Question

Two celestial bodies are separated by some distance. If the mass of any one of the points is doubled while the mass of the other is halved then how far should they be taken so that the gravitational force between them is one-fourth?

Answer 1

Hint – In this question let $m_1$ be the mass of the first body and the mass of the second celestial body be$m_2$. Use the direct relationship between the force, masses and the distance between two bodies that is $F_g=G{\displaystyle \frac{m_1{m}_2}{r^2}}$. Since the masses are now changed therefore application of the same formula will help yielding two different equations. Use them to get the answer.
Step by step answer:

Let the first celestial body have mass $m_1$ and the second celestial body having mass$m_2$.
Let them are separated by the distance r as shown in the figure.
As we know that the gravitational force ($F_g$) working between these two bodies is given as
$\Rightarrow F_g=G{\displaystyle \frac{m_1{m}_2}{r^2}}$...................... (1)
Where, G = universal gravitational constant.
Now it is given that the mass of any one body is doubled and the mass of the other body is halved and the gravitational force working between them is one-fourth of the previous one.
Let mass of first body = $m_3$
And mass of other body = $m_4$
Let the force between them = $F_g^{\prime }$
Therefore, $m_3=2m_1$
And $m_4={\displaystyle \frac{1}{2}}{m}_2$
And, $F_g^{\prime }={\displaystyle \frac{1}{4}}{F}_g$
Let the distance between them is (d).
So the gravitational force between them is
$\Rightarrow F_g^{\prime }=G{\displaystyle \frac{m_3{m}_4}{d^2}}$
Now substitute the values we have,
$\Rightarrow {\displaystyle \frac{F_g}{4}}=G{\displaystyle \frac{2m_1\left({\displaystyle \frac{1}{2}}{m}_2\right)}{d^2}}$
$\Rightarrow {\displaystyle \frac{F_g}{4}}=G{\displaystyle \frac{m_1\left(m_2\right)}{d^2}}$............. (2)
Now divide equation (1) from equation (2) we have,
$\Rightarrow {\displaystyle \frac{F_g}{{\displaystyle \frac{F_g}{4}}}}={\displaystyle \frac{G{\displaystyle \frac{m_1{m}_2}{r^2}}}{G{\displaystyle \frac{m_1\left(m_2\right)}{d^2}}}}$
Now simplify this equation we have,
$\Rightarrow 4={\displaystyle \frac{d^2}{r^2}}$
Now take square root on both sides we have,
$\Rightarrow \sqrt{4}=\sqrt{{\displaystyle \frac{d^2}{r^2}}}$
$\Rightarrow {\displaystyle \frac{d}{r}}=2$
$\Rightarrow d=2r$
So the separation between the bodies should be doubled.
So this is the required 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 $6.674\times {10}^{-11}{m}^{3}K{g}^{-1}{s}^{-2}$. It is advised to remember the direct formula for the force of gravitation between two masses that is $F_g^{\prime }=G{\displaystyle \frac{m_3{m}_4}{d^2}}$.