# If there were two equal masses with one mass in the middle which way would gravity pull it?

If there were two masses with a third exactly in the middle of those two, which way would the middle mass be pulled?

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Asking which way the middle mass will be pulled is essentially like asking whether zero is less than zero or greater than zero. If the masses on the sides are equal, they pull equally hard in opposite directions, and so the net force on the middle mass is zero.

By Newton's 1st law, if the middle mass is initially at rest then it will remain at rest for all times. Now, there is admittedly a loophole here. If the middle mass's initial velocity has a nonzero component in the direction of the line determined by the outer masses, then in the next instant the middle mass will be closer to one of the outer masses, resulting in a greater force in the direction of that mass.

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don't the masses also have to be of identical geometry? I mean, a sphere and a cylinder of the same mass would have different gravitational fields, thus exerting a different force on the mass in the center, even if they are at the same distance from it. –  udiboy1209 Jul 1 '13 at 6:38

The net force on the middle mass would be zero:

$$\sum F=F_{13}+F_{23}=\frac{Gm^2}{r^2}-\frac{Gm^2}{r^2}=0$$

Here I'm writing $F_{13}$ for the force of the first mass on the third (middle mass). All the masses are equal to $m$, and I've just set the distance equal to $r$. The negative sign comes from the first mass pulling in one direction while the second pulls in the other.

Since the net force is zero, the mass is not accelerating (Newton's first law). If the middle mass had a relative velocity with respect to the other masses of $v=0$, then the entire system would be stationary. However, if the middle mass had some initial velocity $v\neq 0$, and if it moved closer to either mass 1 or 2 (so not in a direction perpendicular to their separation), it would then begin to accelerate in that direction.

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