Kepler's second law states that if a planet is revolving around the sun, then its radial vector from the sun sweeps equal areas in equal intervals of time. This is also stated as an equivalent of the law of conservation of angular momentum.

To prove it, we consider a small strip, having area $dA$, and a small angle $d\theta$ , and we relate both of these by$$\dfrac{dA}{dt} = \dfrac{1}{2}r^2\dfrac{d\theta}{dt}$$

This proof is almost used in all sources, and so I am not including a diagram.

Now, we know that the orbit of the planet is an elliptical path. However, further the expression $r\dfrac{d\theta}{dt}$ is replaced by the velocity of the planet $v$.

For angular momentum to be conserved, the net torque relative to the sun on any planet should be zero. If we assume that only sun and planet are interacting, then by the law of gravitation, only a gravitational force acts on the planet, which is directed towards the sun. So, the torque produced at all points is zero, and the angular momentum, which is $$m_{\text {planet}}rv$$ is conserved. This implies that $\dfrac{dA}{dt}$ is a constant, and so both laws are considered to be equivalent to each other.

My question is that how, if the path is not elliptical, can we say that $r\omega$ is the magnitude of the velocity vector at that point? That angular momentum is conserved is immediate, but unless the path is considered to be circular, I can't see why this proof is valid.

My guess, along with reading about this on the internet, says that there is another type of angular momentum, which is different from the rotational one, but I haven't been able to find anything about this. Is this just an assumption to simplify the problem, or is it valid? I certainly know that the equation $$v = r\omega$$

can't be true in a general path.

PS: There is a page on wikipedia talking about the orbital angular momentum used in celestial mechanics, but even its definition is just the normal angular momentum per unit mass. So I don't consider this as a different type of angular momentum.

  • $\begingroup$ Angular momentum $\equiv$ moment of linear momentum. That is $\mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times m \mathbf{v}$. $\endgroup$ Oct 30, 2017 at 18:07

1 Answer 1


If the path is not circular, $v$ given by $v=r\omega$ is actually the tangential component of the velocity, namely $$v_t=r\omega.$$

On the other hand, in general $L=mrv$ is not true. For arbitrary paths we have $$L=|\vec r\times\vec p|=rmv\sin\theta=rmv_t,$$ where $\theta$ is the angle between $\vec r$ and $\vec p$.

Therefore, even for arbitrary paths, $$L=mr^2\omega.$$

There is no other kind of angular momentum. In classical mechanics, it is always $\vec L=\vec r\times\vec p$ for a particle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.