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  • A body falls freely towards the earth from a height , above the surface of the earth, where init

A body falls freely towards the earth from a height \mathrm{2 R}, above the surface of the earth, where initially it was at rest. If \mathrm{R} is the radius of the earth then its velocity on reaching the surface of the earth is
 

Option: 1

\mathrm{\sqrt{\frac{4}{3} g R}}

 


Option: 2

\mathrm{\sqrt{\frac{2}{3} g R}}


Option: 3

\mathrm{\frac{4}{3}gR}


Option: 4

\mathrm{2gR}


Answers (1)

best_answer

Initial energy of the body \mathrm{=\frac{G M m}{(R+2 R)}}

Final energy of the body =\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^2

where \mathrm{m} is the mass of the body and \mathrm{M} is the mass of the earth.

Applying conservation of mechanical energy

\mathrm{\frac{\mathrm{GMm}}{3 \mathrm{R}}=\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^2 }

\mathrm{\Rightarrow \quad \frac{1}{2} \mathrm{mv}^2=\frac{2 \mathrm{GMm}}{3 \mathrm{R}} }

\mathrm{\Rightarrow \quad \mathrm{V}^2=\frac{4 \mathrm{GM}}{3 \mathrm{R}}}            \Rightarrow \mathrm{\mathrm{V}=\sqrt{\frac{4 \mathrm{GM}}{3 \mathrm{R}}} }

\mathrm{ =\sqrt{\frac{4 \mathrm{GMR}}{3 \mathrm{R}^2}}=\sqrt{\frac{4}{3} \mathrm{gR}}}

Hence option 1 is correct.

 

 





 

Posted by

himanshu.meshram

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