Question

Two particles are placed at some distance. If the mass of each of the two particles is doubled and keeping the distance between them unchanged, the value of gravitational force between them will be

• A. unchanged
• B. $\frac{1}{4}\text{times}$
• C. $\frac{1}{2}\text{times}$
• D. $4\text{times}$

Answer 1

The correct option is D $4\text{times}$

Force between two particles of mass $m_1$ and $m_2$ is given as,
$F=\frac{G{m}_1{m}_2}{r^2}\to \left(1\right)$
where G- universal gravitational constant
r- distance between two mass $m_1$ and $m_2$

let $F^{\prime }$ be the new force between two masses i.e $2m_1$ and $2m_2$ and also distance is fixed
$F^{\prime }=\frac{G2m_{1}2m_2}{r^2}\to \left(2\right)$

On dividing the equation 1 with equation 2 we get,
$\frac{F}{F^{\prime }}=\frac{G{m}_1{m}_2}{r^2}\ast \frac{r^2}{G2m_{1}2m_2}$

$\Rightarrow \frac{F}{F^{\prime }}=\frac{1}{4}$

$\Rightarrow F^{\prime }=4F$
so option d is correct.