Съдържание

IOAA 2021, Теория — Задача 9. Pluto Satellites (15 points).

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Всички материали от IOAA 2021

Theory — Q9-1 English (Official) · 15 т.

Условие

9.1 The mass of Charon, the biggest satellite of Pluto, is 1/8th the mass of Pluto. Both bodies move in a circular orbit around a common center of mass. In addition, they both are tidally-locked.
The distance between the center of the planet and the center of the satellite is $R = 19\,640\ \mathrm{km}$ and radius of the satellite is $r = 593\ \mathrm{km}$ .

Let $g_0$ be the gravitational acceleration on the surface of Charon due only to its mass.Let A be the point on Charon surface directly facing Pluto, and B the point diametrically opposite. Compute the percentage difference between gravitational acceleration at A and B respect to $g_0$ . [15 т.]

Отговори

Решение

Pluto and Charon, common centre of mass CM, surface points A and B, and radial unit vector directed towards Pluto.

Let P and C be the centers of Pluto and Charon respectively, and CM their center of mass. Clearly $\overline{CM-P}=\frac19R$ and $\overline{CM-C}=\frac89R$ . The effective gravitational field at points A and B are given by:

[By recognizing the vectorial nature of gravities/forces/fields] [4 points]

\vec g_A=\frac{GM_P}{(R-r)^2}\hat r-\frac{GM_C}{r^2}\hat r-\omega^2\left(\frac89R-r\right)\hat r

[1 point]

\vec g_B=\frac{GM_P}{(R+r)^2}\hat r+\frac{GM_C}{r^2}\hat r-\omega^2\left(\frac89R+r\right)\hat r

[1 point]

where $\omega^2=9\frac{GM_C}{R^3}$ is the angular speed common to both Pluto and Charon. Thus, the gravitational accelerations at these points are:

[by finding angular speed] [2 points]

g_A=\frac{GM_P}{(R-r)^2}-\frac{GM_C}{r^2}-\frac{GM_C}{R^2}\left(8-9\frac rR\right)

g_B=\frac{GM_P}{(R+r)^2}+\frac{GM_C}{r^2}-\frac{GM_C}{R^2}\left(8+9\frac rR\right)

Factorizing the expression to obtain formulas proportional to Charon’s mass [1 point]:

g_A=-\frac{GM_C}{r^2}\left[1-\frac{M_P}{M_C}\left(\frac rR\right)^2\frac1{(1-\frac rR)^2}+\left(\frac rR\right)^2\left(8-9\frac rR\right)\right]

g_B=\frac{GM_C}{r^2}\left[1+\frac{M_P}{M_C}\left(\frac rR\right)^2\frac1{(1+\frac rR)^2}-\left(\frac rR\right)^2\left(8+9\frac rR\right)\right]

Replacing the numerical value of the ratio of the two masses:

g_A=-\frac{GM_C}{r^2}\left[1-\left(\frac rR\right)^2\frac8{(1-\frac rR)^2}+\left(\frac rR\right)^2\left(8-9\frac rR\right)\right]

g_B=\frac{GM_C}{r^2}\left[1+\left(\frac rR\right)^2\frac8{(1+\frac rR)^2}-\left(\frac rR\right)^2\left(8+9\frac rR\right)\right]

Noting that the expressions in square brackets are dimensionless, it is helpful to replace the given values of $r$ and $R$ , resulting [2 points by expressions of $g_A$ and $g_B$ ]:

g_A=-\frac{GM_C}{r^2}(0{,}99929)

g_B=\frac{GM_C}{r^2}(0{,}99933)

The percentage difference with respect to Charon’s normal gravity $g_0=\frac{GM_C}{r^2}$ is

\frac{\left||g_A|-|g_B|\right|}{g_0}\cdot100\%=0{,}004\%=4\times10^{-3}\%

[1 point]

Official source discrepancy

The supplied marking annotations sum to 12 points although the task and key heading state 15 points. All printed annotations are retained; no missing allocation is invented.

Оригинал в Архива: TQ-9-Q.pdf · официални решения: TQ9.pdf · допълнителни данни: TQ-9-A.pdf · допълнителни данни: Standard-Table-of-constants.docx.pdf · допълнителни данни: TQ-G.pdf · допълнителни данни: data.pdf