2024年IOAA理论第7题-漂流者
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英文题目
T7. Castaway (20 points)
After surviving a shipwreck and reaching a small island in the southern hemisphere, a sailor had to estimate the island's latitude using the Sun.
However, due to poor eyesight, the sailor couldn't see the night stars very well, so his best option was to rely on the Sun. He had no information about the date, but he realized the days were longer than the nights.
(a) (7 points) The sailor noticed that on his first day on the island, the angle between the positions of sunrise and sunset on the horizon was \(120^\circ\). With this piece of information, determine the range of possible values for the latitude of the island. Neglect the daily motion of the Sun across the ecliptic.
(b) (13 points) The angle between the positions of sunrise and sunset kept increasing daily. After 40 days, this angle was equal to \(163^\circ\). Estimate the latitude of the island. You may neglect the eccentricity of the Earth's orbit.
中文翻译
T7. 落难者（20分）
一名水手在遭遇海难后到达南半球的一个小岛，需利用太阳估算该岛的纬度。
由于视力不佳，水手无法清晰看到夜间的星星，因此只能依赖太阳。他没有关于日期的信息，但注意到白天长于夜晚。
(a) (7分) 水手注意到在岛上第一天时，日出和日落位置在地平线上的夹角为\(120^\circ\)。利用这一信息，确定该岛纬度的可能取值范围。忽略太阳在黄道上的日常运动。
(b) (13分) 日出与日落位置之间的夹角逐日增大。40天后，该夹角达到\(163^\circ\)。估算该岛的纬度。可忽略地球轨道的偏心率。
官方解答
解答：
(a) 第一步是推导日出时天体方位角的表达式：
$$ \cos(90^\circ + \delta) = \cos(90^\circ - a) \cos(90^\circ + \phi) + \sin(90^\circ - a) \sin(90^\circ + \phi) \cos(180^\circ - A) $$
由于日出时高度角\(a = 0^\circ\)，简化为：
$$ -\sin(\delta) = -\cos(\phi) \cos(A) $$
通过对称性分析，当太阳赤纬为\(-23.5^\circ\)时，纬度下限为：
$$ \phi = -37.1^\circ $$
当太阳赤纬接近\(0^\circ\)时，纬度理论上接近南极。因此可能纬度范围为：
$$ -90^\circ < \phi \leq -37.1^\circ $$
(b) 结合赤纬变化与时间关系，建立方程：
$$ \cos(\phi) = \frac{\sin(\delta_0)}{\cos(A_0)} = \frac{\sin(\delta_{40})}{\cos(A_{40})} $$
通过轨道运动角速度关系：
$$ \theta_{40} = \theta_0 - \frac{40}{365.2564} \times 360^\circ = \theta_0 - 39.4^\circ $$
求解得：
$$ \theta_0 = 53.1^\circ \quad \Rightarrow \quad \delta_0 = -18.6^\circ $$
最终纬度计算为：
$$ \phi = -50.4^\circ $$
