JEE MAIN - Physics (2008 - No. 26)
A planet in a distant solar system is $$10$$ times more massive than the earth and its radius is $$10$$ times smaller. Given that the escape velocity from the earth is $$11\,\,km\,{s^{ - 1}},$$ the escape velocity from the surface of the planet would be
$$1.1\,\,km\,{s^{ - 1}}$$
$$100\,\,km\,{s^{ - 1}}$$
$$110\,\,km\,{s^{ - 1}}$$
$$0.11\,\,km\,{s^{ - 1}}$$
Explanation
Let Me is mass of earth then mass of planet Mp = 10Me.
And let Re is radius of earth then radius of planet Rp = $${{{R_e}} \over {10}}$$
Escape velocity of earth, $${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}} $$
Escape velocity of planet, $${v_p} = \sqrt {{{2G{M_p}} \over {{R_p}}}} $$
$$\therefore$$ $${{{{ {{v_p}}}}} \over {{{ {{v_e}} }}}} = {{\sqrt {{{2G{M_p}} \over {{R_p}}}} } \over {\sqrt {{{2G{M_e}} \over {{R_e}}}} }}$$
$$ = \sqrt {{{{M_p}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}} $$
$$ = \sqrt {{{10{M_e}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{{\mathop{\rm R}\nolimits} _e}/10}}} = 10$$
$$\therefore$$ $${{v_p}} = 10 \times {{v_e}}$$
$$ = 10 \times 11 = 110\,km/s$$
And let Re is radius of earth then radius of planet Rp = $${{{R_e}} \over {10}}$$
Escape velocity of earth, $${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}} $$
Escape velocity of planet, $${v_p} = \sqrt {{{2G{M_p}} \over {{R_p}}}} $$
$$\therefore$$ $${{{{ {{v_p}}}}} \over {{{ {{v_e}} }}}} = {{\sqrt {{{2G{M_p}} \over {{R_p}}}} } \over {\sqrt {{{2G{M_e}} \over {{R_e}}}} }}$$
$$ = \sqrt {{{{M_p}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}} $$
$$ = \sqrt {{{10{M_e}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{{\mathop{\rm R}\nolimits} _e}/10}}} = 10$$
$$\therefore$$ $${{v_p}} = 10 \times {{v_e}}$$
$$ = 10 \times 11 = 110\,km/s$$
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