The gravitational force between two objects is 'F'. If masses of both objects are halved without changing distance between them, then the gravitational force would become

\frac{F}{4}

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\frac{F}{2}

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2 F

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Solution

The correct option is A $\frac{F}{4}$
From the universal law of gravitation we have
$F = G \frac{m_{1} m_{2}}{r^{2}}$
If both the masses are halved the masses are
$F^{'} = \frac{G \frac{m_{1}}{2} \frac{m_{2}}{2}}{r^{2}} = \frac{F}{4}$

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Q. Question 3
The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become

(a)

\frac{F}{4}

(b)

\frac{F}{2}

(c)

F

(d)

2 F

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. If both of their masses are halved without altering the distance between them, the new gravitational force between them would be:

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The gravitational force of attraction between two objects is x. Keeping the masses of the objects unchanged, if the distance between the objects is halved, then the magnitude of gravitational force between them will become :

(a) $x / 4$
(b) $x / 2$
(c) $2 x$
(d) $4 x$

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The gravitational force between two objects is 'F'. If masses of both objects are halved without changing distance between them, then the gravitational force would become

The correct option is A F4From the universal law of gravitation we have F=Gm1 m2r2 If both the masses are halved the masses are F′=Gm12m22r2=F4

The correct option is A $\frac{F}{4}$
From the universal law of gravitation we have
$F = G \frac{m_{1} m_{2}}{r^{2}}$
If both the masses are halved the masses are
$F^{'} = \frac{G \frac{m_{1}}{2} \frac{m_{2}}{2}}{r^{2}} = \frac{F}{4}$