A 280kg object and a 580kg object are separated by 490 m a F

A 280-kg object and a 580-kg object are separated by 4.90 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 48.0-kg object placed midway between them. ___ N (b) At what position (other than an infinitely remote one) can the 48.0-kg object be placed so as to experience a net force of zero from the other two objects? ___ m from the 580 kg mass toward the 280 kg mass

a.)

b.)

Solution

M1 = 280 Kg
M2 = 580Kg
R = 4.90 /2 = 2.45 m

Gravitaion Force is given by, F = (G* M1*M2)/R^2

For 48.0-kg object placed midway between them -

Fnet = (G* M2*M3)/R^2 - (G* M1*M3)/R^2
Fnet = G* M3/R^2 (M2 - M1)
Fnet = (6.67408*10^-11 * 48.0)/ 2.45^2 * (580 - 280)
Fnet =  1.6*10^7 N

(b)
Let it be placed x m from 580 Kg Mass
At point of Equilibrium -
Fnet = 0
(G* M2*M3)/x^2 - (G* M1*M3)/(4.9-x)^2 = 0
M2/x^2 = M1/(4.9-x)^2
580/x^2 = 280/(4.9-x)^2
x = 2.89 m

Distance of the object from the 580 Kg mass = 2.9 m