## Mastering Physics Solutions: Gravitational Force of Three Identical Masses

Gravitational Force of Three Identical Masses

Part A = 0.0000172 N
Part A = -x direction

Solution Below:

Three identical masses of 700 kg each are placed on the x axis. One mass is at x1 = -130 cm, one is at the origin, and one is at x2 = 390 cm.

Part A

What is the magnitude of the net gravitational force Fgrav on the mass at the origin due to the other two masses?
Take the gravitational constant to be = 6.67×10-11 N ⋅ m2 / kg2.

The formula to use is F = GMm/r2. Find the force from x1 and x3, one at a time. Remember that the first force will be negative (going to the left) and the second force will be positive (going to the right). But the overall force will be positive, because Mastering Physics only wants the magnitude:

F1 = GMm/r2
F1 = (6.67*10^-11) * 700 * 700 / -(1.30^2)
F1 = -0.00001934 N

F2 = GMm/r2
F2 = (6.67*10^-11) * 700 * 700 / (3.90^2)
F1 = 0.000002149 N

Fgrav = F1 + F2
Fgrav = -0.00001934 + 0.000002149
Fgrav = -0.0000172

Make this overall force positive, since Mastering Physics only wants the magnitude:

Fgrav = 0.0000172 N

0.0000172 N

Part B

What is the direction of the net gravitational force on the mass at the origin due to the other two masses?

• +x direction
• -x direction
• The answer just depends on which force from Part A was larger (whether mass 1 or mass 2 was closer to the origin). Since the force from the mass on the left was larger, the object at the origin will move to the left. This is in the -x direction.

-x direction

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