Gravitation Question 243
Question: In a satellite if the time of revolution is T, then K.E. is proportional to [BHU 1995]
Options:
A) $ [\frac{1}{T}] $
B) $ [\frac{1}{T^{2}}] $
C) $ [\frac{1}{T^{3}}] $
D) $ [{{T}^{-2/3}}] $
Show Answer
Answer:
Correct Answer: D
Solution:
v $ [=\sqrt{\frac{GM}{r}}] $ \ $ [K.E.\propto v^{2}\propto \frac{1}{r}] $ and $ [T^{2}\propto r^{3}] $ $ [,K.E.,\propto {{T}^{-2/3}}] $
