Съдържание

IOAA 2022, Теория — Задача 9. Solar Retrograde Motion on Mercury

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2022

International Olympiad on Astronomy and Astrophysics 2022 Georgia — Theoretical Competition · 16 август 2022 г. · 20 т.

Условие

Mercury’s orbit has an unusually high eccentricity. Further, its sidereal rotational period is $\frac{2}{3}$ of its sidereal year. As a consequence of these factors, the Sun will exhibit retrograde motion in Mercury’s sky, when Mercury is near perihelion.

Calculate the total duration of this apparent solar retrograde motion during one orbit of Mercury around the Sun. Express your answer in Earth days. (20pt)

Отговори

Решение

From Kepler’s 3 $^{\mathrm{rd}}$ Law, $T^2 = a^3$ in units of Earth years and au respectively.

\therefore T_{\mathrm{Mercury}} = \sqrt{(0.387)^3} = 0.240\,75\,\mathrm{yr} = 87.934\,\mathrm{days} \qquad \mathbf{2.0}

We know that $\omega_{\mathrm{orb}} = v/r$ so will vary throughout the orbit (since $r$ varies within an ellipse) whilst $\omega_{\mathrm{rot}}$ is constant.

\omega_{\mathrm{rot}} = \frac{2\pi}{T_{\mathrm{rot}}} = \frac{2\pi}{\frac{2}{3} \times 87.934 \times 86400} = 1.24 \times 10^{-6}\,\mathrm{rad/s} \qquad \mathbf{2.0}

For retrograde motion, we need $\omega_{\mathrm{orb}} \geq \omega_{\mathrm{rot}}$ .

Using the vis-viva equation for the critical value of $r$ for when $\omega_{\mathrm{rot}} = \omega_{\mathrm{orb}}$ ,

\begin{aligned} \omega_{\mathrm{orb}}^2 = \frac{v^2}{r^2} = \frac{GM_{\odot}}{r^2}\left(\frac{2}{r} - \frac{1}{a}\right) &= \omega_{\mathrm{rot}}^2 \\ \frac{2GM_{\odot}}{r^3} - \frac{\mathrm{GM}_{\odot}}{ar^2} &= \omega_{\mathrm{rot}}^2 \\ \therefore \frac{\omega_{\mathrm{rot}}^2}{GM_{\odot}} r^3 + \frac{1}{a} r - 2 &= 0 \\ (1.160 \times 10^{-32})r^3 + (1.725 \times 10^{-11})r - 2 &= 0 \quad (\text{if } r \text{ is in meters}) \\ 38.99r^3 + 2.584r - 2 &= 0 \quad (\text{if } r \text{ is in au}) && \mathbf{4.0} \end{aligned}

Solving the cubic equation (by any valid method) gives only one non-imaginary root

One possible way would be to use iterations for this polynomial $f(r)$ , using perihelion distance as the starting guess.

$r$	$f(r)$
0.3073	-0.07482
0.3100	-0.03747
0.3120	-0.00967
0.3140	0.01841
0.3130	0.00433
0.3127	0.00012

\therefore r = 0.3127\,\mathrm{au} = 4.684 \times 10^{10}\,\mathrm{m} \qquad \mathbf{4.0}

From knowledge of ellipses, if $E$ is the eccentric anomaly

\begin{aligned} r &= a(1 - e\cos E) \\ \therefore E &= \cos^{-1}\left[\frac{1}{e}\left(1 - \frac{r}{a}\right)\right] = \cos^{-1}\left[\frac{1}{0.206}\left(1 - \frac{0.3127}{0.387}\right)\right] \\ E &= 0.3706\,\mathrm{rad} = 21.23^{\circ} = 21^{\circ}14' && \mathbf{3.0} \end{aligned}

Using Kepler’s Equation we can find the mean anomaly, $M$

\begin{aligned} M &= E - e\sin E = 0.3706 - 0.206 \times \sin 0.3706 \\ &= 0.2960\,\mathrm{rad} = 16.96^{\circ} = 16^{\circ}58' && \mathbf{2.0} \end{aligned}

This is relative to the perihelion, so symmetry demands that the total time the sun is in retrograde corresponds to

\begin{aligned} \Delta M &= 2M \\ \therefore T_{\odot,retro} &= T_{\mathrm{Mercury}}\frac{\Delta M}{2\pi} = 87.934 \times \frac{2 \times 0.296}{2\pi} \\ T_{\odot,retro} &= \boxed{8.28\,\mathrm{days}} && \mathbf{3.0} \end{aligned}

[Accept alternative methods making use of the true anomaly e.g. with the relation $\cos E = \frac{e + \cos\nu}{1 + e\cos\nu}$ . ]

Оригинал в Архива: Theory Problems.pdf · официални решения: Theory Solutions.pdf