Question

The mass of the earth is 80 times that of the moon and its diameter is double that of the moon. If the value of acceleration due to gravity on earth is $9.8\,{\text{m}}\,{{\text{s}}^{ - 2}}$ , then the value of acceleration due to gravity on moon will be:
(A) $0.98\,{\text{m}}\,{{\text{s}}^{ - 2}}$
(B) $0.49\,{\text{m}}\,{{\text{s}}^{ - 2}}$
(C) $9.8\,{\text{m}}\,{{\text{s}}^{ - 2}}$
(D) $4.9\,{\text{m}}\,{{\text{s}}^{ - 2}}$

Answer 1

First of all, we will use the expression which gives the acceleration due to gravity on any heavenly object, which is a function of gravitational constant, mass and radius. We will use this expression to find acceleration due to gravity on the earth and the moon. After that we will simplify these two expressions and substitute the required values followed by manipulation to obtain the result.

Complete step by step solution:
In the given question, we are supplied with the following data:
The mass of the earth is $80$ times that of the moon.The diameter of the earth is double the diameter of the moon.We are also given that the acceleration due to gravity on earth is $9.8\,{\text{m}}\,{{\text{s}}^{ - 2}}$.We are asked to find the value of acceleration due to gravity on the moon.
To begin with, we know that the mass of the moon is far smaller than that of the earth. This implies that the gravitational pull of the moon is also far less than that of the earth. We have already learnt that gravitational force is less significant in case of smaller bodies like rocks, buildings, etc. However, this force is of great significance in case of heavenly bodies, as they have more mass.Let us proceed to solve the numerical.For this, we will apply the formula which gives the acceleration due to gravity, as given below:
$g = \dfrac{{G \times M}}{{{R^2}}}$ …… (1)
Where,
$g$ indicates the acceleration due to gravity.
$G$ indicates the gravitational constant.
$M$ indicates mass of the heavenly body.
$R$ indicates the radius of the body.
Now, for the acceleration due to gravity on the moon, the equation (1) can be written as:
${g_{\text{M}}} = \dfrac{{G \times {M_{\text{M}}}}}{{R_{\text{M}}^2}}$ …… (2)
Now, for the acceleration due to gravity on the earth, the equation (1) can be written as:
${g_{\text{E}}} = \dfrac{{G \times {M_{\text{E}}}}}{{R_{\text{E}}^2}}$ …… (3)
According to the question, we can write:
${M_{\text{E}}} = 80 \times {M_{\text{M}}}$
And,
${R_{\text{E}}} = 2 \times {R_{\text{M}}}$
Now, we divide the equation (2) by the equation (3) and we get:
$\dfrac{{{g_{\text{M}}}}}{{{g_{\text{E}}}}} = \dfrac{{\dfrac{{G \times {M_{\text{M}}}}}{{R_{\text{M}}^2}}}}{{\dfrac{{G \times {M_{\text{E}}}}}{{R_{\text{E}}^2}}}} \\\ \Rightarrow \dfrac{{{g_{\text{M}}}}}{{9.8}} = \dfrac{{\dfrac{{{M_{\text{M}}}}}{{R_{\text{M}}^2}}}}{{\dfrac{{80 \times {M_{\text{M}}}}}{{{{\left( {2 \times {R_{\text{M}}}} \right)}^2}}}}} \\\ \Rightarrow {g_{\text{M}}} = \dfrac{4}{{80}} \times 9.8\,{\text{m}}\,{{\text{s}}^{ - 2}} \\\ \therefore{g_{\text{M}}} = 0.49\,{\text{m}}\,{{\text{s}}^{ - 2}}$
Hence, the value of acceleration due to gravity on the moon is $0.49\,{\text{m}}\,{{\text{s}}^{ - 2}}$.

The correct option is B.

Note: While solving the problem, most of the students tend to make mistake by writing the relation of the radius and mass of the moon and earth as ${M_{\text{M}}} = 80 \times {M_{\text{E}}}$ and ${R_{\text{M}}} = 2 \times {R_{\text{E}}}$ which are completely wrong. It is necessary to seriously understand the relationship. Rather you can write, ${M_{\text{M}}} = \dfrac{{{M_{\text{E}}}}}{{80}}$ and ${R_{\text{M}}} = \dfrac{{{R_{\text{E}}}}}{2}$ , after that you can substitute them in the respective expression. The answer will come out to be the same.