The ratio of escape velocity at earth $$\left( {{v_e}} \right)$$ to the escape velocity at a planet $$\left( {{v_p}} \right)$$ whose radius and mean density are twice as that of earth is
A.
$$1:2\sqrt 2 $$
B.
$$1:4$$
C.
$$1:\sqrt 2 $$
D.
$$1:2$$
Answer :
$$1:2\sqrt 2 $$
Solution :
Since, the escape velocity of earth can be given as
$$\eqalign{
& {v_e} = \sqrt {2gR} = R\sqrt {\frac{8}{3}\pi G\rho } \,\,\left[ {\rho = {\text{density of earth}}} \right] \cr
& \Rightarrow {v_e} = R\sqrt {\frac{8}{3}\pi G\rho } \,......\left( {\text{i}} \right) \cr} $$
As it is given that the radius and mean density of planet are twice as that of earth. So, escape velocity at planet will be
$${v_p} = 2R\sqrt {\frac{8}{3}\pi G2\rho } \,......\left( {{\text{ii}}} \right)$$
Divide, Eq. (i) by Eq. (ii), we get
$$\eqalign{
& \frac{{{v_e}}}{{{v_p}}} = \frac{{R\sqrt {\frac{8}{3}\pi G\rho } }}{{2R\sqrt {\frac{8}{3}\pi G2\rho } }} \cr
& \Rightarrow \frac{{{v_e}}}{{{v_p}}} = \frac{1}{{2\sqrt 2 }} \cr} $$
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