Question
The escape velocity of a sphere of mass $$m$$ is given by ($$G =$$ universal gravitational constant, $${M_e} =$$ mass of the earth and $${R_e} =$$ radius of the earth)
A.
$$\sqrt {\frac{{G{M_e}}}{{{R_e}}}} $$
B.
$$\sqrt {\frac{{2G{M_e}}}{{{R_e}}}} $$
C.
$$\sqrt {\frac{{2GM}}{{{R_e}}}} $$
D.
$$\frac{{G{M_e}}}{{R_e^2}}$$
Answer :
$$\sqrt {\frac{{2G{M_e}}}{{{R_e}}}} $$
Solution :
The binding energy of sphere of mass $$m$$ (say) on the surface of the earth kept at rest is $$\frac{{G{M_e}m}}{{{R_e}}}.$$ To escape it from the earth's surface, this much energy in the form of kinetic energy is supplied to it.
So, $$\frac{1}{2}mv_e^2 = \frac{{G{M_e}m}}{{{R_e}}}$$
or $${v_e} = {\text{escape}}\,{\text{velocity}} = \sqrt {\frac{{2G{M_e}}}{{{R_e}}}} $$
where, $${R_e} =$$ radius of earth,
$${M_e} =$$ mass of the earth.