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., $\text{T} \propto \text{r}^{7 / 4}$ .