The kinetic energy needed to project a body o | vedclass

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Question

$-\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{m}}{\mathrm{R}}+\mathrm{K} \cdot \mathrm{E} \cdot=0+0$

$\mathrm{K.E}=\frac{\mathrm{GM}_{\mathrm{E}} \mathr

Vedclass · Accepted Answer

$mgR$

Vedclass · Answer

$-\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{m}}{\mathrm{R}}+\mathrm{K} \cdot \mathrm{E} \cdot=0+0$

$\mathrm{K.E}=\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{m}}{\mathrm{R}}=\frac{\mathrm{GMe}}{\mathrm{R}^{2}} \cdot \mathrm{mR}=\mathrm{mgR}$

The kinetic energy needed to project a body of mass $m$ from the earth's surface (radius $R$) to infinity is

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