Question

the distance between two particles was reduced from 6-m to 2 m. the force of gravity between these particles will thus be increased by a factor of​

Answer 1

Explanation:

Two particles start moving from same position on a circle of radius 20 cm with speed

401 m/s and 3611 m/s respectively in the same direction. Find the time after which the

particles will meet again.

Answer 2

Given:

The distance between two particles was reduced from 6-m to 2 m.

To find:

Force of gravity is increased by what factor ?

Calculation:

Initial Gravitation Force :

$f1 = \dfrac{G {m}^{2} }{ {6}^{2} } = \dfrac{G {m}^{2} }{36}$

Final Gravitation Force:

$f2 = \dfrac{G {m}^{2} }{ {2}^{2} } = \dfrac{G {m}^{2} }{4}$

Now dividing the equations :

$\dfrac{f2}{f 1} = \dfrac{36}{4}$

$\implies \dfrac{f2}{f 1} = 9$

$\implies f2 = 9(f1)$

So, force increases by a factor of 9.