In orbital mechanics, what does the $ r \dot{\phi}$ represent?

Question

When we write kinetic energy for orbital mechanics, we usually write

$$ K= \frac{1}{2} m \dot{r}^2 + \frac{1}{2}m r^2 \dot{\phi}^2$$

I understand the first term is the tangent velocity along the ellipse but what is the second term? My direct interpretation of it would be the energy in rotating the velocity vector as the particle moves along the path. However, if I recall correctly rotating forces does no work? Or, in other words $$ P = F_{normal} \cdot v = 0 $$ because the normal force to velocity which turns it to allign with path is perpendicular to velocity, wouldn't dot product with it be zero?

Answer 1

If you write a position vector using polar coordinates like $$\vec{u}(t) = r(t) \left( \hat{x}\cos\phi(t) + \hat{y} \sin\phi(t) \right)$$ then the magnitude of the squared velocity is $$ \begin{eqnarray} ||\dot{\vec{u}}||^2 & = & ||\dot{r} \left( \hat{x}\cos\phi + \hat{y} \sin\phi \right) + r \dot{\phi}\left( -\hat{x}\sin\phi + \hat{y} \cos\phi \right)||^2 \\ & = & \left(\dot{r}\cos(\phi)-r\dot{\phi}\sin\phi\right)^2 + \left( \dot{r}\sin\phi + r\dot{\phi}\cos\phi\right)^2 \\ & = & \dot{r}^2+r^2\dot{\phi}^2 \end{eqnarray} $$

Your expression for $K$ is just $\frac{1}{2}m v^2$ with the above expression for $v^2$.

The $\dot{r}$ term represents the motion radially out from the center of the coordinate system, and the $\dot{\phi}$ term is motion in a circle around the center.

Answer 2

It's a consequence of expressing the speed, $v$, in polar coordinates: $$v = \dot{r} \hat{r} + r \dot{\phi} \hat{\phi} \implies v^2 = \dot{r}^2 + r^2 \dot{\phi}^2.$$ If you want to make intuitive sense of what either term means, think of how position is described in polar coordinates: the radial vector denotes how far away from the origin an object is, whereas the polar angle describes its angular orientation about the origin. In the expression for $v^2$, the first term relates to how fast the object is moving out radially -- for instance, in circular motion with constant radius, $\dot{r} = 0$. The second term, on the other hand, describes how fast the $angular$ component is changing. In the same example, if the object has an angular velocity $\dot{\phi} = \omega,$ then $v = r \omega$, which is a familiar relation.

Answer 3

The sketches below should help make this less mysterious.