Does an object falling from very large height as compared to radius of Earth rotate as well due to gravity?

Question

If let us say a rigid body is falling from a height such that value of $g$ remains same then the body does not rotate at all about the centre of mass as centre of mass and gravity coincide hence net gravitational torque is zero but if it falls from a large height such that value of $g$ varies then will the object rotate as well. Assume that only gravity is acting on the object and no other external force ? Eg an astronaut falling towards earth or meteor etc.

Answer 1

Yes, there can be rotation due to tidal forces.

Consider the extreme case of a dumbbell made from two point masses and in orbit round the Earth:

If the centre of mass is a distance $r$ from the centre of the earth then the orbital velocity of the dumbbell as a whole is:

$$ v = \sqrt{\frac{GM}{r}} $$

However the inner mass is closer to the Earth at a distance of $r-d$ and the outer mass is farther away at a distance of $r+d$. That means the velocities the two masses want to orbit at are:

$$ v_1 = \sqrt{\frac{GM}{r-d}} $$

and:

$$ v_2 = \sqrt{\frac{GM}{r+d}} $$

so $v_1 \gt v_2$. That means the dumbbell will rotate as it orbits.

This is something of an extreme example, and I suspect you were thinking of some object falling radially rather than orbiting, but the same argument applies. Since different parts of the object will be at different distances from the centre of the Earth they will experience different forces. If these forces exert a net torque then the object can rotate as it falls.

Answer 2

If the gravitational field is considered to be uniform, there is no torque on the object and it does not rotate. See Torque on a falling object and Will a free-falling rod (without drag) rotate?

In a non-uniform field there can be a torque on an object. It depends on the rate $\frac{dg}{dr} \propto \frac{1}{r^3}$ at which $g$ varies with distance $r$ from the Earth. This is greater for smaller distances $r$. (Torque also depends on the size of the object, which is likely to be very much smaller than $r$, and the distribution of mass in it.) The torque causing the object to turn increases as the object gets closer to the Earth. We don't notice any such tendency at the surface of the Earth - this shows how weak the torque is compared with other forces acting on us.

There is no net torque on a spherically symmetric object even if it rotates : it is in neutral equilibrium. An object without spherical symmetry can be placed in stable or unstable equilibrium. A small perturbation from equilibrium causes a torque.

Torque on an object in stable or unstable equilibrium causes it to oscillate about the its centre of mass like a compound pendulum. The frequency of the oscillation increases with $\frac{dg}{dr}$ as the object gets closer to Earth.

Because torque is gradually increasing this pumps energy into the oscillator (see parametric pumping). This is equivalent to increasing the strength of gravity acting on a pendulum. The amplitude of oscillation increases; at some point the oscillator has sufficient energy to rotate in complete circles.