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Two equal masses separated by a particular distance experience a force of $100 N$ . Now, if the half of one of the masses is removed and the distance between two masses becomes double of the initial distance, the new force (in units of $Newton$ ) between two masses will be [[1]] $Newton$ .

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Detailed Solution

Two equal masses separated by a particular distance experience a force of

100 N

. Now, if the half of one of the masses is removed and the distance between two masses becomes double of the initial distance, the new force (in units of

Newton

) between two masses will be

12.5 N .

According to Newton's law of gravitation, the force between two masses is given by,

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

…………….(1)
Where,

F = original force

G = gravitational constant

m_{1} and m_{2} = masses of object

r = distance between objects

Given, initial force,

F = 100 N

According to given data,

m_{1}^{'} = {\frac{m_{1}}{2}}_{}

and

r_{1} = 2 r

Now, new force will be

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

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

\Rightarrow = \frac{G m_{1} m_{2}}{2 \times 4 r^{2}}

\Rightarrow = \frac{F}{8}

\Rightarrow = \frac{100}{8}

F^{'} = 12.5 N

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