If gravitational attraction between two points masses be given by $F = G \frac{m_{1} m_{2}}{r^{n}}$ . Then, the period of a satellite in a circular orbit will be proportional to

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$r^{\frac{n - 1}{2}}$

independent of $n$

$r^{\frac{n}{2}}$

$r^{\frac{n + 1}{2}}$

answer is B.

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Detailed Solution

For a satellite to move in a circular orbit, the centripetal force should be equal to the gravitational force i.e,

$\frac{m v^{2}}{r} = \frac{G m_{1} m_{2}}{r^{n}}$

or $v \propto r^{(\frac{1 - n}{2})}$

Now, As the time period of a satellite is

$T = \frac{2 π r}{v}$

or $T \propto \frac{r}{v}$

$∵ v \propto r^{(\frac{1 - n}{2})}$

So $T \propto \frac{r}{r^{(\frac{1 - n}{2})}}$

$\Rightarrow T \propto r^{\frac{1 + n}{2}}$

Hence, the period of a satellite in a circular orbit will be proportional to $r^{(\frac{1 + r}{2})}$ .

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