Question

If you decrease the distance between two celestial objects by a factor of 2.5, how does the resultant gravitational force between the two objects compare to the original force? The force is ____ times stronger.

Only enter your numerical answer below 0 do not include units. Report your answer to 2 decimal places.

Answer 1

Answer:  The force is 6.25 times stronger.

Step-by-step explanation:

The formula to find the force of attraction between any two object is given by :-

$F=G\dfrac{Mm}{d^2}$                                 (i)

, where M and m are the masses of the objects , d is the distance between them and G is gravitational constant.

If you decrease the distance between two celestial objects by a factor of 2.5, then , the new distance will become

$d'=\dfrac{1}{2.5}d$

Plug this in the above formula , we get

$F'=G\dfrac{Mm}{(\dfrac{1}{2.5}d)^2}\\\\\Rightarrow\ F'=(2.5)^2(G\dfrac{Mm}{d^2})\\\\\Rightarrow\ F'=6.25(G\dfrac{Mm}{d^2})=6.25F----------\text{from }(i)$

Hence, the force is 6.25 times stronger.