Question 391: Gravitation

Question

The ratio of escape velocity at earth (v
e
) to the escape velocity at a planet (v
p
) whose radius and mean density  are twice as that of earth is

Accepted Answer

As we know, escape velocity,
$${V_e} = \sqrt {{{2GM} \over R}}  = \sqrt {{{2G} \over R}.\left( {{4 \over 3}\pi {R^3}ho } ight)}  \propto R\sqrt ho  $$
$$ 	herefore $$ $${{{V_e}} \over {{V_p}}} = {{{R_e}} \over {{R_p}}}\sqrt {{{{ho _e}} \over {{ho _p}}}} $$
$$ \Rightarrow {{{V_e}} \over {{V_p}}} = {{{R_e}} \over {2{R_e}}}\sqrt {{{{ho _e}} \over {2{ho _e}}}} $$
$$ 	herefore $$ Ration $${{{V_e}} \over {{V_p}}} = 1:2\sqrt 2 $$

Answer

1 : 4

Answer

1 : $$\sqrt 2 $$

Answer

1 : 2

Answer

1 : 2$$\sqrt 2 $$

Gravitation

The ratio of escape velocity at earth (v e ) to the escape velocity at a planet (v p ) whose radius and mean density are twice as that of earth is

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