2020年GeCAA理论第2题-地平论 - 版本历史

2021年6月16日 (三) 14:50 Jingsong Guo

2021-06-16T14:50:27Z

2021年6月16日 (三) 14:49 Jingsong Guo

2021-06-16T14:49:17Z

2021年6月16日 (三) 14:48 Jingsong Guo

2021-06-16T14:48:42Z

2021年6月16日 (三) 14:48 Jingsong Guo

2021-06-16T14:48:07Z

2021年6月16日 (三) 14:47 Jingsong Guo

2021-06-16T14:47:43Z

2021年6月16日 (三) 14:40 Jingsong Guo

2021-06-16T14:40:36Z

2021年6月16日 (三) 14:40 Jingsong Guo

2021-06-16T14:40:04Z

2021年6月16日 (三) 14:39 Jingsong Guo

2021-06-16T14:39:18Z

2021年6月16日 (三) 14:38 Jingsong Guo

2021-06-16T14:38:30Z

2021年6月16日 (三) 13:29 Jingsong Guo

2021-06-16T13:29:10Z

@@ 第80行： / 第80行： @@
 By the spherical Law of Cosines on angle O of spherical triangle OXZ, we have
-$$cos θ1 = cos ϵ cos ϵ + sin ϵ sin ϵ cos(O)$$
+$$cos θ_1 = cos ϵ  cos ϵ + sin ϵ  sin ϵ  cos(O)$$
 $$O = cos^{−1}
+(

@@ 第85行： / 第85行： @@
 $$O = cos^{−1}
+(
-dfrac{cos θ_1 − cos^2 ϵ}
+\dfrac{cos θ_1 − cos^2 ϵ}
 {sin^2 ϵ}
 )$$
@@ 第91行： / 第91行： @@
-$$\therefore Δt = dfrac{\angle O}
+$$\therefore Δt = \dfrac{\angle O}
 {2π}
 × P $$

@@ 第99行： / 第99行： @@
 {2π}
 ×cos^{−1}
+(\dfrac{cos θ_1 − cos^2 ϵ}
 {sin^2 ϵ}
 )$$
@@ 第107行： / 第106行： @@
 $$=
-dfrac{25800}
+\dfrac{25800}
 {2π}
 × cos^{−1}
+(
-dfrac{0.9880 − cos^2 23.5°}
+\dfrac{0.9880 − cos^2 23.5°}
 {sin^2 23.5°}
 )$$$$≈ 1606 yr$$
 Thus, the average temperature of the earth will go down by 1 °C in just over 1600 years. <div style="float:right">(''1.0 points'')</div>

←上一版本		2021年6月16日 (三) 14:48的版本
第115行：		第115行：
	)$$$$≈ 1606 yr$$		)$$$$≈ 1606 yr$$

−	Thus, the average temperature of the earth will go down by 1 °C in just over 1600	+	Thus, the average temperature of the earth will go down by 1 °C in just over 1600 years. <div style="float:right">(''1.0 points'')</div>
−	years. <div style="float:right">(''1.0 points'')</div>

@@ 第113行： / 第113行： @@
 dfrac{0.9880 − cos^2 23.5°}
 {sin^2 23.5°}
-)$$
+)$$$$≈ 1606 yr$$
 Thus, the average temperature of the earth will go down by 1 °C in just over 1600
 years. <div style="float:right">(''1.0 points'')</div>

←上一版本		2021年6月16日 (三) 14:40的版本
第74行：		第74行：
	the celestial sphere, let O be the center of precession, Z be the current direction of		the celestial sphere, let O be the center of precession, Z be the current direction of
	the axis, and X be direction of the axis when it makes an angle of $$θ_1$$ with the sun,		the axis, and X be direction of the axis when it makes an angle of $$θ_1$$ with the sun,
−	3	+
	so $$\angle ZCX = θ_1$$. If ϵ is the radius of precession, then$$ \angle OCZ = \angle OCX = ϵ$$. <div style="float:right">(''1.0 points'')</div>		so $$\angle ZCX = θ_1$$. If ϵ is the radius of precession, then$$ \angle OCZ = \angle OCX = ϵ$$. <div style="float:right">(''1.0 points'')</div>

