201903310540 Solution to 1980-AL-PHY-I-2

The original gravitational force $F$ is

$F=G \displaystyle{\frac{(m)(m)}{(2r)^2}}=G \displaystyle{\frac{m^2}{4r^2}}$ .

The final gravitational force $F'$ is

$F'=G \displaystyle{\frac{(m)(m)}{(2r+4r)^2}}=G \displaystyle{\frac{m^2}{36r^2}}=\displaystyle{\frac{1}{9}}\bigg( G \displaystyle{\frac{m^2}{4r^2}} \bigg) = \displaystyle{\frac{1}{9}}F$ .

Thus $F:F'=9:1$ .

And the answer is B.

M	T	W	T	F	S	S
				1	2	3
4	5	6	7	8	9	10
11	12	13	14	15	16	17
18	19	20	21	22	23	24
25	26	27	28	29	30	31

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