Question 301: Gravitation

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 '$R^{-5 / 2}$', then '$T$' would be proportional to

Accepted Answer

In this scenario, we are given that the gravitational force between the planet and the star is proportional to $R^{-5/2}$. The gravitational force $F$ provides the necessary centripetal force for the planet's circular motion. Therefore, we can write:
$$ F = \frac{k}{R^{5/2}} $$
where $k$ is a proportionality constant.
The centripetal force required to keep the planet in a circular orbit of radius $R$ with a period $T$ is given by:
$$ F_{	ext{centripetal}} = \frac{m v^2}{R} = m \left( \frac{2\pi R}{T} ight)^2 \frac{1}{R} $$
Simplifying the expression:
$$ F_{	ext{centripetal}} = \frac{4 \pi^2 m R}{T^2} $$
Equating the two expressions for the force, we have:
$$ \frac{k}{R^{5/2}} = \frac{4 \pi^2 m R}{T^2} $$
Solving for $T^2$:
$$ T^2 = \frac{4 \pi^2 m R^{7/2}}{k} $$
Taking the square root to solve for $T$:
$$ T = \sqrt{\frac{4 \pi^2 m}{k}} \cdot R^{7/4} $$
Thus, the period $T$ is proportional to $R^{7/4}$.
Therefore, the correct answer is:
Option D: $R^{7/4}$
.

Answer

$\mathrm{R}^{3 / 2}$

Answer

$\mathrm{R}^{3 / 5}$

Answer

$\mathrm{R}^{7 / 2}$

Answer

$\mathrm{R}^{7 / 4}$

Gravitation

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^{-5 / 2}', then 'T' would be proportional to

Choose the correct answer: