(1994, 89)
Solution:
Centripetal force is provided by the given gravitational force: $$\frac{mv^2}{R} = \frac{k}{R}$$.
Solving for $v$, we get $v^2 = \frac{k}{m}$.nSince $k$ and $m$ are constants, $v$ is independent of $R$, meaning $v \propto R^0$.
If the gravitational force between two objects were proportional to $\frac{1}{R}$ (and not as $\frac{1}{R^2}$), where R is the distance between them, then a particle in a circular path (under such a force) would have its orbital speed v, proportional to:
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