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# Suppose the gravitational force varies inversely as the n-th power of distance. Then the time period of a planet in circular orbit of radius 'r' around the sun will be proportional to

$(a)\;r^{\bigg(\Large\frac{n+1}{2}\bigg)} \quad (b)\;r^{\bigg(\Large\frac{n-1}{2}\bigg)} \quad (c)\;r^n \quad (d)\;r^{\bigg(\Large\frac{n-2}{2}\bigg)}$

Can you answer this question?

## 1 Answer

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$F \; \alpha \; \large\frac{1}{r^n}$
$F= \large\frac{K}{r^n}=\large\frac{mv^2}{v}$
$v=\sqrt {\large\frac{R}{m} \bigg(\frac{1}{r^{n-1}}\bigg)}$
Time period $=T=\large\frac{2 \pi r}{v}$
$\qquad = 2 \pi r \sqrt {\large\frac{m}{R} \normalsize r^{n-1}}$
$\qquad = 2 \pi r \sqrt {\large\frac{m}{R} \normalsize r^{n+1}}$
$T \; \alpha \; r^{\bigg(\Large\frac{n+1}{2}\bigg)}$
Hence a is the correct answer.

answered Aug 24, 2013 by
edited Feb 17, 2014 by meena.p

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