答案（英文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
== 标准答案（中文）==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
在700μs内，电磁波经过距离为$$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$。脉冲发出的区域的半径一定不大于光在脉冲持续时间内经过的距离。所以我们用r作为源的半径进行计算。（$$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$）&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    在1660Mhz频率处，流量密度为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    在观测者处，源所张立体角为&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    流量密度与总光度有以下关系&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    在该频率下，总光度可以用瑞利-金斯公式近似得到：&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    其中T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;为亮温度。于是我们有&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2184</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2184"/>
		<updated>2022-03-18T13:13:32Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{\nu}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{c}^{2} / 2 \mathrm{k\nu}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2183</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2183"/>
		<updated>2022-03-18T13:12:43Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{v}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt; is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{C}^{2} / 2 \mathrm{kv}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2182</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2182"/>
		<updated>2022-03-18T13:12:25Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{v}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where T&amp;lt;sub&amp;gt;b&amp;lt;/sub&amp;gt;is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{C}^{2} / 2 \mathrm{kv}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2181</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2181"/>
		<updated>2022-03-18T13:09:15Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
&lt;br /&gt;
'''part1'''&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part2'''&lt;br /&gt;
&lt;br /&gt;
* ''(30%)''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
'''part3'''&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(20%)''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{v}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* ''(10%)''&lt;br /&gt;
&lt;br /&gt;
    where $$T_{b}$$ is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{C}^{2} / 2 \mathrm{kv}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2180</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2180"/>
		<updated>2022-03-18T13:07:26Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
&lt;br /&gt;
# part1&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
   &lt;br /&gt;
    During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
# part2&lt;br /&gt;
&lt;br /&gt;
* '''(30%)'''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
# part3&lt;br /&gt;
&lt;br /&gt;
* '''(20%)'''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* '''(20%)'''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{v}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
&lt;br /&gt;
    where $$T_{b}$$ is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{C}^{2} / 2 \mathrm{kv}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2179</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2179"/>
		<updated>2022-03-18T13:06:39Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
&lt;br /&gt;
# part1&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
   &lt;br /&gt;
    During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
# part2&lt;br /&gt;
&lt;br /&gt;
* '''(30%)'''&lt;br /&gt;
    &lt;br /&gt;
    The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
# part3&lt;br /&gt;
&lt;br /&gt;
* '''(20%)'''&lt;br /&gt;
&lt;br /&gt;
    The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* '''(20%)'''&lt;br /&gt;
&lt;br /&gt;
    while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{v}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
where $$T_{b}$$ is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{C}^{2} / 2 \mathrm{kv}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2178</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2178"/>
