Question

What happens to the forces between two objects, if

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?

Answer 1

STEP 1:

The gravitational force between two objects

$\Rightarrow F=\raisebox{1ex}{$GMm$}\!\left/ \!\raisebox{-1ex}{$r^2$}\right.$

G= Gravitational constant.

r=Distance between two bodies.

M and m are the masses of objects.

STEP 2:

When M becomes 2M

Then, the force also gets doubled.

$$\begin{gathered} \Rightarrow F^{\text{'}}=\frac{G2Mm}{{\left(r\right)}^2} \\ \Rightarrow F^{\text{'}}=\frac{2GMm}{r^2}=2F \end{gathered}$$

STEP 3:

When r becomes 2r

$$\begin{gathered} \Rightarrow F^{\text{'}}=\frac{GMm}{{\left(2r\right)}^2} \\ \Rightarrow F^{\text{'}}=\frac{1}{4}\frac{GMm}{r^2} \end{gathered}$$

Hence, force becomes one-fourth.

When r becomes 3r

$$\begin{gathered} \Rightarrow F^{\text{'}}=\frac{GMm}{{\left(3r\right)}^2} \\ \Rightarrow F^{\text{'}}=\frac{1}{9}\frac{GMm}{r^2} \end{gathered}$$

Hence, force becomes one-ninth.

STEP 4:

When M becomes 2M and m becomes 2m

$$\begin{gathered} \Rightarrow F^{\text{'}}=\frac{G\left(2M\right)\left(2m\right)}{r^2} \\ \Rightarrow F^{\text{'}}=\frac{4\left(GMm\right)}{r^2} \end{gathered}$$

Hence, force becomes four times.