The force of gravitation between two bodies of mass 1 kg each separated by a distance of 1 m in vaccum is

6.67 \times 10^{- 11} N

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6.67 \times 10^{- 10} N

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6.67 \times 10^{- 9} N

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6.67 \times 10^{11} N

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Solution

The correct option is A $6.67 \times 10^{- 11} N$
Given : $m_{1} = m_{2} = 1 k g$ , $r = 1 m$
Force of gravitation $F = \frac{G m_{1} m_{2}}{r^{2}}$
$F = \frac{6.67 \times 10^{- 11} \times 1 \times 1}{1^{2}} = 6.67 \times 10^{- 11} N$

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The force of gravitation between two bodies of mass 1 kg each separated by a distance of 1 m in vaccum is

The correct option is A 6.67×10−11NGiven : m1=m2=1 kg, r=1 mForce of gravitation F=Gm1m2r2 F=6.67×10−11×1×112=6.67×10−11N

The correct option is A $6.67 \times 10^{- 11} N$
Given : $m_{1} = m_{2} = 1 k g$ , $r = 1 m$
Force of gravitation $F = \frac{G m_{1} m_{2}}{r^{2}}$
$F = \frac{6.67 \times 10^{- 11} \times 1 \times 1}{1^{2}} = 6.67 \times 10^{- 11} N$