Question

Show that if a body be projected vertically upward from the surface of the earth so as to reach a height nR above the surface (i) the increase in its potential energy is ((n)/(n + 1)) mgR, (ii) the velocity with it must be projected is sqrt((2ngR)/(n + 1)), where r is the radius of the earth and m the mass of body.

Answer 1

Given :

The height upto which the projected body  reaches above the  surface of the earth :

=nR

Radius of earth = R

Mass of the body = m

To prove :

(1) . The increase in its potential energy is :

  = $\frac{n}{n+1}\times mgR$

(2).The velocity with it must be projected is :

 $= \sqrt{2gR\times \frac{n}{n+1}}$

Solution :

We can write the change in potential energy as :

$\Delta U = \frac{-GMm}{(n+1)R} + \frac{GMm}{R}$

Here , M = mass of the  earth and m =  mass of the projected  body

So , $\Delta U = \frac{GMm}{R}(1-\frac{1}{n+1})$

             $= \frac{GMm}{R}\times \frac{n}{n+1}$

Now multiplying numerator and denominator by R :

$\Delta U = \frac{GMm\times R}{R\times R}\times \frac{n}{n+1}$

      $=\frac{GM}{R^{2} } \times mR\times \frac{n}{n+1}$

       $=mgR \times \frac{n}{n+1}$

So the 1st part is proved i.e. increase or change in its potential energy is $=mgR \times \frac{n}{n+1}$

Now change in potential energy should be equal to change in kinetic energy or increase in potential energy should be equal to decrease in kinetic energy , so :

$\Delta U= \Delta KE$

or, $mgR \times \frac{n}{n+1} = \frac{1}{2}mv^{2}$

So , $v = \sqrt{2gR\frac{n}{n+1}}$

Hence , the 2nd part is also proved i .e . the velocity with which the body must be projected is $v = \sqrt{2gR\frac{n}{n+1}}$ .