		<updated>2022-03-18T13:05:02Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
&lt;br /&gt;
# part1&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
    During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
# part2&lt;br /&gt;
&lt;br /&gt;
* '''(30%)'''&lt;br /&gt;
The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
# part3&lt;br /&gt;
&lt;br /&gt;
* '''(20%)'''&lt;br /&gt;
The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* '''(20%)'''&lt;br /&gt;
while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{v}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
where $$T_{b}$$ is the brightness temperature. Then we have:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{C}^{2} / 2 \mathrm{kv}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2177</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2177"/>
		<updated>2022-03-18T13:04:00Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 英文解答 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
&lt;br /&gt;
# part1&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
    During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
    Flux density observed at 1660 MHz is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
# part2&lt;br /&gt;
&lt;br /&gt;
* '''(30%)'''&lt;br /&gt;
The solid angle subtended by this source of radiation is&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\Omega=\pi(r / R)^{2} &amp;amp;=3.14\left(2.1 \times 10^{7} / 2.3 \times 10^{3} \times 3.086 \times 10^{18}\right)^{2} \\&lt;br /&gt;
&amp;amp;=2.75 \times 10^{-29} \mathrm{sr}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
# part3&lt;br /&gt;
&lt;br /&gt;
* '''(20%)'''&lt;br /&gt;
The flux density is related to the total brightness by the relation:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}}=\mathrm{B}_{1660 \mathrm{MHz}} \Omega&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* '''(20%)'''&lt;br /&gt;
while at this frequency, the total brightness can be approximated from the Rayleigh-Jeans formula:&lt;br /&gt;
&lt;br /&gt;
$$&lt;br /&gt;
\mathrm{B}_{1660 \mathrm{MHz}}=2 k \mathrm{~T}_{\mathrm{b}} \mathrm{v}^{2} / \mathrm{c}^{2}&lt;br /&gt;
$$&lt;br /&gt;
&lt;br /&gt;
* '''(10%)'''&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\mathrm{S}_{1660 \mathrm{MHz}} \mathrm{C}^{2} / 2 \mathrm{kv}^{2} \Omega \\&lt;br /&gt;
\mathrm{T}_{\mathrm{b}} &amp;amp;=\left[3.5 \times 10^{-21} \times\left(3 \times 10^{10}\right)^{2}\right] /\left[2 \times\left(1.38 \times 10^{-16}\right) \times\left(1660 \times 10^{6}\right)^{2} \times\left(2.75 \times 10^{-29}\right)\right] \\&lt;br /&gt;
&amp;amp;=1.5 \times 10^{26} \mathrm{~K}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2176</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2176"/>
		<updated>2022-03-18T12:57:51Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;br /&gt;
&lt;br /&gt;
==中文翻译==&lt;br /&gt;
一个天体发出强烈的连续谱射电信号脉冲，持续时间为700$$\mu s$$。在频率为1660Mhz出测得其流量密度为0.35kJy。如果这个射电源的距离为2.3kpc，估计这个源的亮温度&lt;br /&gt;
&lt;br /&gt;
==英文解答==&lt;br /&gt;
&lt;br /&gt;
# '''(10%)'''&lt;br /&gt;
During 700μs,the radio wave travels about $$r = ct = 3 × 10^{10} × 700 × 10^{-6} = 2.1 × 10^{7} cm$$. The region from where the burst originates must be no larger than the distance that light can travel during the duration of the burst. So we estimate that r is the size of the source. ($$r = 2.1 × 10^{7} cm, R = 2.3 kpc$$)&lt;br /&gt;
&lt;br /&gt;
# '''(10%)'''&lt;br /&gt;
Flux density observed at 1660 MHz is&lt;br /&gt;
$$&lt;br /&gt;
\begin{aligned}&lt;br /&gt;
\mathrm{S}_{1660 \mathrm{MHz}} &amp;amp;=0.35 \mathrm{kJy} \\&lt;br /&gt;
&amp;amp;=0.35 \times 10^{3} \times 10^{-23} \mathrm{ergs} \mathrm{s}^{-1} \mathrm{~cm}^{-2} \mathrm{~Hz}^{-1} \\&lt;br /&gt;
&amp;amp;=3.5 \times 10^{-21} \text { ergs s } \mathrm{cm}^{-2} \mathrm{~Hz}^{-1}&lt;br /&gt;
\end{aligned}&lt;br /&gt;
$$&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2175</id>
		<title>2015IOAA理论短问题第8题-亮温度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC8%E9%A2%98-%E4%BA%AE%E6%B8%A9%E5%BA%A6&amp;diff=2175"/>
