logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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 planet in circular orbit of Radius 'R' around the sun will be proportional to

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

Can you answer this question?
 
 

1 Answer

0 votes
$F=\large\frac{K}{R^n}$
Centripetal force $=MRw^2$
$MRw^2=\large\frac{K}{R^n}$
$w^2=\large\frac{K}{M}$$R^{-(n+1)}$
$ w\; \alpha\; R^{-\bigg(\Large\frac{n+1}{2}\bigg)}$
$w=\large\frac{2 \pi}{T}$
$\therefore \large\frac{2 \pi}{T} \;$$\alpha\; R^{-\bigg(\Large\frac{n+1}{2}\bigg)}$
$T \;\alpha\; R^{\bigg(\Large\frac{n+1}{2}\bigg)}$
Hence c is the correct answer.

 

answered Aug 19, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

Related questions

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