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IOAA 2019, Теория — Задача 9. Distance of the Lagrangian point L2 of the Earth–Moon system

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Всички материали от IOAA 2019

Theory problem sheet · 20 т.

Условие

On 3 January 2019, the Chinese spacecraft Chang'e-4 landed on the far side of the Moon in the area of the von Kármán crater, which was named after the world famous Hungarian-born physicist Theodore von Kármán.

As the Earth remains below the horizon of the spacecraft all the time, a relay station is also necessary for the communication with mission control on the Earth. For this purpose, The Chinese Space Agency launched a spacecraft, Queqiao, which was placed into a halo orbit around the outer Lagrangian point of the Earth–Moon system, $L_2$ , the far side of the Moon.

Calculate the distance ( $h$ ) of this satellite above the surface of the Moon. The Moon's orbit should be considered as a perfect circle with a radius of $R = 384\ 400\ \mathrm{km}$ . Neglect perturbations from the Sun and other planets.

Hint: You can use the following approximation: $1/(1+x)^2 \approx 1 - 2x$ , if $|x| \ll 1$ .

Решение

The relay satellite revolves on a circular orbit, due to the gravitational forces of the Earth and the Moon. Its net acceleration is:

(9.1)\quad \left(r+\frac{M}{m+M}R\right)\omega^2=\frac{GM}{(R+r)^2}+\frac{Gm}{r^2},

[4 p]

where $R$ and $r$ denote the (constant) distances between Earth and Moon, and between Moon and the satellite, respectively, while $M$ and $m$ stand for the masses of Earth and Moon, respectively. Using Kepler's third law, the angular velocity of the satellite, which is equal to the angular velocity of the Moon can be written as

(9.2)\quad \omega^2=\frac{4\pi^2}{P^2}=\frac{G(M+m)}{R^3}.

[2 p]

Substituting Eq. (9.2) into Eq. (9.1), eliminating the gravitational constant ( $G$ ), and introducing the dimensionless quantity $x=r/R$ , we obtain

(9.3)\quad \frac{(M+m)x+M}{R^2}=\frac{M}{R^2(1+x)^2}+\frac{m}{R^2x^2}

[2 p]

(9.4)\quad (M+m)x+M=\frac{M}{(1+x)^2}+\frac{m}{x^2}.

[2 p]

Assuming that $x=r/R\ll 1$ we can apply the approximation $(1+x)^{-2}\approx 1-2x$ , which leads to the following expression:

(9.5)\quad (M+m)x+M=M(1-2x)+\frac{m}{x^2},

[2 p]

from which we get

(9.6)\quad x^3=\frac{m}{3M+m}.

[2 p]

Substituting the mass ratio $m/M=0.0123$ of the Earth-Moon system one can obtain

(9.7)\quad x=0.159\,83,

[2 p]

and therefore

(9.8)\quad r=xR_{\mathrm{Moon}}=0.159\,83\times 384\,400\ \mathrm{km}=61\,440\ \mathrm{km}

[2 p]

From this result we should subtract the radius of the Moon, and such a way we obtain that

(9.9)\quad h=r-R_{\mathrm{Moon}}=61\,440\ \mathrm{km}-1737\ \mathrm{km}=59\,703\ \mathrm{km}

[2 p]

Here is another solution, which is erroneous in principle (or say, "approximative") but numerically gives an almost equivalent result.

One might say, that $m\ll M$ and therefore $m$ can be neglected either in the left hand side of Eq. (9.1) and in the r.h.s. of Eq. (9.2), i.e. one can use the following approximation:

(9.10)\quad M+m\approx M

In this case instead of Eq. (9.6) one can arrive at

(9.11)\quad x^3=\frac{m}{3M},

which leads to

(9.12)\quad x=0.160\,05,

and therefore,

r=xR_{\mathrm{Moon}}=0.160\,05\times 384\,400\ \mathrm{km}=61\,524\ \mathrm{km}

and, finally

h=r-R_{\mathrm{Moon}}=61\,524\ \mathrm{km}-1737\ \mathrm{km}=59\,787\ \mathrm{km}

This latter should be accepted as a full point solution, too. If, however, $m$ is dropped out only from one of the two equations, then at least 2 points should be subtracted.

Оригинал в Архива: TH-problem-final.pdf · официални решения: TH-solution-final.pdf · допълнителни данни: TH-anssheet.pdf