Question 330: 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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