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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.

User PradeepK
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Answer: The force is 6.25 times stronger.

Explanation:

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


F=G(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'=(1)/(2.5)d

Plug this in the above formula , we get


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

Hence, the force is 6.25 times stronger.

User Shien Hong
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