# Can very large objects have a gravitational moment?

Consider the Sears Tower. Its CG is about 1 millimeter below its CM. The reason why is because the base of the tower is closer to the center of the Earth than the top of the tower (by 442 m), and therefore receiving a slightly higher pull of gravity than the top of the tower. As a result, the CG is closer the the ground than the CM, because the part of the tower below the CM is being pulled by gravity (slightly) harder than the part of the tower above the CM

For an object standing up straight, it's quite intuitive and obvious that the CG will a distance away from the CM in a direction aligned with the direction of the gravitational pull. For an object that is at angle relative to the gravitational field (think of a very tall Leaning Tower of Pisa), it's not obvious if the CG would be "lower" in a direction aligned with gravity or along the geometric center of the tower. If the latter, it seems there would be a moment induced by the gravitational force that would tend to align the tower with the gravitational field (see graphic).

Which case, if either, is right?

• Please don't restrict yourself to rods or constant densities either. That's a very interesting case that I want considered, but the general case is what I'm really wanting to understand. – William Grobman Dec 11 '14 at 6:16
• Right case is correct. Without delving deep into proving it, let me just give you two examples: first, tidal locking - there's a reason the Moon is locked with its one side to Earth, - and the second is Integral Trees, a sci-fi novel by Larry Niven. – Orc JMR Dec 11 '14 at 6:56
• Hold on there, @OrcJMR. Tidal locking involves deformation of a non-ridged body and a time-dependent strain. While the book sounds interesting, I'd like something more than cool sci-fi. – William Grobman Dec 11 '14 at 6:59
• The deformation gives an outlet for the energy. But the torque is present without it. For a development of the moment, see dept.aoe.vt.edu/~cdhall/courses/aoe4140/SatDy.pdf – BowlOfRed Dec 11 '14 at 7:09
• @WilliamGrobman That wiki article explains how gravitational force is distributed. The very same force that deforms a non-rigid body will rotate a rigid one. The novel I am not asking you to read either, rather just to consider those trees as an emphasized illustration, a thought experiment on gravity and tidal locking. – Orc JMR Dec 12 '14 at 4:09

The case on the left is correct.

Gravitational gradient goes in the direction of the source of gravitational force.

Think about the extreme case: if the object was leaning almost all the way over at 89.9$^o$, the center of gravity would necessarily be almost directly below the center of mass.

• I'm not convinced by just looking at the two edge cases. Maybe the direction of the CG's displacement is influenced by both the mass distribution and the gravitational field orientation. Symmetry could cancel out the mass distribution effect at the edges but it could be off the CM's vertical at, say, 10deg. – William Grobman Dec 11 '14 at 6:19

If the density is constant, it is guarantedd that M and G will fall into the same vertical line. However, if the density is not constant, it is not difficult to find a counterexample that will have G and M on diffrerent vertical lines.

• If it's not too difficult, would you give an example? What you're proposing is very interesting. – William Grobman Dec 11 '14 at 3:40
• @WilliamGrobman I added an example – Wolphram jonny Dec 11 '14 at 4:26
• This is quite interesting. Assuming you rotated the object as shown in figure 2 and released it to free fall, how would angular momentum be conserved? It seems to me the object would rotate until CM and CG lie on a vertical line (or act as a pendulum in the absence of damping). Assuming the gravitational force is produced by a point mass seems to preclude another body making up for it. – William Grobman Dec 11 '14 at 5:11
• I don't think you even need non-uniform density, I tried working out a case where you have a very long uniform density toothpick at an angle to the Earth, and no matter how I try to work it out it always turns out there will be a torque on it. – Joshua Lin Dec 12 '14 at 22:57
• yes, I now think you might be correct, but havent given enough thought. If you find an example I encourage you to write your own answer. (and I will correct mine) – Wolphram jonny Dec 12 '14 at 23:02