Two satellites A and B having ratio of masses 3 : 1 are revolving in circular orbits of radii ' r ' and ' 4 r '. The ratio of total energy of satellites A to that of B is

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TestHub · Accepted Answer

The total energy of a satellite of mass $m$ revolving in a circular orbit of radius $r$ around a planet of mass $M$ is given by $E = - \frac{GM m}{2 r}$ .

The ratio of total energies of satellites A and B is

$\frac{E _{A}}{E _{B}} = \frac{m _{A}}{m _{B}} \times \frac{r _{B}}{r _{A}}$

Given $\frac{m _{A}}{m _{B}} = \frac{3}{1}$ , $r_{A} = r$ , and $r_{B} = 4 r$ .

Substituting the values, we get

$\frac{E _{A}}{E _{B}} = \frac{3}{1} \times \frac{4 r}{r} = 12$

The ratio of total energy of satellites A to that of B is $12 : 1$ .

Answer: $12 : 1$

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