If the distance between two suspended masses 10kg each is tripled, the gravitational force of attraction between them is reduced by
A
one half
B
one third
C
one quarter
D
one ninth
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Let the distance apart the two masses = m.
From the equation F & M1 m2/ϒ2
F1 α 10 x 10/ϒ2 = F α 100/ ϒ2
F2 α 10 X 10/(3ϒ)2 => F2 α 100/9ϒ2
therefore F2/F1 = 100/9ϒ2/100/ϒ2 = 100/9ϒ2 ϒ2/100 = 1/9
Therefore F 2 : F 1 = 1 : 9
Source: Myschool Past Questions Archive (Record #9672)