The Earth is assumed to be a sphere of radius | vedclass

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Question

According to question and by using $COME$

$-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{R}}+\frac{1}{2} \mathrm{m}(\mathrm{fv})^{2}=0+0$

$\Rightarrow

Vedclass · Accepted Answer

$\frac{1}{{\sqrt 2 }}$

Vedclass · Answer

According to question and by using $COME$

$-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{R}}+\frac{1}{2} \mathrm{m}(\mathrm{fv})^{2}=0+0$

$\Rightarrow \mathrm{fv}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$ but $\mathrm{v}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$

Therefore $f \sqrt{\frac{2 G M}{R}}=\sqrt{\frac{G M}{R}} \Rightarrow f=\frac{1}{\sqrt{2}}$

The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$, where $v$ is its escape velocity from the surface of the Earth. the value of $f$ is

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