MCQ
Two particles are placed at some distance from each other. If, keeping the distance between them unchanged, the mass of each of the two particles is doubled, the value of gravitational force between them will become:
  • A
    $\frac{1}{4}\text{times}$
  • B
    $\frac{1}{2}\text{times}$
  • $4\text{ times}$
  • D
    $2\text{ times}$

Answer

Correct option: C.
$4\text{ times}$
Here,$m_ 1$ and $m_2$ are doubled and r remains constant.
$\text{F} = \text{G}\times\text{m}_1\times\frac{\text{m}_2}{\text{r}^2}$
$=\text{G}\times2\text{m}_1\times\frac{2\text{m}_2}{\text{r}^2}$
$= 4\text{G}\times\text{m}_1\times\frac{\text{m}^2}{\text{r}^2}$

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