Two satellites A and B , having ratio of masses 3 : 1 are in circular orbits of radius r and 4r . Ca

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7.Gravitation

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Two satellites $A$ and $B$ , having ratio of masses $3 : 1$ are in circular orbits of radius $r$ and $4r$ . Calculate the ratio of total mechanical energy of $A$ and $B$

A

$3 : 4$

B

$12 : 1$

C

$4 : 3$

D

$1 : 12$

Solution

$\therefore $ total mechanical energy $E = – \frac{{GMm}}{{2r}}$

$\therefore $ $\frac{{{E_1}}}{{{E_2}}} = \frac{{{m_1}}}{{{m_2}}} \times \frac{{{r_2}}}{{{r_1}}} = \frac{3}{1} \times \frac{{4r}}{r} = \frac{{12}}{1}$

Standard 11

Physics

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