		<updated>2022-03-18T12:45:42Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：创建页面，内容为“==英文题目== A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density…”&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A strong continuum radio signal from a celestial body has been observed as a burst with a very short duration of 700 μs. The observed flux density at a frequency of 1660 MHz is measured to be 0.35 kJy. If the distance from the source is known to be 2.3 kpc, estimate the brightness temperature of this source.&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC7%E9%A2%98-%E6%98%9F%E5%9B%A2%E5%AF%86%E5%BA%A6&amp;diff=2173</id>
		<title>2015IOAA理论短问题第7题-星团密度</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC7%E9%A2%98-%E6%98%9F%E5%9B%A2%E5%AF%86%E5%BA%A6&amp;diff=2173"/>
		<updated>2022-03-18T12:30:35Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：创建页面，内容为“==英文题目== A galaxy at the boundary of a galaxy cluster of radius 10 Mpc is expected to escape from the cluster if it has an initial velocity of at least 700 k…”&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
A galaxy at the boundary of a galaxy cluster of radius 10 Mpc is expected to escape from the cluster if it has an initial velocity of at least 700 km/s relative to the center of the cluster. Calculate the density of the cluster.&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC6%E9%A2%98-%E5%A4%A7%E6%B0%94%E5%85%89%E6%B7%B1&amp;diff=2172</id>
		<title>2015IOAA理论短问题第6题-大气光深</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC6%E9%A2%98-%E5%A4%A7%E6%B0%94%E5%85%89%E6%B7%B1&amp;diff=2172"/>
		<updated>2022-03-18T12:29:59Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：创建页面，内容为“==英文题目== At the start of every observation, a radio telescope is pointed at a point-source calibrator that has a known flux density of 21.86 Jy outside the E…”&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
At the start of every observation, a radio telescope is pointed at a point-source calibrator that has a known flux density of 21.86 Jy outside the Earth’s atmosphere. However, on a certain date, the measured flux density of the calibrator source was 14.27 Jy. If the calibrator source was at an altitude of 35 degrees, estimate the zenith atmospheric optical depth, $$\tau_{z}$$.&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC5%E9%A2%98-%E5%8D%AB%E6%98%9F%E8%BD%A8%E9%81%93&amp;diff=2171</id>
		<title>2015IOAA理论短问题第5题-卫星轨道</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC5%E9%A2%98-%E5%8D%AB%E6%98%9F%E8%BD%A8%E9%81%93&amp;diff=2171"/>
		<updated>2022-03-18T12:29:04Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：创建页面，内容为“==英文题目== An observer is trying to determine an approximate value of the orbital eccentricity of a man-made satellite. When the satellite was at apogee, it wa…”&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
An observer is trying to determine an approximate value of the orbital eccentricity of a man-made satellite. When the satellite was at apogee, it was observed to have moved by $$∆𝜃_{1}$$ = 2′44&amp;quot; in a short time. When the radius vector connecting Earth and the satellite is perpendicular to the major axis (true anomaly is equal to 90o), within the same duration of time, it was observed to have moved by $$∆𝜃_{2}$$ = 21′17&amp;quot;. Assume that the observer is located at the center of the Earth. Find an approximate value of the eccentricity of the satellite’s orbit.&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC4%E9%A2%98-%E8%A7%82%E6%B5%8B%E9%BB%91%E6%B4%9E&amp;diff=2170</id>
		<title>2015IOAA理论短问题第4题-观测黑洞</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC4%E9%A2%98-%E8%A7%82%E6%B5%8B%E9%BB%91%E6%B4%9E&amp;diff=2170"/>
		<updated>2022-03-18T12:26:50Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：创建页面，内容为“==英文题目== Let us assume that an observer using a hypothetical, far-infrared, Earth-sized telescope (wavelength range 20 to 640 $$\mu m$$) found a static and n…”&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==英文题目==&lt;br /&gt;
Let us assume that an observer using a hypothetical, far-infrared, Earth-sized telescope (wavelength range 20 to 640 $$\mu m$$) found a static and neutral supermassive black hole with a mass of $$2.1 \times 10^{10} M_{\odot}$$. Determine the maximum distance at which this black hole can be resolved by the observer.&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=IOAA2015&amp;diff=2169</id>
		<title>IOAA2015</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=IOAA2015&amp;diff=2169"/>
