Escape Velocity And Planetary Density

Question

Two rocky planets A and B are made of material of exactly the same uniform density, but planet B has a radius twice that of planet A. What is the ratio of the escape velocity from planet A to the escape velocity from planet B?

RankGuru · Accepted Answer

Escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$, but for planets of identical density the mass must be written in terms of radius using $M = \frac{4}{3}\pi R^3 \rho$. Substituting gives $v_e = \sqrt{\frac{2G}{R}\cdot\frac{4}{3}\pi R^3 \rho} = \sqrt{\frac{8\pi G \rho}{3}}\,R$, showing that for fixed density the escape velocity is directly proportional to the radius. Therefore $\frac{v_{e,A}}{v_{e,B}} = \frac{R_A}{R_B} = \frac{R}{2R} = \frac{1}{2}$, giving the ratio 1 : 2. The option 1 : 4 wrongly uses the square of the radius ratio. The option 1 : √2 incorrectly takes a square root, ignoring that mass grows as the cube of radius. The option 1 : 8 mistakenly applies the full cube. This extends the NCERT escape-velocity result to bodies of equal density. The relation $v_e = \sqrt{\frac{8\pi G \rho}{3}}\,R$ is a useful diagnostic because it cleanly separates the two ways a planet can have a high escape velocity: by being made of denser material, or by being physically larger. For planet B, the radius doubles and the mass grows eightfold as $R^3$, yet because escape velocity depends on the mass-to-radius combination $M/R$, the net scaling reduces to simple proportionality with radius. A plausibility check confirms that the larger planet, packing eight times the mass into twice the radius, demands a higher escape speed, so planet A's escape velocity being half that of B is consistent.

Escape Velocity And Planetary Density

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