Question

Two particles A and B of mass m1 and m2 respectively are placed at some distance.If mass of each of the two bodies is doubled keeping distance between them unchanged. Value of gravitational force between them will be?​

Answer 1

Answer:

$\boxed{\bold{\mathfrak{F' = 4F}}}$

Given:

Mass of particle A = $\sf m_1$

Mass of particle B = $\sf m_2$

To Find:

Gravitational force between particle A & B when mass of each of them is doubled and distance between them is unchanged

Explanation:

Gravitational force of attraction between two particles:

$\boxed{ \bold{F = \frac{Gm_1 m_2}{r^2} }}$

As the distance between particles is kept constant;

$\therefore$

$\bold{ F \propto m_1 m_2} \: \: \: \: \: \: ...eq_1$

So, when mass of body is doubled gravitational force of attraction will be:

$\implies F ' \propto (2m_1)( 2m_2) \\ \\ \implies \bold{F ' \propto 4m_1 m_2} \: \: \: \: \: ...eq_2$

Dividing $eq_2$ by $eq_1$ we get:

$\implies \frac{F '}{F } = \frac{4 \cancel{m_1 m_2}}{\cancel{m_1 m_2}} \\ \\ \implies \frac{F '}{F } = 4 \\ \\ \implies \bold{F '= 4F}$

So,

Gravitational force between particle A & B becomes four times when mass of each of them is doubled and distance between them is unchanged

Answer 2

Answer:

By using gravitational force of attraction i.e.

F = GMm/r²

We get gravitational force become 4 times when mass of the both particle is double and distance kept unchanged