# notation to describe an elliptical orbit? (3000km x 40000km)

For a project a professor assigned, he described an elliptical orbit by using the notation: (3000km x 40000km). Ive never seen this notation before and couldn't find much help online. I just assumed that... rp = 3000km & ra = 40000km, or am I wrong? Thanks!

That's a common convention. The smaller number is the periapsis altitude, the larger is the apoapsis altitude, normally both given as height above a planetary surface rather than radius from barycenter.

• Important to note that this notation is just as likely to indicate radii as altitudes, so it's should be stated explicitly. In this example, the perigee value is less than the radius of the Earth, so it's (probably) altitude values as you said. Apr 25 '16 at 17:11
• @Chris or a very short-lived orbit... :-) Apr 25 '16 at 19:50

Edit -- I was wrong. I've asked around. John Schilling and Henry Spencer tell me altitude of apoapsis and periapsis is what's usually used to describe elliptical orbits about a planet. These guys know a lot more about space than I do. Please disregard what I said earlier.

I would chastise your prof for ambiguous notation. Most instances I've seen the first number is periapsis distance from body center and second number is apoapsis distance to body center.

But it could also be periapsis and apoapsis altitudes, the distances from body surface rather than center.

If the 3000 km x 40000 km ellipse is an earth orbit, it would seem to be using altitude -- since a 3000 km distance from body center would put the periapsis below earth's surface.

• Is the altitude vs. radius assumption possibly dependent on whether the context is spacecraft versus natural bodies? I feel like I've almost always seen this notation describing altitudes. Apr 25 '16 at 19:38
• @RussellBorogove But if the 3000 km x 40000 km ellipse is taken to mean altitude above a body's surface, then even if we ignore inclination that does not by itself fully describe the orbit; you need additional context to figure out what's going on. Bottom line, it seems potentially ambiguous either way. Depending on the specific assignment (which the OP didn't describe in the question) this could be a major difference. I think HopDavid's point holds.
– user
Apr 26 '16 at 12:31
• @RussellBorogove you may be right. I tried to find examples of (6678 x --) ellipses since 6678 would be a common ellipse with perigee in low earth orbit. Most of the hits are my own posts in NasaSpaceflight forum. Maybe I have been abusing an established convention by using radii instead of altitudes. In a lot of ways using radii is more convenient. It's more straightforward to calculate eccentricity, for example. Apr 26 '16 at 17:18

It is common to describe an orbit using apoapsis and periapsis radii. One reason is that those two parameters are easy to visualize, and provide substantial information about the orbit's shape and energy.

We can, for example, immediately calculate the semimajor axis:

$$a = \frac{r_a + r_p}{2}$$

and the orbital eccentricity:

$$e = \frac{r_a - r_p}{r_a + r_p}.$$

This will give us a fairly accurate idea about the orbit's shape regardless of the body it is orbiting. But we often know the body and its gravitational parameter, $\mu$. With this information we can calculate the orbital energy:

$$\mathcal{E}=-\frac{\mu}{2a},$$

which lets us estimate the orbital velocity at any point in the orbit, because we know that

$$\mathcal{E} = \frac{v^2}{2}-\frac{\mu}{r}$$

and we can obtain $r$ from the orbit's equation:

$$r = \frac{p}{1 + e\cos\nu},$$ where $$p = a(1-e^2)$$ and $v\in[0^{o}, 360^{o})$. At once, we can also calculate the orbital period:

$$\mathcal{P} = 2\pi\sqrt{\frac{a^3}{\mu}},$$

and the specific angular momentum

$$h^2 = \mu p.$$

In my experience, most people use altitude at periapsis and apoapsis (as opposed to radius). To be sure, it is best to always specify, e.g., this is a 40-by-400 kilometer altitude orbit.

Once you have described the kinematic properties of the orbit, you can move on to discuss the orientation properties, where you would need to know the inclination, nodes, and even current orbital location.

• Not sure if you are interested, but here is a harder orbital math question.
– uhoh
Mar 15 '17 at 5:12
• @uhoh I added an answer - thanks for pointing out that interesting question. Mar 17 '17 at 17:49