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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.

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 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

Posted by

Devendra Khairwa

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