What happens to the gravitational force between two objects when:
(i) The distance between them is doubled.
(ii) Masses of both the objects are doubled.

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Solution

We know that, $F_{G} = \frac{G m_{1} m_{2}}{R^{2}}$

(i) $F_{G}^{'} = \frac{G m_{1} m_{2}}{(2 R)^{2}} = \frac{G m_{1} m_{2}}{4 R^{2}} = \frac{F_{G}}{4}$
The force becomes one-fourth of the original value.

(Ii) $F_{G}^{'} = \frac{G (2 m_{1}) (2 m_{2})}{R^{2}} = \frac{4 G m_{1} m_{2}}{R^{2}} = 4 F_{G}$
The force becomes four times the original value.

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