Q96 / 125 · BITSAT 2010 · Physics
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius ' $\mathrm{R}$ ' around the sun will be proportional to
- A$\mathrm{R}^{\mathrm{n}}$
- B$\mathrm{R}^{\left(\frac{\mathrm{n}-1}{2}\right)}$
- Correct answerC$\mathrm{R}\left(\frac{\mathrm{n}+1}{2}\right)$
- D$\mathrm{R}^{\left(\frac{\mathrm{n}-2}{2}\right)}$
Solution
$$
\mathrm{F}=\mathrm{KR}^{-\mathrm{n}}=\mathrm{MR} \omega^{2} \Rightarrow \omega^{2}=\mathrm{KR}^{-(\mathrm{n}+1)}
$$
or $\omega=\mathrm{K}^{\prime} \mathrm{R}^{\frac{-(\mathrm{n}+1)}{2}}$
[where $\mathrm{K}^{\prime}=\mathrm{K}^{1 / 2}$, a constant $]$
$\frac{2 \pi}{T} \alpha R^{\frac{-(n+1)}{2}} \quad \therefore T \alpha R^{\frac{(n+1)}{2}}$
Hence, the correct answer is option: 'C'.