Let us consider two small, identical masses, each of mass *m*, attached by a rigid, massless rod, so that their centers of mass are separated by the fixed distance 2*d*. We then place this “dumbbell” so that it’s center is a distance *R*>*d* from the center of mass of an object of mass *M*, and so that the line connecting the identical masses forms an angle *θ* with the line connecting the center of the dumbbell to the mass *M*. Treating the three masses as point masses, what is the gravitational potential energy of this arrangement, how does it vary with *θ*, and what does that imply physically?

For two bodies, the gravitational potential energy (with zero energy being defined as at infinite separation) is , with *M*_{1} and *M*_{2} being their masses, and r their separation. Here we have three bodies, and the energy is the sum of the energies for the three pairings:

.

First, we note that the third term is independent of both *R* and *θ*; it is the gravitational potential energy in the dumbbell assembly alone. Thus, what is of interest to us is

,

the portion of the potential energy due to the dumbbell’s placement relative to the third mass.

Basic trigonometry tells us that

and

.

So we see:

where .

Examining this over the range of meaningful angles, we see that the extrema of *U*_{v} are at and , with the minimum energy state for a fixed *R* being the former, and the maximum energy state being the latter. Thus, the stable equilibrium will be when the dumbbell has its axis aligned toward the gravitating body; and if it is placed at an angle , it will experience a torque rotating it toward that alignment.

This holds not just for the dumbbell, but qualitatively holds for any object with an elongated axis (even when that bulge is induced by the gradient in the other mass’s gravity (a “tidal bulge”)); thus the phenomenon of tidal locking, the most notable example of which is Earth’s Moon.

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Tags: Friday Physics, Gravitation, physics, Tidal Locking

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