Escape Velocity

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The escape velocity from the surface of the Earth is approximately 11.2 km/s. What would the escape velocity be from a planet having the same density as Earth but twice its radius?

RankGuru · Accepted Answer

NCERT Class 11, Chapter 8 (Gravitation) gives escape velocity as $v_e = \sqrt{\dfrac{2GM}{R}}$, the speed at which an object's total mechanical energy just becomes zero so it can barely reach \infty. When the density $ho$ is held constant, the mass is tied to the size through $M = ho \cdot \dfrac{4}{3}\pi R^3$. Substituting, $v_e = \sqrt{\dfrac{2G}{R} \cdot ho \cdot \dfrac{4}{3}\pi R^3} = \sqrt{\dfrac{8\pi G ho}{3}}\,R$, which shows that at fixed density the escape velocity is directly proportional to the radius, a less obvious result than the bare square-root scaling first suggests. Doubling the radius therefore doubles the escape velocity: $v_e' = 2 	imes 11.2 = 22.4\,	ext{km/s}$. The option $11.2\,	ext{km/s}$ wrongly assumes the radius has no net effect once density is fixed. The option $15.8\,	ext{km/s}$ applies a $\sqrt{2}$ scaling, which would hold only if the mass were unchanged. The option $44.8\,	ext{km/s}$ scales by four, confusing this with an area dependence. As a plausibility check, a larger same-density planet packs far more mass that grows as the cube of the radius, so a clean doubling of the escape speed is consistent with the derived proportionality.

Escape Velocity

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