Question

If gravitational attraction between two point masses be given by $F = G \frac{m_{1} m_{2}}{r^{3}}$ Then, the period of a
satellite in a circular orbit will be proportional to

Easy

Solution

As, $\frac{m v^{2}}{r} = \frac{G m_{1} m_{2}}{r^{3}} or v \propto r^{(\frac{1 - 3}{2})}$
Now, $T = \frac{2 π r}{v} or T \propto \frac{r}{v}$
$∴ T \propto \frac{r}{r^{(\frac{1 - 3}{2})}} \Rightarrow T \propto r^{\frac{1 + 3}{2}} = r^{2}$

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