Question

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles are m_A = 324 kg, m_B = 527 kg, and m_C= 183 kg. Take the positive direction to be to the right. Find the net gravitational force, including sign, acting on (a) particle A, (b) particle B, and (c) particle C.

Answer 1

a] By law of gravitation, Gravitational force on A = GmAmB/rAB^2 +  GmAmC/rAC^2

= 6.67e-11*324*527/0.5^2 + 6.67e-11*324*183/0.75^2

= 0.0000526 N

b] By law of gravitation, Gravitational force on B = -GmAmB/rAB^2 +  GmBmC/rBC^2

= -6.67e-11*324*527/0.5^2 + 6.67e-11*527*183/0.25^2

= 0.0000574 N

C] By law of gravitation, Gravitational force on C = -GmAmB/rAB^2 +  GmBmC/rBC^2

= -6.67e-11*324*183/0.75^2 - 6.67e-11*527*183/0.25^2

= -0.000110 N