JEE MAIN - Physics (2002 - No. 52)
The kinetic energy needed to project a body of mass $$m$$ from the earth surface (radius $$R$$) to infinity is
$$mgR/2$$
$$2mgR$$
$$mgR$$
$$mgR/4$$
Selitys
The required velocity is called escape velocity ($${v_e}$$) to leave the earth surface of a body.
$${v_e} = $$ escape velocity $$ = \sqrt {2gR} $$
Kinetic Energy $$K.E = {1 \over 2}mv_e^2$$
$$\therefore$$ $$K.E = {1 \over 2}m \times 2gR = mgR$$
$${v_e} = $$ escape velocity $$ = \sqrt {2gR} $$
Kinetic Energy $$K.E = {1 \over 2}mv_e^2$$
$$\therefore$$ $$K.E = {1 \over 2}m \times 2gR = mgR$$