Physics

Physics

Problem

Consider a planet that may be treated as a uniform sphere of radius RR and mean density ρ\rho. a. (i) Express the mass MM of the planet in terms of RR, ρ\rho, and fundamental constants. b. (ii) Using energy considerations, derive an expression for the escape velocity from the planet’s surface in terms of GG, ρ\rho, and RR. c. (iii) Hence show that vesc∝Rρv_{\text{esc}} \propto R\sqrt{\rho}. d. (iv) Planet AA has RA=5.0×106 mR_A = 5.0\times10^{6}\,\text{m} and ρA=4500 kg m−3\rho_A = 4500\,\text{kg\,m}^{-3}, while planet BB has RB=7.0×106 mR_B = 7.0\times10^{6}\,\text{m} and ρB=3200 kg m−3\rho_B = 3200\,\text{kg\,m}^{-3}. Calculate the ratio vesc,B/vesc,Av_{\text{esc},B}/v_{\text{esc},A}. Take G=6.67×10−11 N m2kg−2G = 6.67\times10^{-11}\,\text{N\,m}^2\text{kg}^{-2}.
Intuitive explanation

The key insight is that gravity inside a uniform sphere depends only on the mass below you—the shell above contributes zero net force. This means as you move toward the center, the effective mass shrinks, and so does the gravitational field, dropping linearly with distance rather than as 1/r21/r^2. Understanding this shell-by-shell structure lets you find the field (and potential energy) at any depth, and explains why gravity is strongest at the surface, not deeper down.

Intuitive explanation

The key insight is that gravity inside a uniform sphere depends only on the mass below you—the shell above contributes zero net force. This means as you move toward the center, the effective mass shrinks, and so does the gravitational field, dropping linearly with distance rather than as 1/r21/r^2. Understanding this shell-by-shell structure lets you find the field (and potential energy) at any depth, and explains why gravity is strongest at the surface, not deeper down.

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