Question

Find out 'escape velocity of a particle on earth's surface. (Acceleration due to gravity, g- 9.8m/s2
and radius of earth R - 6400 km)​

Answer 1

Answer:

Escape velocity of a particle from earth's surface is the velocity which if provided to the particle, it will escape from earth's gravitational influence.

If that happens, then the total energy of the object will be 0.

Let the particle of mass m is given escape velocity Ve.

Mass of earth = Me

Radius of earth = Re

Total energy = gravitational potential energy + Kinetic energy

$TE = PE + KE\\ \\ \Rightarrow 0 = \frac{-GM_em}{R_e} +\frac{1}{2}\ mV_e^2\\ \\ \Rightarrow \frac{1}{2}\ mV_e^2= \frac{GM_em}{R_e} \\ \\ \Rightarrow V_e^2= \frac{2GM_e}{R_e} \\ \\ \Rightarrow \boxed{V_e= \sqrt{ \frac{2GM_e}{R_e}}}$