Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
0 votes

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

0 votes
$F \; \alpha \; \large\frac{1}{r^n}$
$F= \large\frac{K}{r^n}=\large\frac{mv^2}{v}$
$v=\sqrt {\large\frac{R}{m} \bigg(\frac{1}{r^{n-1}}\bigg)}$
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 meena.p
edited Feb 17, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App