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Which of the following would be the distance between two bodies of mass two kg each so that the force between them is $1 N$ ? (Given gravitation constant $G =$ $6.67 \times 10^{- 11}$ $N m^{2} {kg}^{- 2}$ )

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1.63 \times 10^{- 5} m

1.63 \times 10^{- 6} m

16.3 \times 10^{- 5}

m

0.163 \times 10^{- 5}

m

answer is A.

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Detailed Solution

The distance between two bodies of mass

2 kg

each will be

1.63 \times 10^{- 5}

m

if the force between them is

1 N

.
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}} . . . . . . . . . . . . . (i)

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.
Here know that two bodies are of the same mass

m_{1} = m_{2}

Thus we can write

m = 2 kg

Gravitation constant

G = 6.67 \times 10^{- 11}

N m^{2} {kg}^{- 2}

Force is given

F = 1 N

We are asked to find the distance between the two bodies for the above details.
So we can write the equation

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

F = \frac{Gmm}{r^{2}} . . . . . . . . . . . . . (i)

Substituting all the values in (i)
We get

1 = \frac{(6.67 \times 10^{- 11} Nm 2 kg - 2) \times (2 kg) \times (2 kg)}{r^{2}}

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

\Rightarrow 1 = \frac{26.68 \times 10^{- 11}}{r^{2}}

\Rightarrow r^{2} = \frac{26.68 \times 10^{- 11}}{1}

\Rightarrow r^{2} = \frac{2.668 \times 10^{- 10}}{1}

\Rightarrow r = 1.63 \times 10^{- 5} m

Thus the distance between two bodies of mass

2 kg

will be

1.63 \times 10^{- 5}

m

if the force between them is

1 N

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