Question

A small planet is revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force between the planet and the star were proportional to ${R^{\dfrac{{ - 5}}{2}}}$, then $T$ would be proportional to:
A) ${R^{\dfrac{3}{2}}}$
B) ${R^{\dfrac{3}{5}}}$
C) ${R^{\dfrac{7}{2}}}$
D) ${R^{\dfrac{7}{4}}}$

Answer 1

In this problem, it is given that the planet is revolving around the star, this revolution is due to the gravitational force between the star and the planet. By equating the gravitational force between two masses formula and the motion of the object in circular path formula, the period of revolution is determined.

Useful formula:
Gravitational force between two masses,
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Where, $F$ is the force, $G$ is the gravitational constant, ${m_1}$ is the mass of the first object, ${m_2}$ is the mass of the second object and $r$ is the distance between the two masses.

Motion of the object in circular orbit,
$F = m \times r \times {\omega ^2}$
Where, $F$ is the force, $m$ is the mass of the revolving object, $r$ is the circular radius and $\omega$ is the angular velocity.

Complete step by step solution:
Given that,
The gravitational force is proportional to the ${R^{\dfrac{{ - 5}}{2}}}$.
Then, the gravitational force is written as,
$F = \dfrac{{G \times M \times m}}{{{R^{\dfrac{5}{2}}}}}\,..................\left( 1 \right)$
Here, $M$ is the mass of the star and $m$ is the mass of the planet.
Now,
Motion of the object in circular orbit,
$F = m \times R \times {\omega ^2}\,..............\left( 2 \right)$
Here, $m$ is the mass of the planet.
Now equating the equation (1) and equation (2), then
$\dfrac{{G \times M \times m}}{{{R^{\dfrac{5}{2}}}}} = m \times R \times {\omega ^2}\,................\left( 3 \right)$
By cancelling the mass of the planet $m$ on both sides, then the above equation is written as,
$\dfrac{{G \times M}}{{{R^{\dfrac{5}{2}}}}} = R \times {\omega ^2}$
Now, the angular velocity is also written as $\omega = \dfrac{{2\pi }}{T}$, where $T$ is the period of revolution. And substitute this in the above equation,
$\dfrac{{G \times M}}{{{R^{\dfrac{5}{2}}}}} = R \times {\left( {\dfrac{{2\pi }}{T}} \right)^2}$
By squaring the term $\dfrac{{2\pi }}{T}$ in RHS, then
$\dfrac{{G \times M}}{{{R^{\dfrac{5}{2}}}}} = R \times \dfrac{{4{\pi ^2}}}{{{T^2}}}$
By taking the $R$ from RHS to LHS, then the above equation is written as,
$\dfrac{{G \times M}}{{{R^{\dfrac{5}{2}}} \times {R^1}}} = \dfrac{{4{\pi ^2}}}{{{T^2}}}$
By adding the power of $R$ in the LHS, then
$\dfrac{{G \times M}}{{{R^{\dfrac{5}{2} + 1}}}} = \dfrac{{4{\pi ^2}}}{{{T^2}}}$
On further calculation, then
$\dfrac{{G \times M}}{{{R^{\dfrac{7}{2}}}}} = \dfrac{{4{\pi ^2}}}{{{T^2}}}$
To find the period of revolution, so keep $T$ in one side and other terms in other side, then
${T^2} = \dfrac{{4{\pi ^2} \times {R^{\dfrac{7}{2}}}}}{{G \times M}}$
Assume the term $\dfrac{{4{\pi ^2}}}{{GM}}$ as constant, then
${T^2} \propto {R^{\dfrac{7}{2}}}$
${T} \propto {R^{\dfrac{7}{4}}}$

Hence, the option (D) is correct.

Note: Here the distance between the star and the planet is equal to the radius of the revolution, so in the equation (2), the radius of the planet revolution $r$ is taken as $R$, which is equal to the distance between the star and the planet which is taken in the gravitational force equation.