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# The acceleration due to gravity on the surface of the moon is one sixth that on the surface of earth’s surface and the diameter of the moon is one fourth of that of earth. The ratio of escape velocity on moon and earth will beA. $\dfrac{1}{{\sqrt {24} }}$ B. $\dfrac{2}{{\sqrt {16} }}$ C. $\dfrac{2}{{\sqrt 3 }}$ D. $\sqrt {24}$

Last updated date: 20th Jun 2024
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Hint: The escape velocity is the square root of the product of acceleration due to gravity of earth or moon and diameter of the earth or moon. Then, we will find the ratio of escape velocities for both the earth and moon.

When we throw a body vertically upwards with certain velocity, the body returns to the earth’s surface after some time. However, when the body is thrown with a velocity equal to the escape velocity, the body overcomes the earth’s gravitational pull and also the resistance of the earth’s atmosphere. Escape velocity is defined as the least velocity with which a body must be thrown vertically upwards in order that it may just escape the gravitational pull of earth.

Let Ve be the escape velocity on the earth, then kinetic energy imparted to the body = $\dfrac{1}{2}m{\left( {Ve} \right)^2}$ .so that its magnitude is equal to the gravitational energy i.e., $\dfrac{1}{2}m{\left( {Ve} \right)^2} = \dfrac{{GMm}}{R}$ where m is the mass of the body, M is mass of earth and R is the radius of the earth.
${\left( {V_e} \right)^2} = \dfrac{{2GM}}{R}$ and also $g = \dfrac{{GM}}{{{R^2}}}$ or $GM = g{R^2}$ where g is the acceleration due to gravity.
${\left( {V_e} \right)^2} = \dfrac{{2g{R^2}}}{R}$
$\Rightarrow{V_e} = \sqrt {2gR}$
$\Rightarrow{V_e} = \sqrt {g\left( {2R} \right)}$
$\Rightarrow{V_e} = \sqrt {gD}$ as diameter $\left( D \right) = 2R$ [eqn.1]
Now, let Vm be the escape velocity of the moon, g1 be the acceleration due to gravity on the moon and D1 be the diameter of the moon.
${V_m} = \sqrt {\left( {g1} \right)\left( {D1} \right)}$
$\Rightarrow{V_m} = \sqrt {\dfrac{g}{6} \times \dfrac{D}{4}}$ as diameter of the moon is one-fourth of the earth and acceleration due to gravity on moon is one-sixth of the earth’s gravity.
${V_m} = \sqrt {\dfrac{{gD}}{{24}}}$
$\Rightarrow{V_m} = \dfrac{{V_e}}{{\sqrt {24} }}$ from eqn.1
$\therefore\dfrac{{V_m}}{{V_e}} = \dfrac{1}{{\sqrt {24} }}$

Therefore, option A is correct.

Note:Kindly remember that radius of the earth in the formula of escape velocity should be converted to diameter of the earth as the relationship between diameter of earth and moon is given in the question. This question can be solved directly without doing so many calculations by dividing the product of g and d (diameter) of earth with the product of g1 $\left( {\dfrac{g}{6}} \right)$ and d1$\left( {\dfrac{d}{4}} \right)$ of moon because escape velocity is directly proportional to both g and d.