**Escape Velocity**

When a particle is thrown vertically upwards, it reaches a certain height and comes back to its original position. But when it is given greater initial velocity, it reaches greater height before coming back. A particle can be thrown up with a certain minimum initial velocity so that, the particle goes beyond the earth’s gravitational field and escape from earth, this velocity is known as escape velocity.

**What
is Escape Velocity?**

Escape Velocity is the minimum velocity required by a body to be projected overcomes the gravitational pull of the earth. It is the minimum velocity required by an object to escape the gravitational field that is, escape the land without ever falling back. An object that has this velocity at the earth’s surface will totally escape the earth’s gravitational field ignoring the losses due to the atmosphere.

**Escape Velocity
Formula:**

\(Escape\,\,Velocity({{V}_{e}})=\sqrt{\frac{2GM}{R}}\).

Where,

G = Universal Gravitational Constant = 6.673 x 10⁻¹¹ N.m²/ kg²

M = Mass of the planet (or) moon,

R = Radius from center of gravity.

**How to find the Escape
Velocity?**

**Problem: **Find the escape velocity of the moon,
if mass is 7.35 x 10²² kg and radius is 1.2 x 10⁵ m?

**Solution: **Given,

Mass (m) = 7.35 x 10²² kg,

Radius (R) = 1.2 x 10⁵ m,

G = 6.673 x 10⁻¹¹ N.m²/ kg².

We know that:

\(Escape\,\,Velocity({{V}_{e}})=\sqrt{\frac{2GM}{R}}\).

\(=\sqrt{\frac{2\times 6.673\times {{10}^{-11}}\times 7.35\times {{10}^{22}}}{1.2\times {{10}^{5}}}}\).

∴
Escape Velocity (V_{e}) = 9.04 x 10³ m/ sec.