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MHT CET 2026 (11 Apr, Shift 2) · Acceleration Due to Gravity

Acceleration Due to Gravity question from MHT CET 2026 (11 Apr, Shift 2)

Q1MHT CET 2026 · 11 APR · SHIFT 2Easy

Acceleration Due to Gravity

The density of a planet is three times that of earth and radius $2.5$ times that of earth. If $g_p$ and $g_e$ represents acceleration due to gravity on planet and earth respectively then ratio $g_p$ to $g_e$ is

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Answer: (a)

$g = \frac{GM}{R^2} = \frac{4}{3}\pi G\rho R$ , so $g \propto \rho R$ .
$\frac{g_p}{g_e} = 3 \times 2.5 = 7.5$
So the answer is (a) 7.5.

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