Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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:
(A) $1:4$
(B) $1:\sqrt 2 $
(C) $1:2$
(D) $1:2\sqrt 2 $

seo-qna
Last updated date: 17th Apr 2024
Total views: 34.8k
Views today: 0.34k
Answer
VerifiedVerified
34.8k+ views
Hint: From the question we can see that the radius and mean density of the planet is given. We can use it to calculate escape velocity of both earth and the planet.
Formula Used:
${v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} $\[\]

Complete step by step answer:
As the name suggests, escape velocity is an initial velocity at which, a body when thrown will leave the gravitational field of earth and never come back and its formula is
${v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} $\[\] where G is the gravitational constant, ${M_e},{R_e}$ are the mass and radius of earth
Since density is given, we can write mass as the product of volume and density
Hence $M = V\rho = \dfrac{{4\pi {R^3}}}{3}\rho $ (the shape of earth and planet is spherical)
V is the volume and $\rho $ is the density, now put it in the escape velocity equation,
$v = \sqrt {\dfrac{{2G \times 4\pi {R^3} \times \rho }}{{3R}}} = \sqrt {\dfrac{{2G \times 4\pi {R^2} \times \rho }}{3}} $
Escape velocity of earth${v_e} = \sqrt {\dfrac{{2G \times 4\pi {R_e}^2 \times {\rho _e}}}{3}} $
Similarly escape velocity of planet ${v_p} = \sqrt {\dfrac{{2G \times 4\pi {R_p}^2 \times {\rho _p}}}{3}} $
The ratio is \[\dfrac{{{v_e}}}{{{v_p}}} = \sqrt {\dfrac{{2G \times 4\pi {R_e}^2 \times {\rho _e}}}{3}} \Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{{R_e}}}{{{R_p}}}\sqrt {\dfrac{{{\rho _e}}}{{{\rho _p}}}} \]
Since it is given in the question that the radius and mean density of planet is two times to that of earth
Putting this value in above equation, it becomes

\[\dfrac{{{v_e}}}{{{v_p}}} = \dfrac{1}{2}\sqrt {\dfrac{1}{2}} \Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{1}{{2\sqrt 2 }}\]

Hence, the correct option is D

Additional information:
Mathematically, gravitational constant is the force of attraction of two particles which are of unit mass and are kept at a distance of a unit. It is not affected by the presence of any other body or medium. It is the same in every condition. The SI unit of G is \[N{m^2}k{g^{ - 2}}\]

Note:
The escape velocity of smaller planets like mars, is less and there is no atmosphere and bigger planets like Jupiter, Saturn have very large escape velocity hence they have denser atmospheres in these planets.