Question

The acceleration due to gravity on the moon is only one sixth that of the earth. If the earth moon is assumed to have the same density. The ratio of the radii of moon and the earth will be
$
  (a){\text{ }}\dfrac{1}{6} \\
  (b){\text{ }}\dfrac{1}{{6\left( {1/3} \right)}} \\
  (c){\text{ }}\dfrac{1}{{36}} \\
  (d){\text{ }}\dfrac{1}{{6\left( {2/3} \right)}} \\
 $

Answer 1

Hint: In this question use the direct formula that acceleration due to gravity varies with the radius of the earth that is ${g_e}$ = (GM/$R_e^2$). Since earth is spherical in shape thus volume of earth can be written as $\dfrac{4}{3}\pi R_e^3$ and using this the mass can be obtained as density is the product of mass and volume. Apply this same concept to the moon as well. This will help formulation of two equations. Solve them up properly to get the required ratio of the radii.

Complete step-by-step answer:
As we know (g) is known as acceleration of gravity and the value of (g) is constant at the earth's surface or at sea level.
g = 9.8 ${m^2}$/sec (at earth surface).
But as we know g is dependent on the radius of the earth which is given as (i.e. gravitational force of the earth)
${g_e}$ = (GM/$R_e^2$)……………….. (1), where G = universal gravitational constant.
                                                                   M = mass of earth.
                                                                  ${R_e}$ = radius of the earth
Now as we know that the mass of the earth is the volume of the earth multiplied by density ($\rho $).
As we know earth is in the shape of a sphere so the volume (V) of the earth is $\dfrac{4}{3}\pi R_e^3$.
So the mass (M) of the earth is
M = $\dfrac{4}{3}\pi R_e^3 \times \rho $
Now substitute this value in equation (1) we have,
${g_e}$ = (GM/$R_e^2$) = $\dfrac{{G \times \dfrac{4}{3}\pi R_e^3 \times \rho }}{{R_e^2}} = G \times \dfrac{4}{3}\pi {R_e} \times \rho $.............. (1)
Now let the radius of the moon be ${R_m}$, so the gravitational force on the moon is
${g_m}$ = $\dfrac{{G \times \dfrac{4}{3}\pi R_m^3 \times \rho }}{{R_m^2}} = G \times \dfrac{4}{3}\pi {R_m} \times \rho $................... (2)
Now divide equation (1) and (2) we have,
$ \Rightarrow \dfrac{{{g_e}}}{{{g_m}}} = \dfrac{{G \times \dfrac{4}{3}\pi {R_e} \times \rho }}{{G \times \dfrac{4}{3}\pi {R_m} \times \rho }} = \dfrac{{{R_e}}}{{{R_m}}}$................ (3)
Now it is given that the acceleration due to gravity on the moon is only one sixth of the earth.
$ \Rightarrow {g_m} = \dfrac{1}{6}{g_e}$
$ \Rightarrow \dfrac{{{g_e}}}{{{g_m}}} = 6$
Now substitute this value in equation (3) we have,

$ \Rightarrow \dfrac{{{g_e}}}{{{g_m}}} = \dfrac{{{R_e}}}{{{R_m}}} = 6$
$ \Rightarrow \dfrac{{{R_m}}}{{{R_e}}} = \dfrac{1}{6}$
So the ratio of the radii of the moon to earth is (1/6).
So this is the required answer.
Hence option (A) is the correct answer.

Note – There is often a confusion between g and G. g is the acceleration due to gravity whose value is 9.8 at the surface of the earth however G is the proportionality constant and has a default value of $6.674 \times {10^{11}}mK{g^{ - 1}}{s^{ - 2}}$ . The trick point here was that the moon is default taken as spherical in shape. It is advised to consider the basic shape of the earth and the moon as spherical only while solving problems of this kind.