# Q: 8.18 The escape speed of a projectile on the earth’s surface is  $\small 11.2\hspace {1mm}km\hspace {1mm}s^-^1$. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Let us assume the speed of the body far away from the earth is  $v_f$.

Total energy on earth is :

$=\ \frac{1}{2}mv_p^2\ -\ \frac{1}{2}mv_{esc}^2$

And the total energy when the body is far from the earth is :

$=\ \frac{1}{2}mv_f^2$

(Since the potential energy at far from the earth is zero.)

We will use conservation of energy :  -

$\frac{1}{2}mv_p^2\ -\ \frac{1}{2}mv_{esc}^2\ =\ \frac{1}{2}mv_f^2$

or                                          $v_f\ =\ \sqrt{\left ( v_p^2\ -\ v_{esc}^2 \right )}$

or                                                  $=\ \sqrt{\left ( \left ( 3v_{esc} \right )^2\ -\ v_{esc}^2 \right )}$

or                                                  $=\ \sqrt{8}v_{esc}$

or                                                 $=\ 31.68\ Km/s$

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