# A body is projected vertically upwards from the surface of the earth with a velocity equal to...

## Question:

A body is projected vertically upwards from the surface of the earth with a velocity equal to {eq}\frac{3}{4} {/eq} escape velocity of earth. If {eq}R {/eq} is the radius of earth, what is the maximum height attained by the body?

## Escape Velocity

The escape velocity on Earth is given by:

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

where

{eq}G = gravitational \ constant \\ M = mass \ of \ Earth \\ {/eq}

Using the fact that the the acceleration due to gravity g is equal to:

{eq}g = \frac{GM}{R^2} {/eq}

we can express the escape velocity as:

{eq}v_e = \sqrt{2gR} {/eq}

Let m be the mass of the body. The initial kinetic energy of the body is just equal to zero. Using conservation of mechanical energy

{eq}KE_i + PE_i = KE_f + PE_f \\ 0 - \frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{GMm}{R+h} {/eq}

To solve for h, first we combine the potential energy equations then simplify to get:

{eq}\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{R+h} \\ \frac{1}{2}mv^2 = \frac{GMmR +GMmh - GMmR}{R(R+h)} \\ \frac{1}{2}mv^2 = \frac{GMmh}{R(R+h)} {/eq}

We can express this equation in terms of g and the given escape velocity.

{eq}\frac{1}{2}mv^2 = \frac{GMmh}{R(R+h)} \times \frac{R}{R} \\ \frac{1}{2}m(\frac{3v_e}{4})^2 = \frac{GM}{R^2} \frac{mRh}{(R+h)} \\ \frac{1}{2}m(\frac{3\sqrt{2gR}}{4})^2 = \frac{mgRh}{(R+h)} \\ \frac{9mgR}{16} = \frac{mgRh}{(R+h)} \\ \frac{9}{16} = \frac{h}{(R+h)} \\ 9R = 16h - 9h \\ h = \frac{9R}{7} {/eq}

Thus the maximum height attained by the body is 9/7 of Earth's radius. 