TITLE: How is angular momentum conserved for the orbiting body if the centripetal force disappears?
QUESTION [13 upvotes]: When the centripetal force on an orbiting body disappears (e.g. if it the body is a ball and the force was exerted by a string and the string rips, or, more unrealistically, if the body is the earth and the sun suddenly disappeared), the body continues in linear motion. How is angular momentum conserved here?

REPLY [34 votes]: If the Sun were to magically disappear then the Earth would fly off at a tangent to its orbit. The trajectory would look like this:

The green dot shows the position of the Earth at the instant the Sun disappears. The distance from the Sun, $d$, is the Earth's orbital distance and the velocity $v$ is the Earth's orbital velocity.
When the Sun disappears the Earth heads off in a straight line at constant velocity as shown by the horizontal dashed line, so after some time $t$ it has moved a distance $x = vt$ as I've marked on the diagram. The question is now how the angular momentum can be conserved.
The answer is that angular momentum is given by the vector equation:
$$ \mathbf L = \mathbf r \times m\mathbf v $$
where $\mathbf r$ is the position vector, $\mathbf v$ is the velocity vector and $\times$ is the cross product. We are going to end up with the vector $\mathbf L$ pointing out of the page and the magnitude of $L$ is given by:
$$ |\mathbf L| = m\,|\mathbf r|\,|\mathbf v|\,\sin\theta \tag{1} $$
but looking at our diagram we see that:
$$ \sin\theta = \frac{d}{|\mathbf r|} $$
and if we substitute this into our equation (1) for the angular momentum we get:
$$ |\mathbf L| = m\,|\mathbf r|\,|\mathbf v|\,\frac{d}{|\mathbf r|} = m\,|\mathbf v|\,d \tag{2} $$
And this equation tells us that the angular momentum is constant i.e. it depends only on the constant velocity $\mathbf v$ and the original orbital distance $d$.
Although it initially seems odd an object doesn't have to be moving in a circle to have a constant angular momentum. In fact for any system the angular momentum is always constant unless some external torque is acting on the system.

REPLY [2 votes]: Angular momentum is measured relative to some origin. So the angular momentum of a system is different for different reference frames. However for a closed system in a given reference frame the total angular momentum of the system is constant over time. 
If we set our origin at the center of rotation, the angular momentum of the orbiting body is 
$$L=mr_0v$$
where $r_0$ is the distance between the body and the center of rotation.
After the force disappears, the body continues to move along a straight line tangetially to its former orbit. It's trajectory can be given by 
$$\vec{r}(t)=\begin{pmatrix} -vt \\ r_0  \end{pmatrix} $$
and the angular momentum is 
$$L=m \;\vec{r}\times \vec{v}=m\begin{pmatrix} -vt \\ r_0  \end{pmatrix}\times \begin{pmatrix} -v \\ 0  \end{pmatrix}=mr_0v$$
As you can see, angular momentum is conserved.
However by doing this we are treating the centripetal force on the body as an external force (we are neglecting the reaction force from newtons third law on another object). Since angular momentum is only conserved for closed systems, angular momentum is not conserved in general in this approximation. This approximation only works because we have chosen our origin such that the position vector $\vec{r}$ is always parallel to the force $\vec{F}$ on the body and therefore the force doesn't produce torque $\vec{r}\times\vec{F}$, such that angular momentum is again conserved. If we would choose any other origin, it wouldn't work.
If we don't make such an assumption and calculate the angular momentum of two objects which orbit each other in a closed system (such as earth and the sun), we can choose any origin we want and angular momentum is always conserved.
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