Gravitation Question 128
Question: The gravitational potential energy of a body of mass ?m? at the earth’s surface $ [-mgR_{e}] $ . Its gravitational potential energy at a height $ [R_{e}] $ from the earth’s surface will be (Here $ [R_{e}] $ is the radius of the earth) [AIIMS 2000; MP PET 2000; Pb. PMT 2004]
Options:
A) $ [-2,mgR_{e}] $
B) $ [2,mgR_{e}] $
C) $ [\frac{1}{2}mgR_{e}] $
D) $ [-\frac{1}{2}mgR_{e}] $
Show Answer
Answer:
Correct Answer: D
Solution:
$ [\Delta U=U_{2}-U_{1}=\frac{mgh}{1+\frac{h}{R_{e}}}=\frac{mgR_{e}}{1+\frac{R_{e}}{R_{e}}}=\frac{mgR_{e}}{2}] $ $ [,\Rightarrow ,U_{2}-(-mgR_{e})=\frac{mgR_{e}}{2}] $ Þ$ [U_{2}=-\frac{1}{2}mgR_{e}] $
