Escape speed from the Earth is close to 11.2 km s⁻¹. On another planet

Escape speed from the Earth is close to 11.2 km s⁻¹. On another planet whose radius is half of the Earth’s radius and whose mass density is four times that of the Earth, the escape speed in km s⁻¹ will be close to :

11.2
15.8
5.6
7.9
This question was previously asked in
UPSC NDA-1 – 2024
The correct option is A. The escape speed from the new planet will be close to 11.2 km s⁻¹, the same as Earth’s.
The escape speed (v_e) from a spherical body of mass M and radius R is given by v_e = sqrt(2GM/R). The mass M can be expressed in terms of density (rho) and volume (V = 4/3 * pi * R^3) as M = rho * (4/3 * pi * R^3). Substituting this into the escape speed formula: v_e = sqrt(2G * (rho * 4/3 * pi * R^3) / R) = sqrt(8/3 * pi * G * rho * R²) = R * sqrt(8/3 * pi * G * rho). This shows that v_e is proportional to R * sqrt(rho).
Let Earth’s radius, density, and escape speed be R_e, rho_e, and v_e_e respectively. So, v_e_e ∝ R_e * sqrt(rho_e) = 11.2 km/s. For the new planet, R_p = R_e / 2 and rho_p = 4 * rho_e. The escape speed from the new planet is v_e_p ∝ R_p * sqrt(rho_p) = (R_e / 2) * sqrt(4 * rho_e) = (R_e / 2) * 2 * sqrt(rho_e) = R_e * sqrt(rho_e). Thus, v_e_p is proportional to the same value as v_e_e, meaning v_e_p = v_e_e = 11.2 km/s.
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