Two satellites A and B having masses in the ratio 4 : 3 are revolving in circular orbits of radii 3r and 4r respectively around the earth. The ratio of total mechanical energy of A to B is :

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9 : 16

16 : 9

1 : 1

4 : 3

answer is B.

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Detailed Solution

Given that $\frac{m_{1}}{m_{2}} = \frac{4}{3}, \frac{r_{1}}{r_{2}} = \frac{3}{4}$

Now $TE = \frac{1}{2} {mv}^{2} + (\frac{- GMm}{r})$

but $\frac{{mv}^{2}}{r} = \frac{GMm}{r^{2}} \Rightarrow {mv}^{2} = \frac{GMm}{r}$

$\begin{array}{l} \Rightarrow TE = - \frac{GMm}{2 r} \propto \frac{m}{r} \\ \frac{{TE}_{1}}{{TE}_{2}} = \frac{m_{1}}{m_{2}} \cdot \frac{r_{2}}{r_{1}} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9} \end{array}$

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