Question

The separation between two 1 kg masses is (a) decreased by 2/3, and (b) increased by a factor of 3. How is the mutual gravitational force affected in each case?

Answer 1

(a)

`F'=F* (9)/(4)`

(b)

`F'=F* (1)/(9)`

Step-by-step explanation:

The gravitational force between the two masses is directly proportional to the product of two masses and inversely proportional to the square of distance between them.

This force is always attractive in nature.

Here, masses are of 1 kg each.

(a) Let initially the distance between them is d and the force is F.

Now the distance is 2/3 d , then the force is F'.

So,
`F \alpha (1)/(d^(2))` .... (1)

F' \alpha \frac{9}{4d^{2}} ...(2)

Divide equation (1) by equation (1), we get

`(F')/(F)=(9)/(4)`

`F'=F* (9)/(4)`

(b) Let initially the distance between them is d and the force is F.

Now the distance is 3 d , then the force is F'.

So,
`F \alpha (1)/(d^(2))` .... (1)

F' \alpha \frac{1}{9 d^{2}} ...(2)

Divide equation (1) by equation (1), we get

`(F')/(F)=(1)/(9)`

`F'=F* (1)/(9)`