**Gravitational Force of Three Identical Masses**

Part A = 0.0000172 N

Part A = -x direction

**Solution Below:**

_{1}= -130 cm, one is at the origin, and one is at x

_{2}= 390 cm.

Part A

_{grav}on the mass at the origin due to the other two masses?

Take the gravitational constant to be = 6.67×10

^{-11}N ⋅ m

^{2}/ kg

^{2}.

Express your answer in newtons.

The formula to use is F = GMm/r^{2}. Find the force from x_{1} and x_{3}, 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:

F_{1} = GMm/r^{2}

F_{1} = (6.67*10^-11) * 700 * 700 / -(1.30^2)

F_{1} = -0.00001934 N

F_{2} = GMm/r^{2}

F_{2} = (6.67*10^-11) * 700 * 700 / (3.90^2)

F_{1} = 0.000002149 N

F_{grav} = F_{1} + F_{2}

F_{grav} = -0.00001934 + 0.000002149

F_{grav} = -0.0000172

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

F_{grav} = 0.0000172 N

0.0000172 N

Part B

- +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