# Kepler's second law implies angular momentum is constant?

My textbook says that we can infer from Kepler's second law that angular momentum is conserved for a planet, and therefore gravity is a central force.

Now I understand how constant angular momentum implies that gravity is a central force. However, I don't see how we know that angular momentum is conserved, based on Kepler's second law.

My textbook describes Kepler's second law as follows: $$\int_{t_1}^{t^2}rv_\phi\,\mathrm dt=C\int_{t_1}^{t_2}\mathrm dt=C(t_2-t_1),$$ where $C$ is a constant.

We see that $rv_\phi=r^2\dot\phi=C$. We also know that $|\vec{L}|=|\vec{r}\times\vec{p}|=rmv\sin\theta=mr^2\omega\sin\theta.$

Right, so we can assume $m$ is constant, and $r^2\omega$ as well, by Kepler's second law. What about $\theta$ though? How do we know $\theta$ is constant?

For circular orbits, I can see that $\theta=\frac{1}{2}\pi$, but how about elliptical orbits?

EDIT

Okay, I think I got it. We're considering a solid object (planet) rotating about a fixed axis of rotation, so technically, we should be using $\vec L=I\vec\omega$. But I guess we can approximate the moment of inertia for a planet as $mr^2$, considering the spatial dimensions we're working with. And therefore we get $|\vec L|=I|\vec \omega|=mr^2\omega=$ constant. Given that a planet doesn't 'turn' suddenly, we can also assume the direction of $\vec \omega$ being constant.

• $v$ does not equal $r\omega$. There is also a radial component to the velocity. You were correct until the second-to-last step. After that I suggest you draw two components to the veolcity vector. One in direction of position and one perpendicular. The parallel component cancels out in the cross product and the perpendicular component equals $r\omega$(no $sin\theta$ here because this component is exactly perpendicular) . – Abhijeet Melkani Feb 11 '17 at 15:54
• @A.Melkani Ahhh, great! Amazing, thank you very much. So we get: $$|\vec L|=m|\vec r\times\vec v|=m|\vec r\times[\vec v_\phi+\vec v_r]|=m|\vec r\times\vec v_\phi|+m|\vec r\times\vec v_r|=m|\vec r\times\vec v_\phi|=mrv_\theta=mr^2\omega$$ – Sha Vuklia Feb 11 '17 at 15:59
• Yes, and in fact the $C$ itself is $r^2\omega$ as you have rightfully mentioned. So, angular momentum is $m*C$ which is conserved as both these quantities are conserved. – Abhijeet Melkani Feb 11 '17 at 16:32

My textbook describes Kepler's second law as follows: $$\int_{t_1}^{t^2}rv_\phi\,\mathrm dt=C\int_{t_1}^{t_2}\mathrm dt=C(t_2-t_1),$$ where $C$ is a constant.

That alone says that the magnitude of angular momentum is constant.

Your textbook's $v_\phi$ is the component of the velocity vector that is normal to the radial vector: $\vec v = v_r \hat r + v_\phi \hat \phi$. Thus $\vec L = m \vec r \times \vec v = m r v_\phi\,\hat r \times \hat \phi$. Since since $||\hat r \times \hat \phi|| \equiv 1$, the magnitude of a planet's angular momentum vector is $||\vec L|| = m r v_\phi$. Since mass is constant and since $\int_{t_1}^{t_2} r v_\phi dt = C(t_2-t_1)$, the magnitude of the angular momentum vector is constant.

To arrive at the angular momentum vector being constant, we need to know that it's direction is constant as well. This is a consequence of orbits being planar, which is part of Kepler's first law.

The Kepler's second law states that the radius vector from the Sun to the planet sweeps equal areas in equal times. In another words, the rate of change $\frac{dA}{dt}$ is constant. Consider the figure below,

The are element is $dA=\frac{1}{2}r^2d\theta$ so in the time interval $dt$ we have $$\frac{d\theta}{dt}=\frac{2}{r^2}\frac{dA}{dt},$$ On the other hand the angular momentum magnitude (with respect to $O$) is $L=mr^2\dot\theta$. Thus, $$L=2m\frac{dA}{dt},$$ which is constant.

However this does not prove that the vector $\vec L$ is constant. To prove that the vector does not change its direction one has to assumeeither the first Keppler's law (which implies the orbit lies in a plane) or that the force is central (which automatically implies in the angular momentum conservation).
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