54.6: Escape Velocity

The escape velocity is the initial velocity a particle must have to escape the gravity of its parent body. Typically it refers to the initial velocity a particle must have at the surface of a planet in order to leave the planet forever, and never be pulled back by the planet's gravity. If a particle leaves the surface of a planet with an initial velocity equal to the escape velocity, then the body will move more and more slowly as the particle moves farther from the planet, finally reaching a velocity of zero at $r=\infty$. (We assume only the particle and the planet are present, and ignore all other bodies.)

To compute the escape velocity, consider running the problem with time running backwards: the body starts at $r=\infty$ with zero velocity and falls toward the planet. The impact velocity from infinity will be the same as the escape velocity. Now at $r=\infty$, the potential energy $U=-G M_{p} m / r=0$, where $M_{p}$ is the mass of the planet and $m$ is the mass of the particle. Since the particle is at rest at $r=\infty$, the kinetic energy there is also zero, so the total mechanical energy $K+U=0$. Now let the particle begin falling from $r=\infty$ under the influence of the planet's gravity, until it impacts the planet at $r=R_{p}$, where $R_{p}$ is the radius of the planet. At the point of impact the potential energy is $U=-G M_{p} m / R_{p}$, and its kinetic energy will be $K=m v_{e}^{2} / 2$, where $v_{e}$ is the impact (escape) velocity. By the law of conservation of energy, the total mechanical energy at $r=\infty$ must be the same as it is at $r=R_{p}$ :

$$K+U=\frac{1}{2} m v_{e}^{2}-\frac{G M_{p} m}{R_{p}}=0$$

Solving for the escape velocity $v_{\boldsymbol{e}}$, we find

$$v_{e}=\sqrt{\frac{2 G M_{p}}{R_{p}}}$$

For the Earth, for example, we have (from Appendix L) $G M_{p}=3.986005 \times 10^{14} \mathrm{~m}^{3} \mathrm{~s}^{-2}$ and $R_{p}=$ $6378.140 \times 10^{3} \mathrm{~m}$; substituting these values into Eq. $\PageIndex{1}$, we find the escape velocity for Earth is $v_{e}=11.2$ $\mathrm{km} / \mathrm{s}$. In other words, if you were to fire a projectile from the surface of the Earth with an initial velocity of $11.2 \mathrm{~km} / \mathrm{s}$, it would be able to escape the Earth's gravity, going more and more slowly the higher it goes, finally coming to rest at $r=\infty$.