Derivation of escape velocity

Problem

Starting from conservation of mechanical energy, show that the escape velocity from the surface of a spherical celestial body of mass $M$ and radius $R$ is $v_f = \sqrt{2GM/R}$. Also show that the relation $v_f = \sqrt{2}\,v_\text{orb}$ holds between escape velocity and circular orbital velocity at the surface.

Solution

Setting up with energy. Escape velocity is the minimum initial velocity that allows a body of mass $m$ to move indefinitely away, reaching infinity with zero velocity. The limiting case corresponds to zero total mechanical energy. Between the surface (A) and infinity (B): $\underbrace{\tfrac{1}{2}m v_f^2}_{E_{\text{cin},A}} \underbrace{-\,\frac{GMm}{R}}_{E_{\text{pot},A}} = \underbrace{0}_{E_{\text{cin},B}} + \underbrace{0}_{E_{\text{pot},B}}$

The mass $m$ of the escaping body cancels out and $v_f$ is isolated: $\tfrac{1}{2}v_f^2 = \frac{GM}{R} \quad\Longrightarrow\quad \ev{v_f = \sqrt{\dfrac{2GM}{R}}}$

Relation with orbital velocity. For a circular orbit at the surface, the gravitational force provides the centripetal force: $\frac{GMm}{R^2} = \frac{m\,v_\text{orb}^2}{R} \quad\Longrightarrow\quad v_\text{orb} = \sqrt{\frac{GM}{R}}$

Comparing the two expressions: $\frac{v_f}{v_\text{orb}} = \frac{\sqrt{2GM/R}}{\sqrt{GM/R}} = \sqrt{2} \quad\Longrightarrow\quad \ev{v_f = \sqrt{2}\,v_\text{orb}}$

So escaping from the surface only requires a velocity about $1{,}41$ times that needed to stay in a grazing orbit.

Links

Topics: Gravitazione Concepts: Velocità di fuga · Conservazione dell’energia meccanica · Energia potenziale gravitazionale universale · Orbite Skills: Conservazione dell’energia · Impostazione simbolica