# Perifocal coordinates and the orbit equation

Given the position $(p,q)$ and velocity $(v_p,v_q)$ of a satellite in perifocal coordinates $(\hat{p},\hat{q})$ where $\hat{p}$ is pointing toward periapsis, I can easily calculate the specific angular momentum $h$ with: \begin{equation} h = (p \times v_q) - (q \times v_p) \end{equation} And I can get the eccentricity $e$ with the orbit equation naturally: \begin{equation} e = \frac{\left(\frac{h^2}{μ r} - 1\right) }{\cos(\theta)} \end{equation} where $\mu$ is the graviational parameter of the body being orbited and the radius $r$ and true anomaly $\theta$ was calculated with: \begin{equation} r = \sqrt{p^2 + q^2}, \mathrm{and} \end{equation}

\begin{equation} \theta = \arccos\left(\frac{p}{r}\right). \end{equation} However, I am having trouble calculating the eccentricity directly using the speed $v$ instead of the specific angular momentum.

Using these equations: \begin{eqnarray} h^2 = \mu r (1 + e \cos(\theta)), \\ h = v_\text{perp} r, \\ v_\text{radi} = \frac{\mu}{h} e \sin(\theta), \\ v^2 = v_\text{perp}^2 + v_\text{radi}^2 \end{eqnarray} where $v_\text{perp}$ and $v_\text{radi}$ are the perpendicular and radial speed relative to the position vector from the orbited body, I derived an equation to solve for the eccentricity: \begin{equation} \theta = \frac{\mu}{r} e^2 + \left[\left(\frac{2\mu}{r} - v^2\right) \cos(\theta)\right] e + \left(\frac{\mu}{r} - v^2\right). \end{equation} This is just a quadratic and the solution looks like this: \begin{equation} e = \frac{- \left[\left(\frac{2\mu}{r} - v^2\right) \cos(\theta)\right] \pm \sqrt{\left[\left(\frac{2\mu}{r} - v^2\right) \cos(\theta)\right]^2 - \frac{4\mu}{r}\left(\frac{\mu}{r} - v^2\right)}}{\frac{2\mu}{r}} \end{equation}

This all looked OK to me, but when I tried to compare the first equation (for $h$) with this last equation (for $e$), I find inconsistent results. For example, consider a satellite with these parameters: \begin{eqnarray} (p,q) = (7000, 9000), \\ (v_p,v_q) = (-5, 7). \end{eqnarray} Using the first equation to find $h$ gives: \begin{equation} h = 94000 \end{equation} Now, here I try to calculate $h$ by first calculating $e$ using $v$, $r$ and $\theta$ (in these units, I'll say $\mu = 398600$): \begin{equation} v = \sqrt{v_p^2 + v_q^2} = 8.602, \end{equation} \begin{equation} r = \sqrt{p^2 + q^2} = 11401, \end{equation} \begin{equation} \theta = \arccos\left(\frac{p}{r}\right) = 0.90975. \end{equation} So we have (taking the positive solution to the quadratic equation above): \begin{equation} e = 1.0932, \end{equation} and working back through the orbit equation, I get $h$ again: \begin{equation} h = \sqrt{\mu r (1 + e \cos(\theta))} = 87149. \end{equation} But this is inconsistent with my previously calculated value for $h$ of 94000. I have checked my math several times and feel I must be making some fundamental error though I don't see it.

For reference, I'm trying to reconcile two examples (2.12 and 3.6) found in Curtis' book "Orbital Mechanics for Engineers," 3rd ed.

The example initial conditions do not result in an orbit. The energy is positive and the eccentricity is greater than one. It is a hyperbola.

Also you over-specified the problem by claiming it is in perifocal coordinates, but then providing initial conditions that result in a periapsis that is not on the p-axis.

The computed eccentricity is close, but not correct. $e\approx 1.10768$.

• Overspecified, of course! Thank you. It looks like example 2.12 in Curtis's book is just completely incorrect and/or misleading. The (x,y), (vx,vy) initial conditions above lead to a true anomaly of 32 deg - not the 52 deg I got assuming it was a a consistent perifocal frame. – Johann Jul 4 '15 at 12:33
• Found the example in the book. (It is in the "Look Inside" pages of the book on Amazon!) In fact, the example makes no sense. It is straightforward to compute the eccentricity vector directly from the position and velocity. In perifocal coordinates, the eccentricity vector is by definition $(e,0)$. However for the position and velocity given, the eccentricity vector is $(1.037,0.390)$. The example correctly computes $h$, but gets the true anomaly and therefore $e$ wrong. This error does not show up in the errata for the book. – Mark Adler Jul 4 '15 at 18:33

First of all the eccentricity is a constant, so it should not change with the true anomaly. You could use it as an temporary variable, bit it is not required to calculate $e$.

If you use different definitions for the angular momentum you can find the eccentricity using only the position and velocity at one given time,

$$h=r v_{perp} = \sqrt{\mu a (1-e^2)}.$$

The semi-major axis can be found using the specific orbital energy,

$$a = \frac{\mu r}{2\mu - r v^2}.$$

Combining these two equations yields,

$$e = \sqrt{1 + \frac{r v_{perp}^2}{\mu} \left(\frac{r v^2}{\mu} - 2\right)}.$$

• In this case, the author of the book in question was using the true anomaly and the distance from the body, a function of the true anomaly, to get a constant: the eccentricity. He was using $r\left(1+e\cos\theta\right)=h^2/\mu$. You are doing the same thing, using $r$ and $v$ at some particular instant, which also change in time, to get the constant $e$. – Mark Adler Jul 6 '15 at 4:26