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One of the answers to this question: How does the center of gravity work? said the follows:

Since we can see that things do not start to rotate when they fall, we must assume gravity when averaged out to be pulling in the point where all torque cancel each other out, which is the center of gravity.

Also in some answers to that question, I learned that gravity isn't applied only at the center of gravity in real life, but rather every single particle of an object. Thus, I am confused why things do not start to rotate when they fall in real life as the torque caused by gravity doesn't cancel out?

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  • $\begingroup$ If all parts/objects/particles are fall with the same speed and acceleration, why would one end fall faster than the other end? Also your quote says the torque DOES cancel out so I do not know why your question then assumes the torque doesn't cancel out. $\endgroup$
    – DKNguyen
    Jul 15 at 13:22
  • $\begingroup$ Do you mean "rotate around a horizontal axis"? $\endgroup$ Jul 15 at 13:43
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    $\begingroup$ This may apply in a uniform gravitational field, for reasonable small, solid objects, etc. Consider the well-understood concept of Tidal Locking as a counterexample. $\endgroup$ Jul 15 at 14:56
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A gravitational field affects every bit of matter equally - for any object on earth, the acceleration due to gravity is 9.8 $m/s^2$. When an object is at rest, there is clearly no relative movement between any parts of the object, which would be observed as a bending, stretching, or rotation of the object. When you drop that object, every bit of it is accelerated due to gravity in the exact same way. The object as a whole is now moving, but since every atom was accelerated the exact same amount, there is still no relative movement between any parts of the object. The net torque is zero even when accounting for the force of gravity acting on every part of the object.

This answer assumes a uniform gravitational field, which is never exactly true but is a very good approximation when on the surface of the earth. A non-uniform gravitational field, (such as one experienced by a very long object near a black hole, or by the moon in earth's gravitational gradient for example) can induce a rotation as different parts of an object get accelerated differently, resulting in stretching, bending, or spinning. Tidal locking is an example of this phenomenon, in which a celestial body induces a rotation in another through the force of gravity alone.

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  • $\begingroup$ Thank you! Just to make sure, is it that when things are falling, they would naturally rotate along an axis through its center of gravity, but since there are downward forces exerted on both sides of the center of gravity, the produced torques actually cancel each other out? $\endgroup$
    – Alexia.
    Jul 15 at 13:31
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    $\begingroup$ @Cheryl Right, if there was some initial spin on the object, it would continue to rotate at a constant rate (since there's no net torque), and it would naturally rotate around the center of gravity. The downward forces on either side of the CoG produce torques that cancel each other out - heavier parts of the object will feel a larger force from gravity, but they'll also tend to be closer to the CoG, and lighter parts of the object will feel smaller force but be farther from the CoG, so the torques wind up canceling out. $\endgroup$ Jul 15 at 13:37
  • $\begingroup$ This is not true. Gravitational force varies with distance. Imagine a 100-km long pole oriented vertically. Now you have to be careful -- distortion of objects most certainly can happen, especially if this pole is made out of some soft stretchy material $\endgroup$ Jul 15 at 14:55
  • $\begingroup$ @CarlWitthoft That's why I said for any object on earth - acceleration due to gravity doesn't vary by more than 1% on the surface of the planet. For any reasonably sized, everyday object which one could feasibly drop and observe, variation in gravity and tidal forces are utterly negligible. Things do get more complicated if you have a planetary-scale object or are near a black hole, though. $\endgroup$ Jul 15 at 15:16
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The statement, "Objects don't rotate when they fall" is false.

I know this as an experimental fact b/c I worked on SRTM (https://www2.jpl.nasa.gov/srtm/). Consider the object in the figure: enter image description here

We can approximate this as an asymmetric dumbbell of length $L=60\,$m and two masses, $M=2\,$Gg (the space shuttle), and $m=20\,$Mg (the outboard radar antenna). The system is a side-looking interferometric synthetic aperture radar (InSAR) with 2 antenna forming a baseline that is perpendicular to the radar wave vector...which means it needs to sit at 45 degrees off-horizontal.

That configuration is unstable. It wants to rotate as the space shuttle orbits [read: falls]. A nitrogen cylinder was rigged to a nozzle on the outboard antenna to apply a smooth torque to stabilize the position; however, it froze in space, and the mission used the shuttle's attitude thrusters (not smooth). (The ensuing oscillations in the antenna arm added 1 year to the data processing, as the relative positions of the two antenna phase centers needs to be know to a small fraction of the radar wavelength, which was 5.6 cm).

The Earth's gravity field is not uniform. It's given by the potential:

$$ V(r)=-\frac{GM_E} r $$

which can be Taylor expanded at LEO ($R=6611137.0\,$m):

$$ V(R+\delta r) = V(R)\Big[1-\frac{\delta r} R +\big(\frac{\delta r} R\big)^2 \Big]$$

The lever arms of the two masses relative to the C-O-M are:

$$ L_M = L\frac m {M+m} $$ $$ L_m = L\frac M {M+m} $$

If you plug that into the Taylor expansion and look at the change in potential energy relative to $\theta=0$ (horizontal), the 0th doesn't contribute, and 1st order terms cancel:

$$\Delta U(\theta) = \frac{V(R)}{R^2}\frac{m^2M+M^2m}{M+m}$$

The result is shown in the figure:

enter image description here

So horizontal is an unstable equilibrium. The system wants to be aligned vertically, and it doesn't matter which mass is higher or lower. The 180 degree symmetry is characteristic of quadrupole moments. [This is a quadrupole moment coupling (hence the mass-squared terms) to a tidal tensor (2nd order potential derivative)].

Note that the magnitude, a hundred Joules, is tiny compared with the many TJ of total potential energy; nevertheless, the torque was enough to require thruster corrections every 20 minutes.

In general, forces and torques are caused by energy gradients. For an arbitrary distribution in an arbitrary conservative field, the various terms that contribute to energy are approximated by a multipole expansions. The zeroth order term in gravity is mass times potential:

$$ m\times -\big(\frac{GM} r)\big) \rightarrow m\times(gh) $$

In EM, a dipole couples to the field:

$$ \vec q \cdot \vec E = \vec q \cdot -\nabla \Phi $$

Note that a dipole moment is a 1st moment of charge, which couples to the gradient of the field. Since gravity is strictly attractive, this term must cancel by symmetry. (There are no gravitational dipoles).

The SRTM example is the quadrupole moment of a mass distribution (a 2nd moment), coupling to the tidal tensor:

$$ Q^{ij}\Phi_{ij} $$

The tidal tensor is a 2nd derivate with isotropic terms removed:

$$ \Phi_{ij} = J_{ij}-\frac 1 3 {\rm Tr}(J) $$

there $J$ is the 2nd derivative tensor of the potential:

$$ J_{ij} = \frac{\partial^2 \Phi(r)}{\partial x_i\partial x_j} $$

On can continue this to higher moments, with a rank-3 octupole tensor coupling to a suitable gradient of the tide (a la $\nabla_i\Phi_{jk}$), and so on, but that is not common. By the time fields and their gradients are strong enough that 3rd derivatives matter, you're into General Relativity and different techniques are used.

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Look into tidal locking.

Ask yourself why the same side of the moon always faces earth. In your search for an answer, you will come to understand how gravity is applied to every material point of a body. In some cases, this can in fact cause a gravitational torque.

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