Question

Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one?A) 16F
B) 8F
C) 4F
D) F
E) F/4

Answer 1

The magnitude of the attractive force on each one will be F.

Explanation:

As two small balls are experiencing gravitational force between them ,then they will obey universal law of gravity. As per the universal law of gravity, the gravitational force acting between two objects will be directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between the two objects.

So if the mass of the two small balls A and B are considered as M and m, respectively, with the distance of separation be considered as r. Then the gravitational force of attraction acting between A and B will be

$F = \frac{GMn}{r^{2} }$ , This is the original or initial gravitational force between A and B.

Now, if the masses of A and B are doubled, then the new masses will be M' = 2 M and m' = 2m, respectively. Similarly, if the separation of the balls is also doubled then r' = 2r. So the new gravitational force exerting between A and B is

$F' = \frac{GM'm'}{r'^{2} } = \frac{G*2M*2m}{(2r)^{2} } =\frac{4GMm}{4r^{2} } = \frac{GMm}{r^{2} }=F$

So after doubling the masses as well as the distance of separation, there will be no change in the gravitational force. So the magnitude of the attractive force on each one will be F.