Seeking expression for radial velocity of an object in an elliptical orbit (with focus at coordinate origin) as function of radius

Question

Consider a planar, elliptical orbit in a simplified two body, $\frac{K}{r^2}$ central attractive force problem (i.e. assume m1 >> m2 so focus $f1$ is effectively at m1, with m2 at point $p\left(x,y\right)$) and $\rho$ being the radius (green line) from m1 to m2 ($f1$ to $p\left(x,y\right)$), as indicated in the following plot ($f1$ at coordinate origin, periapsis at left blue hash, apoapsis at right blue hash):

I am trying to determine the expression for (what I am referring to as) the radial velocity $v_\rho=\frac{d\rho}{dt}$ along the direction of $\rho$ toward $f1$, at each point $p\left(x,y\right)$ on the orbital ellipse, strictly as a function of $\rho$. To be clear, I do not seek the expression for the velocity tangent to or normal to (red lines in plot) the orbital ellipse at point $p\left(x,y\right)$, but rather only the velocity along $\rho$ toward $f1$.

I have tried to come up with this expression using the very illuminating discussion in this item How do we describe the radial velocity in elliptical orbits?, but with no success yet. I also thought this might be readily found in a classical mechanics text (e.g. Symon), but haven't found (or recognized) such.

In line with How do we describe the radial velocity in elliptical orbits?, I expect the plot of this radial velocity to be structurally similar to the following but with a continuous, finite value - unlike the infinity exhibited in this plot - for the latus rectum at $\rho=4$ (note that $\dot{\rho}=0$ for periapsis at $\rho\approx+2.14359$ in the plot below, which differs from periapsis at $x\approx-2.14359$ in the elliptical orbit plot above):

Any reference to an existing solution, or advice on deriving one, would be greatly appreciated.

I think I have provided enough information to fully characterize the problem, but can certainly provide more info if I've missed something.

Thanks.

Answer 1

In this answer, I use the convention where the ellipse has the periapsis on the right, which is the opposite to your diagram.

The polar equation of the ellipse aligned with the axes, with the origin at the focus and the major axis on the X axis, with semi-major axis $a$ and eccentricity $e$ is

$$r = \frac{p}{1 + e\cos\theta}$$

where $p = a(1-e^2)$ is the semi-latus rectum. The periapsis is at $\theta=0$, the apoapsis is at $\theta=\pi=180°$.

The specific angular momentum is the angular momentum per unit mass. Angular momentum is conserved in an isolated system, so $h$, the magnitude of the specific angular momentum vector, is constant. It can be shown that $$h^2 = \mu p$$ and $$h = r^2\omega$$ where $$\mu = G(m_1 + m_2)$$ is the gravitational parameter and $$\omega = \dot\theta = \frac{d\theta}{dt}$$ is the angular velocity.

Differentiating the equation for $r$ with respect to time, we get

$$\begin{align} \dot r & = p(1+e\cos\theta)^{-2}(e\sin\theta)\omega\\ & = \frac{r^2\omega}{p} \left(e\sin\theta\right)\\ \dot r & = \frac{h}{p} \left(e\sin\theta\right)\\ \end{align}$$

We can now create a parametric plot of $r$ and $\dot r$ using $\theta$ as the parameter. But we can express $\dot r$ in terms of $r$ via the Pythagorean identity, $$cos^2\theta + \sin^2\theta = 1$$ After some algebra, we get $$\dot r^2 = \mu\left(\frac2r - \frac{p}{r^2}-\frac1a\right)$$ Note that the vis-viva equation is $$v^2 = \mu\left(\frac2r - \frac1a\right)$$ where $v$ is the (tangential) speed.

Thus $$\dot r^2 = v^2-\mu\left(\frac{\mu p}{r^2}\right)\\ = v^2-\left(\frac{h}{r}\right)^2$$

Incidentally, $$v^2 = \frac{\mu}{p}\left(1+2e\cos\theta+e^2\right)$$

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Here's a parametric plot for $a=5, e=3/5$, using "natural" units, so that an orbit with $a=1$ has an orbital period of $1$. In our Solar System, that corresponds to solar orbits measured in AU (astronomical units) and years. In such units, where the orbiting body has negligible mass, $\mu = 4\pi^2$.

Here's the corresponding orbit plot.

The coloured dots correspond to equal 15° steps of $\theta$, starting at red and going around the hue cycle, through orange, yellow, green, cyan, blue, magenta, and back to red.

Here's the plotting script, in Sage / Python. The script can also plot $\dot r$ as a function of $r$, but that's currently commented out.

Here's the orbit plotting script. And here's a "combo" script that can do both graphs, on the same plot, if desired.

Answer 2

The polar equation of the ellipse

$$\mathbf r=r(\theta)\begin{bmatrix} \cos(\theta) \\ \sin(\theta) \\ \end{bmatrix}\quad, r(\theta)={\frac {p}{1+e\cos \left( \theta \right) }}$$

thus $~\dot{\mathbf{r}}\cdot\mathbf e_r~$ equal to

$$\dot r(\theta)=\frac{\partial r}{\partial \theta}\,\dot\theta=\frac{\partial r}{\partial \theta}\,\frac{h}{r^2}={\frac {e\sin \left( \theta \right) h}{p}}$$

with: $$p=a\,(1-e^2)\quad,h=\sqrt{\mu\,p}\quad,\mu=G\,(m_1+m_2)\quad\Rightarrow\\ \dot r={\frac {e\sin \left( \theta \right) \sqrt {\mu\,a \left( 1-{e}^{2} \right) }}{a \left( 1-{e}^{2} \right) }} $$