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

• A. $R^{3/2}$
• B. $R^{3/5}$
• C. $R^{7/2}$
• D. $R^{7/4}$

Answer 1

Answer: (D)

According to the question, the gravitational force between the planet and the star is $F\propto \frac{1}{R^{5/2}}$
$\therefore F=\frac{GMm}{R^{5/2}}$
Where $M$ and $M$ be masses of star and planet respectively.
For motion of planet in circular orbit,
$\begin{array}{l}mR{\omega }^2=\frac{GMm}{R^{5/2}}\\ mR{\left(\frac{2\pi }{T}\right)}^2=\frac{GMm}{R^{5/2}}\\ \left(\therefore \omega =\frac{2\pi }{T}\right)\\ \frac{4{\pi }^2}{T^2}=\frac{GM}{R^{7/2}}\Rightarrow T^2=\frac{4{\pi }^2}{GM}{R}^{7/2}\\ T^2\propto R^{7/2}\text{ or }{T}^2\propto R^{7/4}\end{array}$