The escape velocity from the surface of the earth is $${v_e}.$$ The escape velocity from the surface of a planet whose mass and radius are three times those of the earth, will be
A.
$${v_e}$$
B.
$$3{v_e}$$
C.
$$9{v_e}$$
D.
$$\frac{1}{{3{v_e}}}$$
Answer :
$${v_e}$$
Solution :
Escape velocity on surface of the earth is given by
$${v_e} = \sqrt {2g{R_e}} = \sqrt {\frac{{2G{M_e}}}{{{R_e}}}} \,\,\left( {\because g = \frac{{G{M_e}}}{{R_e^2}}} \right)$$
where, $${M_e} =$$ mass of earth
$${R_e} =$$ radius of the earth
$$G =$$ gravitational constant
$$\therefore {v_e} \propto \sqrt {\frac{{{M_e}}}{{{R_e}}}} $$
If $${v_p}$$ is escape velocity from the surface of the planet, then $$\frac{{{v_e}}}{{{v_p}}} = \sqrt {\frac{{{M_e}}}{{{R_e}}}} \times \sqrt {\frac{{{R_p}}}{{{M_p}}}} $$
where, $${M_p}$$ is mass of the planet and $${R_p}$$ is radius of the planet.
but $${R_p} = 3{R_e}\,\,\left( {{\text{given}}} \right)$$
and $${M_p} = 3{M_e}$$
$$\eqalign{
& \therefore \frac{{{v_e}}}{{{v_p}}} = \sqrt {\frac{{{M_e}}}{{{R_e}}}} \times \sqrt {\frac{{3{R_e}}}{{3{M_e}}}} = \frac{1}{1} = 1 \cr
& {\text{or}}\,\,{v_p} = {v_e} \cr} $$
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