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Q. 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 $r^{- \left(\frac{5}{2}\right)}$ , then $T$ would be proportional to

NTA AbhyasNTA Abhyas 2020

Solution:

since the gravitational force provides the necessary centripetal force,
$\frac{\text{mv}^{2}}{\text{r}} \propto \text{r}^{- 5 / 2}$
$⇒ \, \, \frac{\text{mv}^{2}}{\text{r}} = \text{kr}^{- 5 / 2}$ , k = constant
$∴ \, \, \text{v} = \sqrt{\frac{\text{kr}^{- 3 / 2}}{\text{m}}}$
$\text{T} = \frac{2 \pi \text{r}}{\text{v}} = 2 \pi \text{r} \times \frac{\sqrt{\text{m}}}{\sqrt{\text{k} \text{r}^{- 3 / 2}}}$
i.e., $\text{T} \propto \text{r}^{7 / 4}$ .