A uniform spherical planet (Radius R) has acc | vedclass

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Question

variation with depth

$g_{d}=g(1-d / R)$

$\frac{g}{4}=g(1-d / R)$

$\frac{d}{R}=3 / 4$

$a=3 R / 4$

variation with height

$g_{h}=g\left

Vedclass · Accepted Answer

$\frac{{9R}}{4}$

Vedclass · Answer

variation with depth

$g_{d}=g(1-d / R)$

$\frac{g}{4}=g(1-d / R)$

$\frac{d}{R}=3 / 4$

$a=3 R / 4$

variation with height

$g_{h}=g\left(1-\frac{2 h}{R}\right)$

$\frac{g}{4}=g\left(1-\frac{2 h}{R}\right)$

$\frac{2 h}{R}=\frac{3}{4}$

$h=\frac{3 R}{8}$

max distance $=a+h=\frac{3 R}{4}+\frac{3 R}{8}$

$=\frac{9 R}{8}$

A uniform spherical planet (Radius $R$) has acceleration due to gravity at its surface $g.$ Points $P$ and $Q$ located inside and outside the planet have acceleration due to gravity $\frac{g}{4} .$ Maximum possible separation between $P$ and $Q$ is

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