At what height above the earth’s surfac | vedclass

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Question

Given $:$ $g_{h}=1 \% g_{s}=0.01 g_{s}$

Using $g_{h}=g_{s}\left(\frac{R}{R+h}\right)^{2}$

Or $\quad 0.01 g_{s}=g_{s}\left(\frac{R}{R+h}\right)^{

Vedclass · Accepted Answer

$9\,R$

Vedclass · Answer

Given $:$ $g_{h}=1 \% g_{s}=0.01 g_{s}$

Using $g_{h}=g_{s}\left(\frac{R}{R+h}\right)^{2}$

Or $\quad 0.01 g_{s}=g_{s}\left(\frac{R}{R+h}\right)^{2}$

or $\quad 0.1=\frac{R}{R+h}$

or $\quad 0.1 R+0.1 h=R$

$\Longrightarrow h=9 R$

At what height above the earth’s surface does the acceleration due to gravity fall to $1\%$ of its value at the earth’s surface ?

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