Considering earth to be a sphere of radius ' R ' having uniform density ' ρ ', then value of acceleration due to gravity ' g ' in terms of R, ρ and G is (a) g = √((3π R)/(ρ G)) (b) g = √(4/3πρ GR) (c) g = 4/3πρ GR (d) g = GMρ R²

Answer: (c). g = GM/R² with M = 4/3π R³ ρ. g = G/R²·4/3π R³ρ = 4/3πρ GR So the answer is (c) g = 4/3πρ GR.

Considering earth to be a sphere of radius ' R ' having uniform… | MHT CET 2023 (14 May, Shift 1)

Q48MHT CET 2023 · 14 MAY · SHIFT 1Easy

Acceleration Due to Gravity

Considering earth to be a sphere of radius ' $R$ ' having uniform density ' $\rho$ ', then value of acceleration due to gravity ' $g$ ' in terms of $R, \rho$ and $G$ is

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Answer: (c)

$g = \frac{GM}{R^2}$ with $M = \frac{4}{3}\pi R^3 \rho$ .
$g = \frac{G}{R^2}\cdot\frac{4}{3}\pi R^3\rho = \frac{4}{3}\pi\rho GR$
So the answer is (c) $g = \frac{4}{3}\pi\rho GR$ .

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