0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ – udiboy1209 Jul 1 '13 at 6:38
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

For this answer, I'll assume that the center mass (absent any influence from the other two masses) is a perfect sphere, and that the other two masses are "above" and "below" the center mass.

The middle mass will actually be pulled in both directions at once, because the top edge of the middle mass is closer to the mass on the top than to the mass on the bottom, so that edge will be pulled upward. The bottom edge of the middle mass will likewise be pulled downward, and the center of the middle mass will remain stationary. Because there's no such thing as a perfectly rigid body, the middle mass will be slightly deformed, in much the same way that the earth's radius is slightly greater at the equator than at the poles, except that in this case, the deformation will be in the opposite direction - the "poles" will have a slightly greater radius than the "equator".

Note that this is still true even if the mass is not spherical.

$\endgroup$
0
$\begingroup$

All three masses will be interacting via gravity. A static equilibrium configuration is not possible, just as in the electrostatic case. Let's assume all three masses are brought in position in empty space, at the same time, away from all nearby influences. If initially they are at rest, and released to attract one another, he whole system has negative total energy and the three masses can move in bound orbits, or one mass can escape and leave the other two in a bound orbit, depending on the exact initial conditions. This is the 3-body problem and it is notoriously difficult to solve (It has no exact analytical solution), but the reduced 3-body problem is more easily solved by approximation technique, assuming the middle mass is much smaller than the other two. (e.g. earth, satellite, sun).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.