First slide

Gravitaional potential energy

Question

The ratio of escape velocity on earth ( $ν_{e}$ ) to the escape velocity on a planet ( $ν_{p}$ ) whose radius and mean density are twice as that of earth is [NEET 2016]

Easy

A

$1 : 2 \sqrt{2}$
B

1 : 4
C

1 : $\sqrt{2}$
D

1 : 2

Solution

Since, the escape velocity of the earth can be given as
$v_{e} = \sqrt{2 g R} = \sqrt{2 (\frac{G M}{R^{2}}) R} = R \sqrt{\frac{8}{3} π G ρ}$
$[∵ mass of earth, m = ρ (\frac{4}{3} π R^{3})]$
$\Rightarrow v_{e} = R \sqrt{\frac{8}{3} π G ρ}$ … (i)
As it is given that, the radius and mean density of planet are twice as that of the earth. So, escape velocity at planet will be
$v_{p} = 2 R \sqrt{\frac{8}{3} π G 2 p}$ …(ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{v_{e}}{v_{p}} = \frac{R \sqrt{\frac{8}{3} π G ρ}}{2 R \sqrt{\frac{8}{3} π G 2 ρ}} \Rightarrow \frac{v_{e}}{v_{p}} = \frac{1}{2 \sqrt{2}}$

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