2019年IOAA理论第9题-地月系统拉格朗日L2点的距离
英文题目
9. Distance of the Lagrangian point L2 of the Earth–Moon system (20 p)
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, L2, 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 $$𝑅=384 400 km$$. Neglect perturbations from the Sun and other planets.
Hint: You can use the following approximation: $$1/(1+𝑥)^2≈1−2𝑥$$, if $$|𝑥|≪1$$.
中文翻译
2019年1月3日，中国的嫦娥四号探测器在月球背面的冯卡门撞击坑着陆。
对于这个探测器来说，地球永远在地平线以下。所以我们需要一个中继站来保持通讯、控制。中国国家航天局发射了鹊桥号卫星来完成这个任务。鹊桥号运行在地月系统远端L2点的晕轨道上。
计算鹊桥卫星在月面上方的高度(ℎ)。月球的轨道考虑为正圆，半径为$$𝑅=384 400 km$$。忽略太阳和其他行星的摄动。
提示：当$$|𝑥|≪1$$时，可以使用简化公式 $$1/(1+𝑥)^2≈1−2𝑥$$
官方解答
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)
$$(r + \dfrac{M} {m +M} R) ω^2 = \dfrac{GM} {(R + r)2} + \dfrac{Gm} {r2}$$ ;
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) $$ω^2 = \dfrac{4{\pi}^2} {P^2} = \dfrac{G(M + m)} {R^3}$$
Substituting Eq. (9.2) into Eq. (9.1), eliminating the gravitational constant (G), and introducing the
dimensionless quantity $$x = r/R ≪1$$, we obtain
(9.3)$$\dfrac{(M + m)x +M} {R^2} = \dfrac{M} {R^2(1 + x)^2} + \dfrac{m} {R^2x^2}$$
(9.4) $$(M + m)x +M =
\dfrac{M}
{(1 + x)^2} +
\dfrac{m}
{x^2}$$
Assuming that $$x = r/R ≪1$$ we can apply the approximation $$1/(1+𝑥)^2≈1−2𝑥$$, which leads to the following expression:
(9.5) $$(M + m)x +M = M(1 - 2x) + \dfrac{m} {x^2}$$;
from which we get
(9.6) $$x^3 = \dfrac{m} {3M + m}$$;
Substituting the mass ratio m=M = 0:0123 of the Earth-Moon system one can obtain
(9.7) $$x = 0.159 83$$;
and therefore
(9.8) $$r = xR_{Moon} = 0.159 83 \times 384 400km = 61 440 km$$
From this result we should subtract the radius of the Moon, and such a way we obtain that
(9.9) $$h = r - R_{Moon} = 61 440 km - 1737 km = 59 703 km$$
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) $$M + m \approx M$$
In this case instead of Eq. (9.6) one can arrive at
(9.11) $$x^3 = \dfrac{m} {3M}$$
which leads to
(9.12) $$x = 0.160 05$$;
and therefore,
$$r = xR_{Moon} = 0.160 05 \times 384 400km = 61 524 km$$
and, finally
$$h = r - R_{Moon} = 61 524 km - 1737 km = 59 787 km$$
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.
