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Question: The gravitational potential energy of a body of mass ‘m’ at the earth’s surface \(- m g R _ { e }\)....

The gravitational potential energy of a body of mass ‘m’ at the earth’s surface −mgRe- m g R _ { e }. Its gravitational potential energy at a height ReR _ { e } from the earth’s surface will be (Here ReR _ { e } is the radius of the earth)

A

−2mgRe- 2 m g R _ { e }

B

2mgRe2 m g R _ { e }

C

12mgRe\frac { 1 } { 2 } m g R _ { e }

D

−12mgRe- \frac { 1 } { 2 } m g R _ { e }

Answer

−12mgRe- \frac { 1 } { 2 } m g R _ { e }

Explanation

Solution

ΔU=U2−U1=mgh1+hRe=mgRe1+ReRe=mgRe2\Delta U = U _ { 2 } - U _ { 1 } = \frac { m g h } { 1 + \frac { h } { R _ { e } } } = \frac { m g R _ { e } } { 1 + \frac { R _ { e } } { R _ { e } } } = \frac { m g R _ { e } } { 2 }

⇒U2−(−mgRe)=mgRe2\Rightarrow U _ { 2 } - \left( - m g R _ { e } \right) = \frac { m g R _ { e } } { 2 } ⇒ U2=−12mgReU _ { 2 } = - \frac { 1 } { 2 } m g R _ { e }