2024年IOAA理论第10题-最大日食

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英文题目

T10. Greatest Eclipse (75 points)

The greatest eclipse is defined as the instant when the axis of the Moon's shadow cone gets closest to the centre of the Earth in a solar eclipse. This problem explores the geometry of this phenomenon, using the solar eclipse of 29\(^\text{th}\) May 1919 as an example, as it has great historical significance for being the first time astronomers were able to observationally verify general relativity. One of the scientific expeditions to observe this eclipse took place in the Brazilian city of Sobral.

The two following tables show the Cartesian and spherical coordinates of the Sun and the Moon at the time of the greatest eclipse. The system used for these coordinates is right-handed and has the origin at the centre of the Earth, the positive \(x\)-axis pointing towards the Greenwich meridian, and the positive \(z\)-axis pointing towards the North Pole. For the rest of this problem, this will be referred to as System I.

Spherical Coordinates:

	Centre of the Sun	Centre of the Moon
Radial Distance (\(r\))	\(1.516 \times 10^{11}\) m	\(3.589 \times 10^8\) m
Polar Angle (\(\theta\))	\(68^{\circ}29'44.1''\)	\(68^{\circ}47'41.6''\)
Azimuthal Angle (\(\varphi\))	\(-1^h11^{m}28.2^s\)	\(-1^h11^{m}22.9^s\)

Cartesian Coordinates:

	Centre of the Sun	Centre of the Moon
\(x\)	\(1.342 \times 10^{11}\) m	\(3.185 \times 10^8\) m
\(y\)	\(-4.327 \times 10^{10}\) m	\(-1.025 \times 10^8\) m
\(z\)	\(5.557 \times 10^{10}\) m	\(1.298 \times 10^8\) m

For this problem, assume that the Earth is a perfect sphere.

Note: The spherical coordinates of a point \(P\) are defined as follows:

- Radial distance (\(r\)): distance between the origin (\(O\)) and \(P\) (range: \(r \geq 0\)). - Polar angle (\(\theta\)): angle between the positive \(z\)-axis and the line segment \(OP\) (range: \(0^{\circ} \leq \theta \leq 180^{\circ}\)). - Azimuthal angle (\(\varphi\)): angle between the positive \(x\)-axis and the projection of the line segment \(OP\) onto the \(xy\)-plane (range: \(-12^h \leq \varphi < 12^h\)).

Part I: Geographic Coordinates (25 points)

(a) (3 points) Determine the declination of the Sun and the Moon during the greatest eclipse for a geocentric observer.

(b) (3 points) Determine the right ascension of the Sun and the Moon at the time of the greatest eclipse for a geocentric observer. The local sidereal time at Greenwich at that same moment was \(5^h32^m35.5^s\).

(c) (4 points) Find a unit vector that indicates the direction of the axis of the Moon’s shadow cone. This vector should point from the Moon to the vicinity of the centre of the Earth.

(d) (15 points) Determine the latitude and the longitude of the point where the axis of the Moon's shadow cone crosses the surface of the Earth during the greatest eclipse.

Part II: Duration of the Totality (50 points)

Precisely determining the duration of totality of a solar eclipse involves complex calculations that would be beyond the scope of this problem. However, it is possible to obtain a reasonable approximation for this value using the two following assumptions:

- The size of the umbra on the surface of the Earth remains roughly constant throughout totality at a given location. - The velocity of the umbra on the surface of the Earth remains roughly constant throughout totality at a given location.

(e) (10 points) Estimate the radius of the umbra during the greatest eclipse. In order to simplify the calculations, assume that the umbra is small enough that it can be considered approximately flat and that the axis of the Moon's shadow cone is extremely close to the centre of the Earth during the greatest eclipse.

(f) (3 points) Calculate the velocity of the Earth's rotation at the latitude of the centre of the umbra.

(g) (4 points) Determine the orbital velocity of the Moon at the instant of the greatest eclipse. Neglect the changes in the semi-major axis of the Moon’s orbit.

For the remaining items of this problem, assume that the tangential velocity of the Moon is roughly the same as the orbital velocity and neglect its radial component.

