Question

Imagine of a light planet revolving around a massive star in a circular orbit of radius $r$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $r^{\frac{5}{2}}$, then the square of the time period will be proportional to

• A. $r^3$
• B. $r^2$
• C. $r^{2.5}$
• D. $r^{3.5}$

Answer 1

The correct option is D $r^{3.5}$

Since the planet is in circular motion around the star, we can write the equation for circular motion where the gravitational force is providing the necessary centripetal force.

Let, $\omega$ be the angular speed of the planet.

According to the problem, the gravitational force of attraction,$$F\propto \frac{1}{r^{5/2}}$$$\Rightarrow F=-\frac{K}{r^{5/2}}$

Where, $K$ is a positive constant.

Since, here the gravitational force is providing the centripetal force. So, we can write

$\frac{K}{r^{5/2}}=m{\omega }^{2}r$

$\Rightarrow {\omega }^2=\frac{K}{m{r}^{7/2}}$

$\Rightarrow {\left(\frac{2\pi }{T}\right)}^2=\frac{K}{m{r}^{7/2}}[\because T=\frac{2\pi }{\omega }]$

$\Rightarrow T^2=\frac{4{\pi }^{2}m{r}^{7/2}}{K}$

$\therefore T^2\propto r^{3.5}$

Hence, option (d) is the correct answer.

Why this question? Note: As per Kepler's third law of planetary motion, $T^2\alpha R^3$. This result is based on the fact that gravitational force follows inverse square law. If a hypothetical force doesn't follow inverse square law, this result will not be applicable.