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A body of mass \mathrm{m}\: \mathrm{kg}. starts falling from a point 2 \mathrm{R} above the Earth's surface, its kinetic energy when it has fallen to a point \mathrm{' R '} above the Earth's surface is

\mathrm{[ \mathrm{R}= Radius \: of\: Earth, \mathrm{M} - mass \: of \: earth, \mathrm{G}= Gravitational \: constant ]}

 

Option: 1

\mathrm{\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}}}

 


Option: 2

\mathrm{\frac{1}{6} \frac{\mathrm{GMm}}{\mathrm{R}}}
 


Option: 3

\mathrm{\frac{2}{3} \frac{\mathrm{GMm}}{\mathrm{R}}}
 


Option: 4

\mathrm{\frac{1}{3} \frac{\mathrm{GMm}}{\mathrm{R}}}


Answers (1)

best_answer

Gravitational potential energy of a body of mass \mathrm{m} at height \mathrm{h} above the surface of earth is given as \mathrm{\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}}

Here, initial gravitational potential energy

\mathrm{ \mathrm{U}_1=\frac{\mathrm{GMm}}{(\mathrm{R}+2 \mathrm{R})}=-\frac{\mathrm{GMm}}{3 \mathrm{R}} }

Final gravitational potential energy,

\mathrm{ \mathrm{U}_2=\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{R})}=-\frac{\mathrm{GMm}}{2 \mathrm{R}} }

\mathrm{ \text { gain in } \mathrm{K} \cdot \mathrm{E}=\text { loss in potential energy }=-\frac{\mathrm{GMm}}{3 \mathrm{R}}-\left(-\frac{\mathrm{GMm}}{2 \mathrm{R}}\right) }

\mathrm{ =\frac{\mathrm{GMm}}{\mathrm{R}}\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\mathrm{GMm}}{6 \mathrm{R}} }

Hence option 2 is correct.



 

Posted by

himanshu.meshram

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