Two planets of masses $M_{1}$ and $M_{2}$ have satellites of masses $m_{1}$ and $m_{2}$ respectively, revolving around them at the same sadius r. The period of the first satellite (of mass $m_{1}$ ) is twice as that of the second. which one of the following relations is correct ?

4 M_{1} = M_{2}

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2 M_{1} = M_{2}

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M_{1} = {2 M}_{2}

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m_{1} M_{1} = m_{2} M_{2}

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Solution

The correct option is A $4 M_{1} = M_{2}$
Time period of satellite of mass $m$ revolving at radius $r$ is given by, $T = \frac{2 π r}{v}$

Now, Centripetal force = Gravitational force

$\frac{m v^{2}}{r} = \frac{G M m}{r^{2}}$

$v = \sqrt{\frac{G M}{r}}$

$T = \frac{2 π r}{\sqrt{G M / r}} = \frac{2 π r^{3 / 2}}{\sqrt{G M}}$

Since r is same for both satellites, $T \propto \frac{1}{\sqrt{M}}$

Given, $T_{1} = 2 T_{2}$

So , $\frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

$\frac{M_{2}}{M_{1}} = 2^{2}$

$M_{2} = 4 M_{1}$

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