Search any question & find its solution
Question:
Answered & Verified by Expert
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 $r^{-\frac{7}{2}}$, then $T^2$ is proportional to
Options:
Solution:
2434 Upvotes
Verified Answer
The correct answer is:
$r^{9 / 2}$
For the planet to orbit around the star, the
centripetal force must be provided by gravitational force. Hence, $\mathrm{F}_{\mathrm{d}}=\mathrm{F}_{\mathrm{a}}$.
$\mathrm{F}_{\mathrm{a}} \propto-\mathrm{r}^{-7 / 2}$
...(Given)
(-ve sign indicates force is towards the centre of orbit)
$$
\begin{array}{ll}
& \text { Hence, } \mathrm{a} \propto-\mathrm{r}^{-7 / 2} \\
\therefore \quad & -\omega^2 \mathrm{r} \propto-\mathrm{r}^{-7 / 2} \\
\therefore \quad & \omega^2 \propto \mathrm{r}^{-9 / 2} \\
\therefore \quad & \frac{4 \pi^2}{\mathrm{~T}^2} \propto \mathrm{r}^{-9 / 2} \\
& \Rightarrow \mathrm{T}^2 \propto \mathrm{r}^{9 / 2}
\end{array}
$$
centripetal force must be provided by gravitational force. Hence, $\mathrm{F}_{\mathrm{d}}=\mathrm{F}_{\mathrm{a}}$.
$\mathrm{F}_{\mathrm{a}} \propto-\mathrm{r}^{-7 / 2}$
...(Given)
(-ve sign indicates force is towards the centre of orbit)
$$
\begin{array}{ll}
& \text { Hence, } \mathrm{a} \propto-\mathrm{r}^{-7 / 2} \\
\therefore \quad & -\omega^2 \mathrm{r} \propto-\mathrm{r}^{-7 / 2} \\
\therefore \quad & \omega^2 \propto \mathrm{r}^{-9 / 2} \\
\therefore \quad & \frac{4 \pi^2}{\mathrm{~T}^2} \propto \mathrm{r}^{-9 / 2} \\
& \Rightarrow \mathrm{T}^2 \propto \mathrm{r}^{9 / 2}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.