Question

Obtain an expression for escape velocity of an object of mass ‘m’ from the surface of planet of mass M and radius R. If the escape velocity of planet is know to be 11.2 km s^-1. How fast will the object move If its velocity of launch is 22.4 km s^-1 from the

Answer 1

The minimum velocity required to escape from the gravitational force of the planet is known as the escape velocity of the planet. Consider the object ‘m’ on the surface of Earth.

Initial PE = \(-\frac {GMm}{R}\)

Initial KE = \(\frac {1}{2}\) m (V_esc)²

Total energy = \(\frac {1}{2}\)m (V_esc)² \(-\frac {GMm}{R}\)

To escape, the KE should greater than or equal to its PE.

Initial velocity of object = 22.4 km s^-1

= 2 V_esc

Total Initial Energy

= \(\frac {1}{2}\) m(2V_esc)² - \(\frac {GMm}{R}\)

= 2 m V²_esc- \(\frac {1}{2}\)m V²_esc [From(1)]

= \(\frac {3}{2}\) m V²_esc .......(2)

Total Final energy = PE (at ∞) + \(\frac {1}{2}\)mV_f²

= 0 + \(\frac {1}{2}\) mV_f²

= \(\frac {1}{2}\)mV_f² .........(3)

By conservation of energy, (2) = (3)

V_f = \(\sqrt 3\)V_esc

= 19.4 km s^-1.