A uniform solid sphere of radius R produces a gravitational acceleration of a o on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes a o 4 is,

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Aicharya · Accepted Answer

On the surface of the uniform solid sphere, gravitational field(also called gravitational acceleration) is given by,
$\frac{- G M}{R^{2}} = a_{0}$ .
Gravitational field at any height $h$ is given by,
$a = a_{0} {[1 + \frac{h}{R}]}^{- 2} \Rightarrow \frac{a_{0}}{4} = a_{0} {[1 + \frac{h}{R}]}^{- 2} \Rightarrow {[1 + \frac{h}{R}]}^{2} = 4 \Rightarrow h = R$
Hence, distance from the centre will be,
$= R + R = 2 R$

A uniform solid sphere of radius R produces a gravitational acceleration of a o on its surface. The distance…

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