# Calculating orbital properties after an instantaneous burn not at apogee/perigee

Logically the problem of a satellite in orbit around the Earth can be fully defined by the position and velocity of the satellite (assuming a perfectly spherical Earth etc.). However when I try and get from a velocity and radius I can't calculate much more than the energy in the orbit and the semimajor axis via the vis-viva equation. Am I missing some underlying issue with this problem or have I just not found the correct equation?

As Yoda once said about the constants of motion, "There is another." He was referring to angular momentum.

The specific angular momentum vector is the cross product of the position and velocity vectors. That vector is a constant through the entire orbit. With that and the semi-major axis $a$ (you figured that part out already), you can get the eccentricity $e$. Here $\mathcal{M}$ is the specific angular momentum vector and $\mu$ is the $GM$ of the central body. Note that for the cross product, $\overrightarrow{r}$ and $\overrightarrow{v}$ are three-dimensional vectors.

$$\overrightarrow{\mathcal{M}}=\overrightarrow{r}\times\overrightarrow{v}$$

$$\overrightarrow{\mathcal{M}}\cdot\overrightarrow{\mathcal{M}}=\mu a\left(1-e^2\right)$$

If you are working in the plane of the orbit, then the angular momentum vector is in the Z direction, and $\overrightarrow{\mathcal{M}}\cdot\overrightarrow{\mathcal{M}}$ is simply $\left(x v_y - y v_x\right)^2$.

Determining the orientation of the orbit, e.g. the argument of periapsis in two dimensions or the three Euler angles in three dimensions, and the current eccentric, true, or mean anomaly position in the orbit is left as an exercise for the reader.

Let $r_o$, $v_o$, and $\gamma_o$ be the instantaneous position, speed, and flight path angle of a satellite in a Keplarian orbit around a perfecly spherical Earth with gravitational parameter $\mu=GM$. The specific relative angular momentum $$h = rv\cos\gamma = r_ov_o\cos\gamma_o$$ is constant anywhere on the trajectory. This can be used to calculate the orbit parameter (semi-latus rectum), $ℓ=h^2/\mu$ of the orbit equations $$r = \frac{ℓ}{1+e\cos\nu}$$ What remains is to determine $e$, the eccentricity of the orbit.

Specific orbital energy is also constant throughout the orbit so that, at periapsis (subscript $p$) $$v^2_p - v^2_o = 2\mu\left(\frac{1}{r_p}-\frac{1}{r_o}\right)$$ At periapsis, $\gamma=0$ and $r=r_p$ so that, from the specific relative angular momentum relationship, $$v_p=\frac{r_ov_o}{r_p}\cos\gamma_o$$ Substitute this into the preceeding equation and re-arrange to obtain a quadratic equation in $r_p/r_o$: $$(2c_o-1)\left(\frac{r_p}{r_o}\right)^2 -2c_o\left(\frac{r_p}{r_o}\right) +\cos^2\gamma_o = 0$$ where $$c_o = \frac{\mu}{r_ov^2_o}.$$ The two solutions of the quadratic equation are $$r_p,r_a = r_o\frac{c_o\pm\sqrt{c^2_o-(2c_o-1)\cos^2\gamma_o}}{ 2c_o-1},$$ The smallest solution is the periapsis distance, $r_p$, and the larger one is the apoapsis distance, $r_a$.

At periapsis, $\nu=0$ so that the eccentricity of the orbit can be obtained from the orbit equations at that point: $$r_p = \frac{ℓ}{1+e}$$ or $$e = \frac{ℓ}{r_p}-1.$$ If $e=0$, the orbit is circular; if $0\leq e<1$, it is elliptical; if $e=1$, it is parabolic; and if $e>1$, the orbit is hyperbolic.

The flight path angle can be calculated using $$\tan\gamma = \frac{e\sin\nu}{1+e\cos\nu}$$

Hope the sheds more light on your problem.

• Alas, you did not say how to get $\gamma$ at the start, in order to get $h$. – Mark Adler Feb 28 '15 at 18:43

Generally speaking, the problem of a satellite in orbit around the Earth is fully defined as long as you take into account the radius and velocity vectors of the satellite, not just their magnitude.