Question 231: Gravitation

Question

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

Accepted Answer

Let M
e
is mass of earth then mass of planet M
p
= 10M
e
.
And let R
e
is radius of earth then radius of planet R
p
= $${{{R_e}} \over {10}}$$
Escape velocity of earth, $${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}} $$
Escape velocity of planet, $${v_p} = \sqrt {{{2G{M_p}} \over {{R_p}}}} $$
$$	herefore$$ $${{{{ {{v_p}}}}} \over {{{ {{v_e}} }}}} = {{\sqrt {{{2G{M_p}} \over {{R_p}}}} } \over {\sqrt {{{2G{M_e}} \over {{R_e}}}} }}$$
$$ = \sqrt {{{{M_p}} \over {{M_e}}} 	imes {{{{\mathop{m R}
olimits} _e}} \over {{R_p}}}} $$
$$ = \sqrt {{{10{M_e}} \over {{M_e}}} 	imes {{{{\mathop{m R}
olimits} _e}} \over {{{\mathop{m R}
olimits} _e}/10}}}  = 10$$
$$	herefore$$ $${{v_p}} = 10 	imes {{v_e}}$$
$$ = 10 	imes 11 = 110\,km/s$$

Answer

$$1.1\,\,km\,{s^{ - 1}}$$

Answer

$$100\,\,km\,{s^{ - 1}}$$

Answer

$$110\,\,km\,{s^{ - 1}}$$

Answer

$$0.11\,\,km\,{s^{ - 1}}$$

Gravitation

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

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