Gravitational Field Intensity

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At a point located at a distance of 2R from the centre of a uniform spherical planet of radius R, the gravitational field strength is measured. How does this value compare with the surface field strength g?

RankGuru · Accepted Answer

NCERT Class 11, Chapter 8 (Gravitation) defines gravitational field intensity as the force per unit mass, $E = \dfrac{GM}{r^2}$, which outside a uniform sphere behaves exactly as if all mass were concentrated at the centre. At the surface, $r = R$, the field equals $g = \dfrac{GM}{R^2}$. At a distance $r = 2R$ from the centre, $E = \dfrac{GM}{(2R)^2} = \dfrac{GM}{4R^2} = \dfrac{g}{4}$. The point lies outside the planet, so the inverse-square law applies and only the radius matters. The option $g/2$ uses a linear distance dependence instead of squaring. The option $2g$ wrongly increases the field with distance. The option $g$ ignores any change with position. As a plausibility check, doubling the distance from the centre must weaken the field by a factor of four because of the inverse-square dependence, giving $g/4$, which is smaller than the surface value as expected.

Gravitational Field Intensity

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