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 is proportional to $r^{-5/2}$ , then $T$ will be proportional to

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

Answer 1

Answer: (C)

Gravitational force provides the necessary centripetal force required by the given planet for circular motion. So
$\frac{{\text{mv}}^2}{\text{r}}\propto {\text{r}}^{-5/2}$
$\Rightarrow \frac{{\text{mv}}^2}{\text{r}}={\text{kr}}^{-5/2}$ , $k$ = constant
$\therefore \text{v}=\sqrt{\frac{{\text{kr}}^{-3/2}}{\text{m}}}$
$\text{T}=\frac{2\pi \text{r}}{\text{v}}=2\pi \text{r}\times \frac{\sqrt{\text{m}}}{\sqrt{\text{k}{\text{r}}^{-3/2}}}$
i.e., $\text{T}\propto {\text{r}}^{7/4}$ .