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The gravitational force between two objects is $4 N$ . Which of the following would be the force between them if the distance between the two bodies is doubled?

(Given gravitation constant $G$ is $6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ )

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1 N

10 N

0.1 N

0.001 N

answer is A.

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Detailed Solution

The force between the two bodies will be

1 N

if the distance between them is doubled.
As per Newton's universal law of gravitation we know that force of attraction between two bodies is given by

F = \frac{G m_{1} m_{2}}{r^{2}} . . . . . . . . . . (a)

G

is the gravitational constant.

m_{1}

is the mass of the first body.

m_{2}

is the mass of the second body.

r

is the distance between two bodies.
Gravitation constant

G = 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}

Force is given

F = 4 N

Substituting all the values in (a)

\Rightarrow 4 = \frac{(6.67 \times 10^{- 11}) \times (m_{1}) \times (m_{2})}{r^{2}}

\Rightarrow r^{2} = \frac{6.67 \times 10^{- 11} m_{1} m_{2}}{4}

\Rightarrow r^{2} = 1.6675 \times 10^{- 11} \times m_{1} m_{2}

If the distance between them is doubled then

r

becomes

2 r

Thus, new force of between the two bodies be

F^{'}

Hence

F^{'} = \frac{G m_{1} m_{2}}{{(2 r)}^{2}}

\Rightarrow F^{'} = \frac{G m_{1} m_{2}}{{4 r}^{2}} . . . . . . . . . . . . . . . . (b)

Substituting the value of

G, m_{1}, m_{2}

and

r^{2}

in (b)

\Rightarrow F^{'} = \frac{6.67}{4 \times 1.6675}

\Rightarrow F^{'} = 1 N

Thus, the force between the two bodies will be

1 N

if the distance is doubled with the given details.

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