A 105kg object and a 405kg object are separated by 410 m a F

A 105-kg object and a 405-kg object are separated by 4.10 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 30.0-kg object placed midway between them. (b) At what position (other than an infinitely remote one) can the 30.0-kg object be placed so as to experience a net force of zero from the other two objects?

Solution

The net force will the vector sum of force due to both. in this case both forces act along same line but in opposite direction.so we simply subtract.

F = (G*m1*m)/r^2 - (G*m2*m)/r^2

F = G*m/r^2 {m1 -m2}

where G = Gravitational constat; m1= 405 kg; m2=105 kg; m=30kg; 2r=4.10m;

F = 6.67*10^(-11)*30*(405 - 105)/(2.05^2) = 1.4284*10^(-7)N

b)

both the forces have to be equal

(G*m1*m)/r^2 = (G*m2*m)/(d-r)^2

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

r/d-r = sqrt(405/105) = 1.9639

r = 1.9639*4.1 - 1.9639*r

r = 2.7166 m