# Is it possible to represent an orbit with just a periapsis position and 2 angular velocities?

I was thinking of how complex the orbital elements are and got to wondering whether you can represent orbits in a more simple way. If you start with an ECI coordinate frame and you know the spherical position of the periapsis. You could then represent the orbit with a latitudinal and longitudinal angular speed at that periapsis.

For example:

A roughly circular Polar orbit could have a periapsis at 0 deg,0 deg 6700km (right over the north pole) and a longitudinal angular speed of 0.065 deg/s, and a latitudinal angular speed of 0 deg/s.

An Equatorial circular orbit could be ((0 deg, 90 deg, 6700km), 0 deg/s, 0.065 deg /s)

To make the orbit more eccentric you would increase one or both of the angular speeds. By using a combination of the two angular speeds you control the inclination of the orbit:

A 45 degree inclined orbit could be ((0 deg, 90 deg, 6700km), .0463 deg/s, .0463 deg /s)

The length of the vector (.0463,.0463) is .065 so you would have the same circular orbit as before even though each angular speed is less.

A retrograde (westward) orbit would have a negative latitudinal angular speed (the longitudinal angular speed is always positive.)

The reason this should work is because at the periapsis the radial velocity is always zero. So you can eliminate 1 element. Is there any reason why this would not work?

• I don't think you can define a longitudinal angular speed when you are exactly over the pole, it doesn't make mathematical sense. If you specify longitude and and the two lat/lon components of velocity when crossing the equator AND an absolute time (i.e. four numbers) it's easy to convince yourself that you could specify an orbit, but step 1 would be to go find an ephemeris for the rotation of the Earth to figure out Earth's rotational orientation at that moment. – uhoh Aug 1 at 3:45
• So I think the answer to your current question is no, there is not enough information in three (or even four) numbers to completely define a Kepler orbit unless you also depend on external ephemerides. – uhoh Aug 1 at 3:45
• @uhoh But the 6 classical elements don't take into account earths rotation either. You are right about the poles being a problem. I just ran a simulation on my computer. The longitudinal angular speed is not affected by being at the poles but the latitudinal angular speed is always 0 when the orbit passes over the poles. Which is a small problem. I don't think the idea is completely debunked because It still works for orbits that don't pass exactly over the pole. – Alex Aug 1 at 4:49
• They don't have to take it into account because they are in an inertial frame. One of the problems with this technique is that it uses a time-dependent rotating coordinate system. With Kepler, longitude of ascending node is pinned to a fixed reference direction and reference plane, which are inertial. But let's wait and see what more authoritative answers say; I'm no expert. – uhoh Aug 1 at 5:05
• @uhoh I'm pretty sure that is pinned to a galactic reference direction (an imaginary line passing through the constellation Ares). Not a direction the earth is pointing. To figure out if an orbit passes over a location on earth with Kepler elements you still need a date and time. I'm also no expert so I am excited to see what one thinks about this. – Alex Aug 1 at 5:43