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The gravitational potential energy of a body of mass ‘m’ at the earth’s surface is –mgRe. Its gravitational potential energy at a height Re from the earth surface will be (Re is the radius of earth)

Option: 1

-2mgRe

 


Option: 2

2mgRe

 


Option: 3

1/2 mgRe

 


Option: 4

-1/2 mgRe


Answers (1)

best_answer

As we learn

Work done against gravity -

W=\Delta U=GMm\left [ \frac{1}{r_{1}}-\frac{1}{r_{2}} \right ]

W\rightarrow work done

\Delta U\rightarrow change in Potential energy

r_{1},r_{2}\rightarrow distances

- wherein

If body is moved from surface of earth to a point h above surface of earth then use the given formula

 

 \Delta U = U_{2}- U_{1} = \frac{mgh}{1+\frac{h}{R_{e}}}

as h = Re

\Delta U =\frac{mgR_{e}}{2}

U_{2} - (-mgR_{e}) = \frac{mgR_{e}}{2}

U_{2} = \frac{mgR_{e}}{2} + (-mgR_{e}) = \frac{-1}{2}mgR_{e}

 

Posted by

jitender.kumar

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