# 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.

• For future reference, please use MathJax to format equations. I have edited your question as an example. – Aaron Stevens Jun 25 '19 at 15:57

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 $$0$$