Question

The mass of the moon is $\dfrac{1}{{81}}$ th of the earth but the gravitational pull is $\dfrac{1}{6}$ th of the earth.
(A) The radius of earth is $\dfrac{9}{{\sqrt 6 }}$ of the moon.
(B) The radius of the moon is $\dfrac{{81}}{6}$ of the earth.
(C) Moon is the satellite of the earth
(D) None of the above.

Answer 1

Hint : The gravitational pull of the body is the acceleration due to gravitational force of the planet. From the law of gravitation, the gravitational pull is inversely proportional to the square of distance between two bodies.
The gravitational pull is expressed as,
$g = \dfrac{{Gm}}{{{r^2}}}$
Here, $m$ is the mass of the body, $r$ is the radius distance and $G$ is the gravitational constant, which is equal to $6.67 \times {10^{ - 11}}\;{\rm{N}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}$ .

Complete Step By Step Answer:
From the given problem, the mass of the moon is $\dfrac{1}{{81}}$ th of the earth, ${m_m} = \dfrac{1}{{81}}{m_e}$ and the gravitational pull of the mass of the moon is $\dfrac{1}{6}$ th of the earth, ${g_m} = \dfrac{1}{6}{g_e}$ .
We know that the gravitational pull is acceleration due to the gravitational force on the planet or satellite.
The gravitational pull of the planet earth,
${g_e} = \dfrac{{G{m_e}}}{{r_e^2}}$
Rewrite the above equation,
${r_e} = \sqrt {\dfrac{{G{m_e}}}{{{g_e}}}}$ ... (I)
Similarly, the gravitational pull of the satellite moon,
${g_m} = \dfrac{{G{m_m}}}{{r_m^2}}$
Rewrite the above equation,
${r_m} = \sqrt {\dfrac{{G{m_m}}}{{{g_m}}}}$ ... (II)
Divide equation (I) and (II), we have,
$\dfrac{{{r_e}}}{{{r_m}}} = \dfrac{{\sqrt {\dfrac{{G{m_e}}}{{{g_e}}}} }}{{\sqrt {\dfrac{{G{m_m}}}{{{g_m}}}} }}\\\ = \sqrt {\dfrac{{{m_e} \times {g_m}}}{{{g_e} \times {m_m}}}}$
Now we substitute the given values in the above equation.
$\dfrac{{{r_e}}}{{{r_m}}} = \sqrt {\dfrac{{{m_e} \times \dfrac{1}{6}{g_e}}}{{{g_e} \times \dfrac{1}{{81}}{m_e}}}} \\\ \dfrac{{{r_e}}}{{{r_m}}} = \dfrac{9}{{\sqrt 6 }}\\\ {r_e} = \dfrac{9}{{\sqrt 6 }}{r_m}$
Thus, the radius of earth is $\dfrac{9}{{\sqrt 6 }}$ times the radius of the moon and option (a) is correct.

Note :
The value of acceleration due to the force of gravity of the planet or satellite is not depending on the mass of the planet, its size or its shape falling under the gravity. The universal law of gravitation holds everywhere. The value of acceleration due to gravity changes with the height and the depth. The angular velocity of rotation of earth affects the value of g.