# What is Escape Velocity?

The minimum velocity that a body must have in order to escape the gravitational attraction of a particular planet or other objects is called escape velocity.

Now let's understand the concept of escape velocity in one shot. If an object is thrown vertically up from the surface of the earth the object rises to a certain height and falls to the earth's surface due to the gravity of the earth.

At maximum height, the velocity of the object is zero. The faster the object is thrown the greater the height of the object it reaches after which its velocity becomes zero.

As the velocity increases gradually a state is reached where the object's velocity is no longer zero. That is it no longer returns to the surface of the earth.

In this case, the object moves further away from the earth and the gravitational attraction of the earth on the object becomes less and less. So we can say, the object reaches an infinite distance from earth. The minimum velocity required for this to occur is the escape velocity

It is notable that the concept of escape velocity applies not only to objects launched from the earth but also to objects launched from the surface of any planet or satellite.

## Measurements of escape velocity

Let us consider the mass of the earth is M and the radius of the earth is R. The gravitational force on an object of mass m at a distance of r from the center of the earth,

$\dpi{110}{\color{Red} F=\frac{GMm}{r^{2}}}$

Now the work is done to move the object to a very small distance dr along the line of action against this gravitational attraction

$\dpi{110}{\color{Red} dW=F.dr}$

$\dpi{110}{\color{Red} Or\;dW=\frac{GMm}{r^{2}}.dr}$

Hence the amount of work done to send the object to an infinite distance from the surface of the earth

$\dpi{110}{\color{Red} W=\int_{R}^{\infty }\frac{GMm}{r^{2}}.dr}$

$\dpi{110}{\color{Red} Or\;W=GMm\left [ -\frac{1}{r} \right ]_{R}^{\infty }}$

$\dpi{110}{\color{Red} Or\;W=\frac{GMm}{R}}$

Again, let's Ve is the escape velocity of the object on the earth. Then the initial kinetic energy of the object

$\dpi{110}{\color{Red} K_{i}=\frac{1}{2}mV_{e}^{2}}$

If this kinetic energy becomes equal to the above work then the object escapes from the gravitational pull of the Earth.

$\dpi{110}{\color{Red} \therefore \;\frac{1}{2}mV_{e}^{2}=\frac{GMm}{R}}$

$\dpi{110}{\color{Red} Or\;V_{e}^{2}=\frac{2GM}{R}}$

$\dpi{110}{\color{Red} Or\;V_{e}=\sqrt{\frac{2GM}{R}}\to (1)}$

We also know the acceleration due to gravity at the surface of the earth

$\dpi{110}{\color{Red} g=\frac{GM}{R^{2}}}$

$\dpi{110}{\color{Red} Or\;GM=gR^{2}}$

Now by substituting the value of GM in equation number one we get

$\dpi{110}{\color{Red} V_{e}=\sqrt{2gR}\to (2)}$

This is the formula for measuring the escape velocity on the surface of the earth.

We know the acceleration due to gravity on the surface of the earth g = 9.8 m/s². And the radius of the earth R = 6400 km = 6.4 × 10⁶ m. Now substituting these values in equation number (2) we get

Ve =11200 m/s

Or Ve = 11.2 km/s

This value is the measure of the escape velocity of any object on the surface of the earth.

This means if an object of any mass is thrown upwards from the earth's surface with a speed of 11.2 km/s, the object goes out of the earth's gravitational field i.e it does not return to the earth.

Watch on