(h) (14 points) Using System II, determine the velocity vector of the Moon during the greatest eclipse. Note that the intersection between the Celestial Equator and the lunar orbit that is closer to the position of the eclipse has a right ascension of \(23^{h}07^{m}59.2^s\).

(i) (10 points) Write the velocity vector of the Moon in System III. Note that in System I, the azimuthal angle difference between the positions of the origins OII and OIII is negligible, so you should only take into account the difference in the polar angles.

(j) (6 points) Calculate the speed of the centre of the umbra along the surface of the Earth at the instant of the greatest eclipse.

(k) (3 points) Estimate the duration of the totality of the eclipse at the location with the coordinates found in part (d).

中文翻译

T10. 最大食分（75分）

最大食分定义为月球本影锥轴最接近地球中心的时刻。本题将探讨这一现象的几何原理，并以1919年5月29日日全食为例（该次日食具有重大历史意义，是天文学家首次通过观测验证广义相对论的契机）。其中一支科学考察队就是在巴西索布拉尔市进行观测的。

以下两个表格分别给出了最大食分时刻太阳和月球的球面坐标与笛卡尔坐标。该坐标系为右手坐标系，原点在地球中心，x轴正方向指向格林尼治子午线，z轴正方向指向北极点。在下文中称为"系统I"。

球面坐标：

	太阳中心	月球中心
径向距离 (\(r\))	\(1.516 \times 10^{11}\) m	\(3.589 \times 10^8\) m
极角 (\(\theta\))	\(68^{\circ}29'44.1''\)	\(68^{\circ}47'41.6''\)
方位角 (\(\varphi\))	\(-1^h11^{m}28.2^s\)	\(-1^h11^{m}22.9^s\)

笛卡尔坐标：

	太阳中心	月球中心
\(x\)	\(1.342 \times 10^{11}\) m	\(3.185 \times 10^8\) m
\(y\)	\(-4.327 \times 10^{10}\) m	\(-1.025 \times 10^8\) m
\(z\)	\(5.557 \times 10^{10}\) m	\(1.298 \times 10^8\) m

注意：点\(P\)的球面坐标定义如下：

- 径向距离 (\(r\))：原点\(O\)到\(P\)的距离（范围：\(r \geq 0\)）

- 极角 (\(\theta\))：z轴正方向与线段\(OP\)的夹角（范围：\(0^{\circ} \leq \theta \leq 180^{\circ}\)）

