Question

The masses and coordinates of three spheres are as follows: 10 kg, x = 2.00 m, y = 1.75 m; 32 kg, x = -1.50 m, y = -1.50 m; 60 kg, x = 0.00 m, y= -0.50 m. What is the magnitude of the gravitational force on a 10 kg sphere located at the origin due to the other spheres?

Answer 1

The radius from the origin to the first mass is:

R1 = sqrt(2.00^2 + 1.75^2) = 2.65 m

The gravitational force of the first mass on the origin is:

F1 = G*m0*m2/R^2 = (6.673 × 10^-11)*10*10/2.65^2 = 9.50*10^-10N

This has x and y components:

F1x = F1*2.00/2.65 = 716*10^-12N

F1y = F1*1.75/2.65 = 6.273*10^-10N

The Radius to mass number 2 is:

R2 = sqrt(1.5^2 + 1.5^2) = 2.12m

The gravitational force caused by the second mass is:

F2 = G*m0*m2/R2^2 = (6.673 × 10^-11)*10*32/2.12^2 = 4.75*10^-9N

This has the following components:

F2x = F2*(-1.5/2.12) = -3.36*10^-9N

F2y = F2*(-1.5/2.12) = -3.36*10^-9N

The Radius to mass number 3 is: 0.50m

The gravitational force due to mass number 3 is:

F3 = G*m0*m3/R3^2 = (6.673 × 10^-11)*10*60/.50^2 = 160.2*10^-9N

The x component of F3 is zero.

The y component of F3 is F3 directed along the negative y axis:

F3y = -160.2*10^-9N

Sum up the components:

Fx = F1x + F2x + F3x = -264.5*10^-12N

Fy = F1y + F2y + F3y = -162.9*10^-9N

|F| = sqrt(Fx^2 + Fy^2) = 4.03*10^-4N