Question

when the distance between the masses M1 and M2 is reduced to one by third of initial distance then ratio of initial to final force is?​

Answer 1

Answer:

Ratio - 1 : 9

Explanation:

Check out the attachment for the solution.

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Answer 2

Answer:

$\large\boxed{\sf{1:9}}$

Explanation:

Given that there are two masses $M_{1}$ and $M_{2}$

Let the distance between them is $\bold{r}$.

Therefore, initial gravitational force is given by,

$\large\bold{=>F=G\dfrac{M_{1}M_{2}}{{r}^{2}}}$

Now, it's said that, the distance is reduced to one third of initial distance.

Therefore, the new distance is $\bold{\dfrac{r}{3}}$

Therefore, final gravitational force will be,

$\large\bold{= > f = G \dfrac{M_{1}M_{2}}{ {( \frac{r}{3} )}^{2} } }\\ \\\\ = > f = G \dfrac{M_{1}M_{2}}{ \frac{ {r}^{2} }{9} } \\\\ \\ = > f = 9(G \dfrac{M_{1}M_{2}}{ {r}^{2} } ) \\ \\ \\ = > f = 9F \\\\ \\ = > \dfrac{F}{f} = \dfrac{1}{9}$

Hence, ratio of initial force to final force is 1:9