6.7 Escape velocity

  Gravitation

The escape velocity is defined as the minimum speed with which a body has to be projected vertically upward from the surface of any planet so that it just crosses the gravitational field of that planet and never returns on its own.

Consider a particle of mass $m$ kept on the surface of any planet of radius $R$ and mass $M$ is as shown in the figure.

Potential energy of the system is, $$U = - \frac{{GMm}}{R}$$

Let the escape velocity be $v_e$

Kinetic energy of the system is, $$K = \frac{1}{2}mv_e^2$$

For the particle to leave the gravitational field of earth then the binding energy should be zero, i.e.

$$B.\,E. = \left| T \right|$$$$B.\,E. = \left| {K + U} \right|$$$$B.\,E. = \left| {\frac{1}{2}mv_e^2 - \frac{{GMm}}{R}} \right|$$ or $$\frac{1}{2}mv_e^2 - \frac{{GMm}}{R} = 0$$$${v_e} = \sqrt {\frac{{2GM}}{R}} $$ where,

$M$: Mass of the planet
$R$: Radius of the planet

As we know, $${\text{Mass}} = {\text{Volume}} \times {\text{Density}}$$$$M = V\rho $$$$M = \left( {\frac{4}{3}\pi {R^3}} \right)\rho $$ So, $${v_e} = \sqrt {\frac{{2G}}{R} \times \frac{4}{3}\pi {R^3}\rho } $$$${v_e} = \sqrt {\frac{{\pi G{R^2}\rho }}{3}} $$
1. Therefore, the escape speed depends upon the mass and radius of the planet from the surface of which the body is to be projected.

2. The escape speed is independent of the mass of the body.

3. For earth the escape velocity is ${v_e} = 11.2\,km/s$.

4. If a body is projected from a planet with a speed $v$ which is smaller than the escape speed $v$ which is smaller than the escape speed $v_e$ (i.e. $v<v_e$) then the body after reaching a certain height may either move in an orbit around the planet or may fall back to the planet.

5. If the speed of projection $u$ of the body from the surface of a planet is greater than the escape speed $(v_e)$ of that planet, the body will escape out from the gravitationall field of that planet and will move in the interstellar space with speed $v$.

From the conservation of energy, $$\frac{1}{2}m{u^2} - \frac{{GMm}}{R} = \frac{1}{2}m{v^2} + 0$$

Potential energy can be written in terms of escape velocity as,
$$\frac{{GMm}}{R} = \frac{1}{2}mv_e^2$$ So, $$\frac{1}{2}m{v^2} = \frac{1}{2}m{u^2} - \frac{1}{2}mv_e^2$$$${v^2} = {u^2} - v_e^2$$ or $$v = \sqrt {{u^2} - v_e^2} $$

Note: Trajectory of a body projected from point $P$ in the direction $PQ$ with different initial velocities.

Let a body be projected from point $P$ with velocity $v$ in the direction $PQ$. For different values of $v$ the paths are different.