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  4. The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be

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Q. The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be

NTA AbhyasNTA Abhyas 2020Gravitation

Solution:

$g′=\frac{g R^{2}}{\left(R + h\right)^{2}}$
$\frac{g}{4}=\frac{g R^{2}}{\left(R + h\right)^{2}}$
$\frac{1}{4}=\frac{R^{2}}{\left(R + h\right)^{2}}$
$\left(R + h\right)^{2}=4R^{2}$
$\left(R + h\right)=2R$
$h=R=R_{e}$