@@ 第28行： / 第28行： @@
 ''Solution''
- Assume the surface area of one side of the flat Earth is $$A$$. Let the angle
+Assume the surface area of one side of the flat Earth is $$A$$. Let the angle
 between the Sun and the flat Earth’s center axis be $$θ$$, where $$θ$$ is initially $$0°$$. As
 the Sun’s rays are parallel, the power delivered to the Earth by the Sun will be
@@ 第94行： / 第94行： @@
 {2π}
 × P $$
 $$=
 \dfrac{P}

@@ 第35行： / 第35行： @@
 At equilibrium, this is the energy radiated away via blackbody radiation, so the
 equilibrium temperature $$T$$ satisfies
-$$S_⊙Acos θ = σ(2A)T^4$$ 2.0
+$$S_⊙Acos θ = σ(2A)T^4$$ <div style="float:right">(''2.0 points'')</div>
 , where the factor of 2 comes from the fact that the flat Earth would radiate energy
 from both sides.
 This yields
-T(θ) = 4
+$$T(θ) = \sqrt[4]{\dfrac{
+S_⊙ cos θ}{
-S⊙ cos θ
+σ}}$$
 σ
-and we wish to find the value of θ1 such that T(θ1) = T(0) − ΔT.
+and we wish to find the value of $$θ_1$$ such that $$T(θ_1) = T(0) − ΔT$$.
 Thus,
-−
+$$\sqrt[4]{\dfrac{S_⊙ cos θ_1}
-S⊙ cos θ1
+{2σ}}
-σ
+= \sqrt[4]{{S_⊙}
-= 4
+{2σ}}
+− ΔT$$ <div style="float:right">(''2.0 points'')</div>
-S⊙
 σ
-− ΔT 2.0
+$$cos θ_1 =
-− ΔT 4
+(1 − ΔT \sqrt[4]{\dfrac{
+σ}
-σ
+{S⊙}}
-S⊙
+)^4$$
-!4
+<div style="float:right">(''1.0 points'')</div>
-.0
+$$=
+(1 − \sqrt[4]{\dfrac{
+× 5.67 × 10^{−8}}
-− 4
+{1366}}
+)^4$$
-× 5.67 × 10−8
+$$cos θ_1 = 0.9880$$ <div style="float:right">(''1.0 points'')</div>
-!4
-cos θ1 = 0.9880 1.0
 Now, we find the time it takes for the axis to make such an angle with the Sun. On
 the celestial sphere, let O be the center of precession, Z be the current direction of
-the axis, and X be direction of the axis when it makes an angle of θ1 with the sun,
+the axis, and X be direction of the axis when it makes an angle of $$θ_1$$ with the sun,
-so ]ZCX = θ1. If ϵ is the radius of precession, then ]OCZ = ]OCX = ϵ. 1.0
+so $$\angle ZCX = θ_1$$. If ϵ is the radius of precession, then$$ \angle OCZ = \angle OCX = ϵ$$. <div style="float:right">(''1.0 points'')</div>
 By the spherical Law of Cosines on angle O of spherical triangle OXZ, we have
-cos θ1 = cos ϵ cos ϵ + sin ϵ sin ϵ cos(^O)
-= cos2 ϵ + sin2 ϵ cos(^O)
+$$cos θ1 = cos ϵ cos ϵ + sin ϵ sin ϵ cos(O)$$
-^O = cos−1
-�
+$$= cos2 ϵ + sin2 ϵ cos(O)$$
-cos θ1 − cos2 ϵ
+$$O = cos^{−1}
-sin2 ϵ
-�
+dfrac{cos θ_1 − cos^2 ϵ}
-.0
+{sin^2 ϵ}
-) Δt = ^O
-π
+<div style="float:right">(''1.0 points'')</div>
 × P =
-−
-π
+$$\therefore Δt = dfrac{\angle O}
-× cos−1
+{2π}
-�
+× P $$
-cos θ1 − cos2 ϵ
+$$=
-sin2 ϵ
+\dfrac{P}
-�
+{2π}
-.0
+×cos^{−1}
-−
+(
+dfrac{cos θ_1 − cos^2 ϵ}
-π
+{sin^2 ϵ}
-× cos−1
+)$$
-�
+<div style="float:right">(''1.0 points'')</div>
-.9880 − cos2 23.5°
-sin2 23.5°
-�
+$$=
-≈ 1606 yr
+dfrac{25800}
 Thus, the average temperature of the earth will go down by 1 °C in just over 1600
-years. 1.0
+years. <div style="float:right">(''1.0 points'')</div>

