Objects with masses of 307 kg and 796 kg are separated by 0.369 m. A 55.5 kg mas

Question

Objects with masses of 307 kg and 796 kg are separated by 0.369 m. A 55.5 kg mass is placed midway between them.

A. Find the magnitude of the net gravitational force exerted by the two larger masses on the 55.5 kg mass. The value of the universal gravitational constant is 6.672 × 1011 N · m2 /kg2 . Answer in units of N

B. Leaving the distance between the 307 kg and the 796 kg masses fixed, at what distance from the 796 kg mass (other than infinitely remote ones) does the 55.5 kg mass experience a net force of zero? Answer in units of m.

Explanation / Answer

here,

m1 = 307 kg
m2 = 796 kg
m3 = 55.5 kg
d = 0.369/2 = 0.184 m
G = 6.672 * 10^11 N·m2/kg2

Gravitational Force is given by :
F = G*m1m2/r^2

A)

F13 = G*m1m2/r^2
F13 = (6.672 * 10^11)*(307*55.5)/0.184^2
F13 = 3.358 * 10^-5 N

Similarly ,

F23 = G*m2m3/r^2
F23 = (6.672 * 10^11)*(55.5*796)/0.184^2
F23 = 8.706 * 10^-5 N

Fnet = 5.348 * 10^-5 N

B)
The resultant force F towards m2 is:
F = G m2 m3 / x^2 - G m1 m3 / (d - x)^2

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

796/(x^2) = 307 / (0.396 - x)^2

796(0.396 - x)^2 = 307x^2

489 x^2 - 630.432*x + 124.826 = 0

x = 0.244 m

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