- 方位角 (\(\varphi\))：x轴正方向与线段\(OP\)在xy平面投影的夹角（范围：\(-12^h \leq \varphi < 12^h\)）

第一部分：地理坐标（25分）

(a)（3分）确定地心观测者所见的最大食分时刻太阳和月球的赤纬。

(b)（3分）确定地心观测者所见的最大食分时刻太阳和月球的赤经。此时格林尼治的本地恒星时为\(5^h32^m35.5^s\)。

(c)（4分）求表示月球本影锥轴方向的单位向量。该向量应从月球指向地球中心附近。

(d)（15分）确定月球本影锥轴在最大食分时刻与地球表面的交点经纬度。

第二部分：全食持续时间（50分）

精确计算日全食持续时间涉及复杂运算，超出本题范围。但通过以下两个假设可获得合理近似值：

- 在特定位置，地球表面本影尺寸在全食期间大致保持不变

- 在特定位置，地球表面本影移动速度在全食期间大致保持不变

(e)（10分）估算最大食分时刻本影半径。为简化计算，可假设本影足够小可视为平面，且此时月球本影锥轴极度接近地球中心。

(f)（3分）计算本影中心所在纬度处的地球自转线速度。

(g)（4分）确定最大食分时刻月球的轨道速度。忽略月球轨道半长轴的变化。

以下问题中，假设月球的切向速度大致等于轨道速度，并忽略其径向分量。

(h)（14分）使用系统II确定月球在最大食分时刻的速度向量。已知月球轨道与天赤道交点中较接近食分位置的那个点赤经为\(23^h07^m59.2^s\)。

(i)（10分）写出系统III中月球的速度向量。注意在系统I中，原点OII与OIII的方位角差异可忽略，只需考虑极角差异。

(j)（6分）计算最大食分时刻本影中心沿地球表面的移动速率。

(k)（3分）估算(d)中坐标位置的全食持续时间。

官方解答

第一部分：地理坐标解答

(a) 赤纬计算：

$$ \delta = 90^\circ - \theta $$

太阳赤纬：

$$ \delta_\odot = 90^\circ - 68^\circ29'44.1'' = 21^\circ30'15.9'' $$

月球赤纬：

$$ \delta_{\text{Moon}} = 90^\circ - 68^\circ47'41.6'' = 21^\circ12'18.4'' $$

(b) 赤经计算：

$$ \alpha = \varphi + \text{LST}_{\text{Greenwich}} $$

太阳赤经：

$$ \alpha_\odot = -1^h11^m28.2^s + 5^h32^m35.5^s = 4^h21^m7.3^s $$

月球赤经：

$$ \alpha_{\text{Moon}} = -1^h11^m22.9^s + 5^h32^m35.5^s = 4^h21^m12.6^s $$

(c) 本影轴单位向量：

$$ \vec{u} = \frac{\vec{M}_{\text{Moon}} - \vec{M}_\odot}{|\vec{M}_{\text{Moon}} - \vec{M}_\odot|} $$

计算结果：

$$ \vec{u} = \left\langle -0.8855, 0.2855, -0.3666 \right\rangle $$

(d) 地表交点坐标：

1. 建立方程：

$$ |\vec{M}_{\text{Moon}} + k\vec{u}|^2 = R_\oplus^2 $$

2. 解得缩放因子：

$$ k = 3.528 \times 10^8 $$

3. 转换球面坐标：

$$ \theta = \arccos\left(\frac{z}{r}\right) = 85^\circ38' $$

$$ \varphi = \arctan\left(\frac{y}{x}\right) = -16^\circ42' $$

最终坐标：

$$ 4^\circ22'N, 16^\circ42'W $$

第二部分：全食持续时间解答

(e) 本影半径估算：

$$ \beta = \arccos\left(\frac{R_\odot - R_{\text{Moon}}}{d_\odot - d_{\text{Moon}}}\right) = 89.74^\circ $$

$$ r_{\text{umbra}} = \frac{R_{\text{Moon}} - (d_{\text{Moon}} - R_\odot)\cos\beta}{\sin\beta} = 1.196 \times 10^5 \text{ m} $$

(f) 地球自转线速度：

$$ v_{\text{rot}} = \frac{2\pi R_\oplus \cos\phi}{T_{\text{sidereal}}} = 463.7 \text{ m/s} $$

(g) 月球轨道速度：

$$ v_{\text{Moon}} = \sqrt{GM_\oplus\left(\frac{2}{d_{\text{Moon}}} - \frac{1}{a_{\text{Moon}}}\right)} = 1088 \text{ m/s} $$

(h) 系统II速度向量：

$$ \kappa = \arctan\left(\frac{\sin\delta_{\text{Moon}}}{\tan(\alpha_{\text{Moon}} - \alpha_{\text{intersection}})}\right) = 4^\circ17' $$

$$ \vec{v}_{\text{II}} = \begin{bmatrix} 1085\cos\kappa \\ 1085\sin\kappa \\ 0 \end{bmatrix} = \begin{bmatrix} 1085 \\ 81.25 \\ 0 \end{bmatrix} \text{ m/s} $$

(i) 系统III速度向量：

$$ \vec{v}_{\text{III}} = \begin{bmatrix} v_x \\ v_y\cos\Delta\theta \\ -v_y\sin\Delta\theta \end{bmatrix} = \begin{bmatrix} 1085 \\ 77.76 \\ -23.54 \end{bmatrix} \text{ m/s} $$

(j) 本影移动速率合成：

$$ \vec{v}_{\text{umbra}} = \sqrt{(v_x - v_{\text{rot}})^2 + v_y^2} = 626.3 \text{ m/s} $$

(k) 全食持续时间：

$$ \Delta t = \frac{2r_{\text{umbra}}}{v_{\text{umbra}}} = 6\text{分}21.4\text{秒} $$