# Acceleration due to conservation of angular momentum in absence of a force

I don't understand why the conservation of angular momentum can imply an acceleration, in absence of a force.

Consider for istance planetary motion. The angular momentum $\vec{L}$ of the planets is conserved and that means $\mid \vec{L} \mid=mr^2 \dot{\theta}=mrv_{\theta}$ is conserved too.

Consider the acceleration in polar coordinates $$\left( \ddot r - r\dot\theta^2 \right) \hat{\mathbf r} + \left( r\ddot\theta+ 2\dot r \dot\theta\right) \hat{\boldsymbol{\theta}} \$$

The second term is zero since $\vec{L}$ is constant. This means that there is no acceleration in the direction of $\hat{\boldsymbol{\theta}}$, which is clear since the gravitational force is a central froce.

But if the distance $r$ decreases $v_{\theta}$ (i.e. the velocity in the direction of $\hat{\boldsymbol{\theta}}$) must increase in order to keep $\mid\vec{L} \mid$ constant.

How can $v_{\theta}$ increase if there is no acceleration in the direction of $\hat{\boldsymbol{\theta}}$?

I understood that it happens because of the conservation of angular momentum but if there is an acceleration, necessarily a force is needed. I don't see where do this force come from.

• @JohnRennie The second term is zero if $L$ is constant since it can be rewritten as $a_{\theta}=\frac{1}{r}[\frac{d}{dt}(r^2 \dot{\theta})]=\frac{1}{r}[\frac{d}{dt}(\frac{L}{m})]$ Commented Apr 9, 2016 at 7:51

First draw a circular path. In order for an object to follow that path at a constant speed, there must be a force acting on it towards the centre of the circle. At any instant, that force has no component in the direction of motion, and so, if that's the only force acting on the object, its speed will stay constant.

Now draw a spiral path (i.e. one in which the radius is decreasing but the position of the centre of curvature is constant). Make it spiral inwards. Now draw a tangent to the path at any point to indicate the direction of motion of an object following that path at the instant it's at that point. Consider the situation in which the only force acting on the object acts towards the centre of curvature - draw a line in that direction. Unlike in the case where the radius is constant, the direction of motion is no longer normal to the force - there is a force component in the direction of motion, and it causes the object to increase its speed.

It's not conservation of angular momentum that causes the object to speed up as the radius is decreased (or slow down as it's decreased), it's the force acting on the object. Conservation of angular momentum (which comes from F=ma) is simply a trick used in certain cases because it makes the calculations easier than using F=ma - it's not magic.

I have tried to illustrate the error which has been made about the second term.

Assume that a mass at position $A$ ($\vec r_1$) is under the influence of an attractive central force which originates from point $C$.

The velocity of the mass is $\vec v_1$ with radial and tangential components of velocity $\vec v_{1r}$ and $\vec v_{1\theta}$.

A little later in time the mass has moved to position $B$ and the radial component of velocity has changed to $v_r$.
The tangential component has not changed.

The mass has a new velocity $v_2$
The new position vector of the particle is $r_2$ and so the radial and tangential components of $v_2$ are as shown in green in the diagram below.

Note that $|\vec v_{1\theta}| \ne |\vec v_{2\theta}|$ and so that tangential acceleration term is not zero.

Although not drawn to scale with the increase in the magnitude of the position vector the tangential component of velocity has decreased and the product would stay constant - so the angular momentum is constant.

• Ref. to "The new position vector ... ". To understand, there is needed basic knowledge on how some known "momentum" (p= m x v) "translates" into "force" (f = m x a). To know 2 p's of two bodies does not suffice to calculate their accelaration, you need two positions of one, some delta. Commented Nov 19, 2022 at 11:47