Question

Four asteroids, of mass M = 3 × 10^6 kg, are positioned at the corners of an imaginary square of side length s = 2.4 × 10^8 m. Two other asteroids of mass m = 2 × 10^6 kg are positioned in the middle of the left and right sides of the square. Mass of spaceship is 2000kg. If a spaceship is positioned at the middle of the bottom side of the square what is the net force on the ship in Newtons?

Answer 1

let m' = 2000 kg

The forces exerted by lower two asteroids on spaceship are equal in magnitude and acts opposite direction.
so, they get cancelled.

magnitude of force exerted by each upper asteroid(M) on spaceship,

F1 = G*M*m'/(s^2 + (s/2)^2)

= 6.67*10^-11*3*10^6*2000/( (2.4*10^8)^2 + (1.2*10^8)^2)

= 5.56*10^-18 N

angle made F1 with vertical, theta1 = tan^-1((s/2)/s)

= tan^(1/2)

= 26.6 degrees

magnitude of force exerted by each upper asteroid(m) on spaceship,

F2 = G*m*m'/((s/2)^2 + (s/2)^2)

= 6.67*10^-11*3*10^6*2000/( (1.2*10^8)^2 + (1.2*10^8)^2)

= 1.39*10^-17 N

angle made F2 with vertical, theta2 = tan^-1((s/2)/(s/2))

= tan^(1)

= 45 degrees

the horizontal components get cancelled.

net froce will be in upward direction.

Net force acting on spaceship, Fnet = 2*F1*cos(theta1) + 2*f2*cos(theta2)

= 2*5.56*10^-18*cos(26.6) + 2*1.39*10^-17*cos(45)

= 2.96*10^-17 N <<<<<<<<<<---------------Answer