Imagine a light planet revolving around a very massive star in a circular orbit of radius $m$ with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to ${r^{-5/2}}$, then the square of the time period will be proportional to:
A) ${r^-3}$
B) ${r^-2}$
C) ${r^{-2.5}}$
D) ${r^{-3.5}}$

Hint: The gravitation law states that every point mass attracts every other point mass by a force acting along the line intersecting the two points. The force is proportional to the product of the two masses, and inversely proportional to the square of the distance between them.
In equilibrium, the centrifugal force and the gravitational force will be equal.

Complete step by step solution:
Newton's law of universal gravitation is usually stated as that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

The force of gravity is a power that draws two mass objects. We call the force of gravity attractive because it constantly tries to draw masses together, never pushes them apart.

Let a planet of mass $m$, revolving in a circular orbit of radius $r$ around a massive star.
For equilibrium,
${F_ {centrifugal}} = {F_ {gravitation}} $
Where $F$ is force

According to the question,
${F_ {gravitation}}$ = $\dfrac{{GMm}}{r^{\dfrac{5}{2}}}$
Where, $G$ is gravitational constant, $M,m$ are the masses and $r$ is the radius.
Now, we know that
${F_ {gravitation}} = \dfrac{{m{v^2}}}{r}$
Where $v$ is the velocity

From the above two equations,
$\dfrac {{m {v^2}}} {r}$ = $\dfrac{{GMm}}{r^{\dfrac{5}{2}}}$
Solving,
${v^2} = \dfrac{{GM}}{{{r^{3.5}}}}$
Now, Time period is defined as
$T = \dfrac{{2\pi r}} {v} $
Therefore,
${T^2} = \dfrac{{4{\pi ^2} {r^2}}} {{{v^2}}} $
We know that
${v^2} = \dfrac{{GM}}{{{r^{3.5}}}}$
Thus,
${T^2} = \dfrac{{\dfrac{{4{\pi ^2} {r^2}}} {{GM}}}} {{{r^ {3.5}}}} $
Hence, ${T^2} $ is proportional to ${r^ {-3.5}} $.

The square of the time period will be proportional to ${r^ {-3.5}}$.

Note: The relation of the distance of objects in free fall to the square of the time taken had recently been confirmed by Grimaldi and Riccioli between 1640 and 1650. They had also made a calculation of the gravitational constant by recording the oscillations of a pendulum.