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

If a planet orbiting the sun in a circular orbit suddenly stops, it will fall onto the sun in a time n(T) where T is the period of planets revolution , then n is

\[(a)\;\sqrt 8 \quad (b)\;\frac{\sqrt 2}{8} \quad (c)\; \sqrt 2 \quad (d)\;\sqrt 6 \]
Can you answer this question?
 
 

1 Answer

0 votes
Let $r$ be the radius of the orbot of the planet around the sun .
Let time taken to reach the sun be $'t'$ .
Time taken for the planet to revolve around the sun of radius $r=T$
We consider the planet to reach the sun is an elliptical path with time period $2t'=T'$
By Kepler's law
$\large\frac{T'}{T} =\bigg( \large\frac{r'}{r}\bigg)^{3/2}$
$r'= \large\frac{1}{2} $$r$
$T'= T\bigg(\large\frac{1}{2}\bigg)^{3/2}$
$t'= T\bigg(\large\frac{1}{2}\bigg)^{3/2}$$ \times \large\frac{1}{2}$
$t'=\large\frac{\sqrt 2}{8}$$T$
Hence b is the correct answer.

 

answered Aug 29, 2013 by meena.p
edited Feb 18, 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
...