Torque and angular acceleration in elliptical orbit I am stuck in a supposedly simple aspect. Consider the Sun-Earth system. The torque is zero and angular momentum is conserved. $L = I\omega$ is constant, but since $I$ changes, $\omega$ should change as well. That means there is a non-zero angular acceleration.
Now consider $\tau = I\alpha$ which should be zero. Since $I$ can't be zero, angular acceleration must be zero.
I can't explain the apparent contradiction. Your answers would be much appreciated. 
 A: The problem here is that your system is not a rigid body. We can think of it as a point particle (the Earth) rotating about another point particle (the Sun). The system is not rigid because the distance between these two point particles changes over time. Therefore, we need to be more careful with how we apply these rotational analogs of Newton's laws.
Typically we have:
$$\tau=\frac{\text dL}{\text dt}$$
and, for constant $I$ this means:
$$\tau=\frac{\text d}{\text dt}\left(I\omega\right)=I\alpha$$
However, as you have pointed out, $I$ is changing, therefore we really have:
$$\tau=\frac{\text d}{\text dt}\left(I\omega\right)=\frac{\text dI}{\text dt}\omega+I\alpha$$
Now I won't go into any specific calculations for elliptical orbits (you can do this if you want), but you can see how if the particle is moving inward then $I$ would be decreasing (negative derivative) and the particle would be speeding up (positive $\alpha$), and if the particle is moving away then $I$ would be increasing (positive derivative) and the particle would be slowing down (negative $\alpha$).
The above qualitative argument shows that the two terms in this equation will have opposite sign, and this shows how it could be that these two terms will always cancel out in the orbit to give a net torque of $0$
