Gravitation Question 257
Question: Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is [g is acceleration due to gravity at Earth’s surface] [MP PMT 2002]
Options:
A) $ [2\pi \sqrt{\frac{2R}{g}}] $
B) $ [4\sqrt{2}\pi \sqrt{\frac{R}{g}}] $
C) $ [2\pi \sqrt{\frac{R}{g}}] $
D) $ [8\pi \sqrt{\frac{R}{g}}] $
Show Answer
Answer:
Correct Answer: B
Solution:
$ [T=2\pi \sqrt{\frac{{{(R+h)}^{3}}}{gR^{2}}}=2\pi \sqrt{\frac{{{(2R)}^{3}}}{gR^{2}}}=4\sqrt{2\pi }\sqrt{\frac{R}{g}}] $
