Question 314: Gravitation

Question

Consider a light planet revolving around a massive star in a circular orbit of radius '$r$' with time period '$T$'. If the gravitational force of attraction between the planet and the star is proportional to $\mathrm{r}^{\frac{7}{2}}$, then $\mathrm{T}^2$ is proportional to

Accepted Answer

For the planet to orbit around the star, the centripetal force must be provided by gravitational force. Hence, $\mathrm{F}_{\mathrm{G}}=\mathrm{F}_{\mathrm{a}}$.
$\mathrm{F}_{\mathrm{a}} \propto-\mathrm{r}^{-7 / 2}\quad$ .... (Given)
($-$ve sign indicates force is towards the centre of orbit)
Hence, $\mathrm{a} \propto-\mathrm{r}^{-7 / 2}$
$\begin{array}{ll}
	herefore & -\omega^2 \mathrm{r} \propto-\mathrm{r}^{-7 / 2} \
	herefore & \omega^2 \propto \mathrm{r}^{-9 / 2} \
	herefore & \frac{4 \pi^2}{\mathrm{~T}^2} \propto \mathrm{r}^{-9 / 2} \
& \quad \Rightarrow \mathrm{T}^2 \propto \mathrm{r}^{9 / 2}
\end{array}$

Answer

$r^{9 / 2}$

Answer

$r^{7 / 2}$

Answer

$r^{5 / 2}$

Answer

$r^{3 / 2}$

Gravitation

Consider a light planet revolving around a massive star in a circular orbit of radius 'r' with time period 'T'. If the gravitational force of attraction between the planet and the star is proportional to \mathrm{r}^{\frac{7}{2}}, then \mathrm{T}^2 is proportional to

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