Escape velocity from Earth (energy)

Problem

Using conservation of energy, calculate the escape velocity from the Earth’s surface. Data: $M_T = 6{,}0\cdot 10^{24}\ \text{kg}$, $R_T = 6{,}4\cdot 10^6\ \text{m}$, $G = 6{,}67\cdot 10^{-11}\ \text{N}\,\text{m}^2/\text{kg}^2$.

Solution

Setting up with energy. Escape velocity is the minimum initial velocity for which the object reaches infinity with zero velocity. The limiting case is zero total mechanical energy. Between the surface (A) and infinity (B): $\tfrac{1}{2}m v_f^2 - \frac{GM_T m}{R_T} = \underbrace{0}_{E_{\text{cin},\infty}} + \underbrace{0}_{E_{\text{pot},\infty}}$

The object’s mass $m$ cancels out: $v_f = \sqrt{\frac{2GM_T}{R_T}}$

Numerical substitution. $v_f = \sqrt{\frac{2\cdot 6{,}67\cdot 10^{-11}\cdot 6{,}0\cdot 10^{24}}{6{,}4\cdot 10^6}} = \sqrt{\frac{8{,}0\cdot 10^{14}}{6{,}4\cdot 10^6}} = \sqrt{1{,}25\cdot 10^8}$

$\ev{v_f \approx 11{,}2\ \text{km/s}}$

Escape velocity does not depend on the mass of the escaping object: a stone and a spacecraft have the same $v_f$.

Links

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