Question

The mass of the moon is 1% of mass of the earth. The ratio of gravitational pull of earth on moon to that of moon on earth will be-
A.) 1:1
B.) 1:10
C.) 1:100
D.) 2:1

Answer 1

Hint: We will first write the 1% given in the questions in terms of numbers and not percentage i.e. 0.01. Then we will find out the force acting on the moon by Earth and vice versa then divide the equations calculated in order to get the ratio.

Complete step-by-step answer:

Refer to the solution below.
Formula used: $F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$.
Let ${m_1}$ be the mass of the earth.
Let ${m_2}$ be the mass of the moon which is 1% of the mass of the earth. So-
$ \Rightarrow {m_2} = 0.01{m_1}$ (as $1\% = \dfrac{1}{{100}} = 0.01$)
Let r be the distance between the moon and the earth and G be the gravitational constant.
Force of Earth acting on Moon will be-
$
   \Rightarrow {F_{em}} = G\dfrac{{{m_1}{m_2}}}{{{r^2}}} \\
    \\
   \Rightarrow {F_{em}} = G\dfrac{{{m_1}0.01{m_1}}}{{{r^2}}} \\
    \\
   \Rightarrow {F_{em}} = G\dfrac{{0.01{m_1}^2}}{{{r^2}}} \\
$
Let the above equation be equation 1, we get-
$ \Rightarrow {F_{em}} = G\dfrac{{0.01{m_1}^2}}{{{r^2}}}$ (equation 1)
Force of Moon on Earth will be-
(It will be same as the force of Earth acting on Moon according to the Third Law of Motion)
$
   \Rightarrow {F_{me}} = G\dfrac{{{m_1}{m_2}}}{{{r^2}}} \\
    \\
   \Rightarrow {F_{me}} = G\dfrac{{{m_1}0.01{m_1}}}{{{r^2}}} \\
    \\
   \Rightarrow {F_{me}} = G\dfrac{{0.01{m_1}^2}}{{{r^2}}} \\
$
Let the above equation be equation 2, we get-
$ \Rightarrow {F_{em}} = G\dfrac{{0.01{m_1}^2}}{{{r^2}}}$ (equation 2)
To find the ratio, we will divide the equations.
Dividing equation 1 and equation 2, we will have-
$
   \Rightarrow \dfrac{{{F_{em}}}}{{{F_{me}}}} = \dfrac{{G\dfrac{{0.01{m_1}^2}}{{{r^2}}}}}{{G\dfrac{{0.01{m_1}^2}}{{{r^2}}}}} \\
    \\
   \Rightarrow \dfrac{{{F_{em}}}}{{{F_{me}}}} = \dfrac{1}{1} \\
    \\
   \Rightarrow {F_{em}}:{F_{me}} = 1:1 \\
$
Hence, option A is the correct option.

Note: Newton’s Third Law of Motion used in the above solution states that- For each action, there is an equivalent and inverse reaction. The statement implies that in each interaction, there is a pair of forces following up on the two collaborating objects.