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Q.
If $ R $ is the radius of the Earth then the height above the Earth's surface at which the acceleration due to gravity decreases by $ 20\% $ is
J & K CETJ & K CET 2016Gravitation
Solution:
If $g$ and $g_{h}$ denote the accelerations due to gravity at the Earths surface and at the height $h$ above the Earths surface respectively, then
$\frac{g_{h}}{g}=\frac{R^{2}}{\left(R+h\right)^{2}}$
But according to the question
$g_{h}=g-\frac{20}{100} $
$g=\frac{80}{100}$
$g=\frac{4}{5} g $
$\therefore \frac{4}{5}=\frac{R^{2}}{\left(R+h\right)^{2}}$ or $\sqrt{\frac{4}{5}}=\frac{R}{\left(R+h\right)}$
or $\frac{2}{\sqrt{5}}=\frac{R}{R+h}$ or $2R+2h=\sqrt{5}R$
or $2h=\left(\sqrt{5}-2\right)R$ or
$h=\frac{\left(\sqrt{5}-2\right)R}{2}$
$=\left(\frac{\sqrt{5}}{2}-1\right)R$