# A particle is at rest at a distance R (earth's radius) above the earth's radius. What is the...

## Question:

A particle is at rest at a distance R (earth's radius) above the earth's radius. What is the minimum speed with which it should be projected so that it should not return?

## Escape Velocity

In order for a rocket to escape the Earth's gravity, it must be launched at a certain speed known as the escape velocity. When an object is launched with this speed, it will attain a height at which gravity cannot pull it back to the surface.

By conservation of mechanical energy:

{eq}KE_i + PE_i = KE_f + PE_f \\ {/eq}

when the object escape the Earth, the sum of its final kinetic and potential energy becomes zero. So our equation becomes:

{eq}\frac{1}{2}mv_0^2 - \frac{GMm}{R} = 0 \\ \frac{1}{2}mv_0^2 = \frac{GMm}{R} \\ v_0^2 = \frac{2GM}{R} {/eq}

So the escape velocity must be equal to:

{eq}v_0 = \sqrt{\frac{2GM}{R}} {/eq}

or in terms of the acceleration due to gravity g:

{eq}g = \frac{GM}{R^2} \\ v_0 = \sqrt{\frac{2GM}{R} \times \frac{R}{R}} \\ v_0 = \sqrt{\frac{2GMR}{R^2}} \\ v_0 = \sqrt{2gR} {/eq} 