←上一版本		2021年6月16日 (三) 13:29的版本
第22行：		第22行：
	Determine how many years it will take for the Earth’s equilibrium temperature to decrease		Determine how many years it will take for the Earth’s equilibrium temperature to decrease
	by 1 °C.		by 1 °C.
		+
		+	==中文翻译==
		+
		+	==官方解答==
		+	''Solution''
		+
		+	Assume the surface area of one side of the flat Earth is $$A$$. Let the angle
		+	between the Sun and the flat Earth’s center axis be $$θ$$, where $$θ$$ is initially $$0°$$. As
		+	the Sun’s rays are parallel, the power delivered to the Earth by the Sun will be
		+	$$S_⊙Acos θ $$at any given point in time.
		+
		+	At equilibrium, this is the energy radiated away via blackbody radiation, so the
		+	equilibrium temperature $$T$$ satisfies
		+	$$S_⊙Acos θ = σ(2A)T^4$$ 2.0
		+	, where the factor of 2 comes from the fact that the flat Earth would radiate energy
		+	from both sides.
		+	This yields
		+	T(θ) = 4
		+	r
		+	S⊙ cos θ
		+	2σ
		+	and we wish to find the value of θ1 such that T(θ1) = T(0) − ΔT.
		+	Thus,
		+	4
		+	r
		+	S⊙ cos θ1
		+	2σ
		+	= 4
		+	r
		+	S⊙
		+	2σ
		+	− ΔT 2.0
		+	cos θ1 =
		+
		+	1 − ΔT 4
		+	s
		+	2σ
		+	S⊙
		+	!4
		+	1.0
		+	=
		+
		+	1 − 4
		+	r
		+	2 × 5.67 × 10−8
		+	1366
		+	!4
		+	cos θ1 = 0.9880 1.0
		+	Now, we find the time it takes for the axis to make such an angle with the Sun. On
		+	the celestial sphere, let O be the center of precession, Z be the current direction of
		+	the axis, and X be direction of the axis when it makes an angle of θ1 with the sun,
		+	3
		+	so ]ZCX = θ1. If ϵ is the radius of precession, then ]OCZ = ]OCX = ϵ. 1.0
		+	By the spherical Law of Cosines on angle O of spherical triangle OXZ, we have
		+	cos θ1 = cos ϵ cos ϵ + sin ϵ sin ϵ cos(^O)
		+	= cos2 ϵ + sin2 ϵ cos(^O)
		+	^O = cos−1
		+	�
		+	cos θ1 − cos2 ϵ
		+	sin2 ϵ
		+	�
		+	1.0
		+	) Δt = ^O
		+	2π
		+	× P =
		+	P
		+	2π
		+	× cos−1
		+	�
		+	cos θ1 − cos2 ϵ
		+	sin2 ϵ
		+	�
		+	1.0
		+	=
		+	25800
		+	2π
		+	× cos−1
		+	�
		+	0.9880 − cos2 23.5°
		+	sin2 23.5°
		+	�
		+	≈ 1606 yr
		+	Thus, the average temperature of the earth will go down by 1 °C in just over 1600
		+	years. 1.0
		+	4