# Escape velocity

We all know that everything thrown vertically upward will eventually fall down. When we throw an object, it moves up to a maximum height and then comes back to the surface. The maximum height attained depends on the velocity of the object. If we talk about the energy, the Kinetic Energy present in the body is used in doing work against gravity. As the body moves upward, its kinetic energy keeps on decreasing and the corresponding amount of potential energy keeps on increasing. When the object is at the maximum height, all kinetic energy is converted into potential energy and the velocity of object becomes zero. Once the velocity is zero, the body starts to fall under the influence of gravity.

So, if we slowly increase the speed of the object, maximum height attained also keeps on increasing and a stage will come when the body will escape the gravitational field of Earth and will never return to the surface.

The minimum velocity of a body that will just take it outside the gravitational field of earth (or any other planet/moon) is called as the Escape velocity.

#### Escape velocity of Earth

To find the escape velocity of an object let’s suppose that $m$ is the mass of the body and $v$ is the initial velocity from the surface of the earth. Now the kinetic energy of the body is given by,

$K.E. = \frac{1}{2}mv^2$

The potential energy of the body at the surface of earth is

$P.E. = -G\frac{M_em}{R_e}$

Total energy of the body at the surface of earth is

$K.E. + P.E. = \frac{1}{2}mv^2 - G\frac{M_em}{R_e}$

Let assume that initial kinetic energy is just good enough to carry object out of the gravitational field. In this situation, the kinetic energy of the body becomes zero when it escapes the gravitational field and all kinetic energy is used up in doing work against gravity and as explained above gets converted into potential energy. Since gravitational potential energy is negative, the maximum P.E. of a body in the earth’s gravitational field in zero which is at infinity. So, the total energy of the body just outside the earth’s gravitational field is zero.

$K.E. + P.E. = 0$

Earth’s gravitational field is conservative force field so the total energy remains constant at any point in the field.

$\frac{1}{2}mv^2 - G\frac{M_em}{R_e} = 0$

$v^2 = \frac{2GM_e}{R_e}$

$v^2 = 2gR_e$

since $g = \frac{GM_e}{R_e^2}$

so, $v_{escape} = \sqrt{2gR_e}$

To calculate the escape velocity of earth, we just have to plug in some numbers like $g = 9.8 m/s^2$ and mean radius of earth is $R_e = 6.368 \times 10^6 m$

$v_{escape} = \sqrt{2 \times 9.8 \times 6.368 \times 10^6} = 11.17 \times 10^3 m/s$

This implies that if a body is thrown with a velocity of 11.2 km/s from the surface of the earth, it will never come back to the surface of the earth.