Gravitation Question 282
Question: Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is [CBSE PMT 1995; RPMT 2003]
Options:
A)$ [v=\frac{1}{2R}\sqrt{\frac{1}{Gm}}] $
B)$ [v=\sqrt{\frac{Gm}{2R}}] $
C) $ [v=\frac{1}{2}\sqrt{\frac{Gm}{R}}] $
D) $ [v=\sqrt{\frac{4Gm}{R}}] $
Show Answer
Answer:
Correct Answer: C
Solution:
Centripetal force provided by the gravitational force of attraction between two particles i.e. $ [\frac{mv^{2}}{R}=\frac{Gm\times m}{{{(2R)}^{2}}}] $ $ [\Rightarrow v=\frac{1}{2}\sqrt{\frac{Gm}{R}}] $
BETA