		<updated>2022-03-18T12:25:35Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 短问题 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==理论考试==&lt;br /&gt;
====短问题====&lt;br /&gt;
PS:以下题目的标题为本页面编辑者自行命名&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第1题-共振轨道]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第2题-行星质量]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第3题-小行星掩星]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第4题-观测黑洞]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第5题-卫星轨道]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第6题-大气光深]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第7题-星团密度]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第8题-亮温度]]&lt;br /&gt;
&lt;br /&gt;
==数据分析考试==&lt;br /&gt;
&lt;br /&gt;
==观测考试==&lt;br /&gt;
&lt;br /&gt;
==中国队参赛名单==&lt;br /&gt;
&lt;br /&gt;
==参赛国家==&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=IOAA2015&amp;diff=2168</id>
		<title>IOAA2015</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=IOAA2015&amp;diff=2168"/>
		<updated>2022-03-18T12:25:26Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 理论考试 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==理论考试==&lt;br /&gt;
====短问题====&lt;br /&gt;
PS:以下题目的标题为本页面编辑者自行命名&lt;br /&gt;
[[2015IOAA理论短问题第1题-共振轨道]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第2题-行星质量]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第3题-小行星掩星]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第4题-观测黑洞]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第5题-卫星轨道]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第6题-大气光深]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第7题-星团密度]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第8题-亮温度]]&lt;br /&gt;
&lt;br /&gt;
==数据分析考试==&lt;br /&gt;
&lt;br /&gt;
==观测考试==&lt;br /&gt;
&lt;br /&gt;
==中国队参赛名单==&lt;br /&gt;
&lt;br /&gt;
==参赛国家==&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC3%E9%A2%98-%E5%B0%8F%E8%A1%8C%E6%98%9F%E6%8E%A9%E6%98%9F&amp;diff=2167</id>
		<title>2015IOAA理论短问题第3题-小行星掩星</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=2015IOAA%E7%90%86%E8%AE%BA%E7%9F%AD%E9%97%AE%E9%A2%98%E7%AC%AC3%E9%A2%98-%E5%B0%8F%E8%A1%8C%E6%98%9F%E6%8E%A9%E6%98%9F&amp;diff=2167"/>
		<updated>2022-03-18T12:19:40Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：创建页面，内容为“==英文题目== On 27 May 2015 at 02:18:49, the occultation of the star HIP 89931 (𝛿 −Sgr) by the asteroid 1285 Julietta was observed from Borobudur temple, wh…”&lt;/p&gt;
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&lt;div&gt;==英文题目==&lt;br /&gt;
On 27 May 2015 at 02:18:49, the occultation of the star HIP 89931 (𝛿 −Sgr) by the asteroid 1285 Julietta was observed from Borobudur temple, which was located at the center of the asteroid shadow path. It lasted for only 6.201 s. Assume that Earth’s orbit is circular and the orbit of Julietta is on the ecliptic plane and revolves in the same direction as Earth. At the occultation, Julietta is near its aphelion. At the time of the occultation the distances of Julietta from the Sun and the Earth are 3.076 AU and 2.156 AU respectively. Find the approximate diameter of asteroid Julietta, if the semi major axis of Julietta is 𝑎 = 2.9914 AU.&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
	<entry>
		<id>https://www.astro-init.top/index.php?title=IOAA2015&amp;diff=2166</id>
		<title>IOAA2015</title>
		<link rel="alternate" type="text/html" href="https://www.astro-init.top/index.php?title=IOAA2015&amp;diff=2166"/>
		<updated>2022-03-18T12:19:16Z</updated>

		<summary type="html">&lt;p&gt;某九流辣鸡民科：/* 短问题 */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;==理论考试==&lt;br /&gt;
====短问题====&lt;br /&gt;
[[2015IOAA理论短问题第1题-共振轨道]]&lt;br /&gt;
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[[2015IOAA理论短问题第2题-行星质量]]&lt;br /&gt;
&lt;br /&gt;
[[2015IOAA理论短问题第3题-小行星掩星]]&lt;br /&gt;
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==数据分析考试==&lt;br /&gt;
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==观测考试==&lt;br /&gt;
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==中国队参赛名单==&lt;br /&gt;
&lt;br /&gt;
==参赛国家==&lt;/div&gt;</summary>
		<author><name>某九流辣鸡民科</name></author>
		
	</entry>
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