MHT CET · Physics · 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 \(r^{-\frac{7}{2}}\), then \(T^2\) is proportional to
- A \(r^{9 / 2}\)
- B \(r^{7 / 2}\)
- C \(r^{5 / 2}\)
- D \(r^{3 / 2}\)
Answer & Solution
Correct Answer
(A) \(r^{9 / 2}\)
Step-by-step Solution
Detailed explanation
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)
\(
\begin{array}{ll}
& \text {Hence, } \mathrm{a} \propto-\mathrm{r}^{-7 / 2} \\
\therefore & -\omega^2 \mathrm{r} \propto-\mathrm{r}^{-7 / 2} \\
\therefore & \omega^2 \propto \mathrm{r}^{-9 / 2} \\
\therefore & \frac{4 \pi^2}{\mathrm{~T}^2} \propto \mathrm{r}^{-9 / 2} \\
& \Rightarrow \mathrm{T}^2 \propto \mathrm{r}^{9 / 2}
\end{array}
\)
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)
\(
\begin{array}{ll}
& \text {Hence, } \mathrm{a} \propto-\mathrm{r}^{-7 / 2} \\
\therefore & -\omega^2 \mathrm{r} \propto-\mathrm{r}^{-7 / 2} \\
\therefore & \omega^2 \propto \mathrm{r}^{-9 / 2} \\
\therefore & \frac{4 \pi^2}{\mathrm{~T}^2} \propto \mathrm{r}^{-9 / 2} \\
& \Rightarrow \mathrm{T}^2 \propto \mathrm{r}^{9 / 2}
\end{array}
\)
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