Question

Part A: Objects with masses of 81 kg and 634 kg are separated by 0.362 m. A 72.9 kg mass is placed midway between them.Find the magnitude of the net gravitational force exerted by the two larger masses on the 72.9 kg mass. The value of the universal gravitational constant is 6.672 × 10−11 N · m2 /kg2 . Answer in units of N.

634 kg 81 kg 72.9 kg 0.362 m

Part B: Leaving the distance between the 81 kg and the 634 kg masses fixed, at what distance from the 634 kg mass (other than infinitely remote ones) does the 72.9 kg mass experience a net force of zero? Answer in units of m

Answer 1

Given,

m1 = 81 kg ; m2 = 634 kg ; m3 = 72.9 kg ; r = 0.362 m

a)We know that,

Fg = G m1 m2/r^2

F = G m2 m3/x^2 - Gm1 m2/(r - x)^2 = G m3 (m2/x^2 - m1/(r - x)^2]

x = r/2, we get

F = 4 G m3(m2 - m1)/r^2

F = 4 x 6.672 x 10^-11 x 72.9 (634 - 81)/0.362^2 = 8.2 x 10^-5 N

Hence, F = 8.2 x 10^-5 N

B)for the force to be zero,

F = G m3 (m2/x^2 - m1/(r - x)^2] = 0

m2/x^2 = m1/(r - x)^2

m2 (r - x)^2 = m1x^2

m2 (r^2 + x^2 - 2 r x) = m1 x^2

m2 r^2 + m2 x^2 - 2 m2 r x - m1 x^2 = 0

x^2 (m2 - m1) - 2 m2 r x + m2 r^2 = 0

x = r (m2 +/- sqrt (m1 m2))/(m2 - m1)

x = 0.362 (634 - sqrt (81 x 634))/(634 - 81) = 0.267 m

Hence, x = 0.267 m