Orb.mec: Is the inclination of an ecliptic-perpendicular orbit 113,4º?

[xpell]

I'm trying to understand this one. Let's imagine we want to launch a satellite to the perpendicular-to-the-plane-of-the-ecliptic orbit ("passing through the North and South ecliptic poles", or perpendicular to the Sun-Earth line, through the Earth). Since the Earth's axial tilt is around 23.4º, would the inclination of this orbit (and the azimuth of the rocket launch) be 113.4º (90+23.4)? (I think so!) Or maybe would it be 66.6º (90-23.4), as I've been suggested elsewhere?

And a couple secondary questions:

1. If it is 113.4º, this is a retrograde orbit, isn't it?
2. Would this be most probably an Earth-centric orbit or an Heliocentric orbit?
3. Have you heard of any real satellite using this orbit?

Sorry for my English, it's not my primary language, hope you'll understand! Thank you in advance for your answers!

 

[mfb]

Inclinations larger than 90° are called retrograde orbits, right.
The direction of the rocket launch is different from the final motion of the satellite, as you always have the rotation of Earth as additional velocity.Assuming you mean an orbit around earth:

Perpendicular to the ecliptic plane is not enough to define the orbit (and with it, its inclination) in a unique way, and the sun-earth line varies within a year. Do you want your orbit to change to keep it Sun-synchronous? That orbit is used by several satellites.

The inclination should be between the two values you calculated, depending on the chosen orbit.If you mean an orbit around sun, the axial tilt of Earth does not matter as the ecliptic is defined in terms of the orbit.

 

[tfr000]

A few things:
You could launch a rocket into an orbit perpendicular to the ecliptic. Its inclination to the ecliptic would be, of course, 90°. Because the Earth spins and revolves around the Sun, the launch angle is always changing. However, once in its orbit, it would more or less maintain a constant angle relative to Earth's equator and poles. The exact angle depends on exactly when and in what direction it was launched. It won't be inclined at (90°-23.4°) to the equator unless you happened to launch it such that its ascending and descending nodes (google them) match the equinoxes.

You could launch a rocket into either a geocentric or a heliocentric orbit like this. Heliocentric requires more energy, in order to escape Earth's gravity.

By the way, you probably have a bad definition for "azimuth" as it applies to this kind of problem.