longitude of ascending/descending node

[darian]

Ok this is something I can't quite figure out. If a person launched a space craft from the Cape with orbit inclination of 90` (polar orbit) then the velocity the space craft would automatically have would be the earths rotational velocity which is ~1000mph. Now it would appear that the LAN would have to never move and the space craft would pass over the Cape on every single orbit because the LAN would also be 1000mph along with the earths rotation.

Now if they launched at the inc of say 80 then the LAN would travel at 1000+ 17,500(cos(80)) right ?

However this is not the case because the LAN is always moving backwards towards the west of the Cape :huh: A little help here.

 

[Urwumpe]

The LAN is measured at the reference time of the orbit - in Orbiters case of using "J2000", 1.1.2000, 12:00.

This way, the LAN and the Inclination give you a constant orientation of the orbit plane.

 

[tblaxland]

To make this simple we will ignore any effects of Earth's non-spherical gravity, any lunar/planetary body perturbations and assume that your mass is insignificant compared to the mass of the Earth.

Once you leave the surface/atmosphere of the Earth the only force that affects your trajectory is Earth's gravity which always points at the centre of the Earth and is therefore always in the same plane as your orbit. This means that the longitude of your ascending node will remain stationary. If you look at MapMFD it looks like the LAN is moving westward because the Earth is rotating underneath you. The only way your LAN will move is if there is a force perpendicular to the plane of your orbit.

One point about launching to a 90deg inclination polar orbit. You actually have to point slightly west of north/south to get your inc to 90deg so you cancel out the velocity you had sitting on the surface. If you launch directly north/south you will end up with an inclination slightly less than 90deg.

Hope this helps.

 

[darian]

So what happens to the ~1000mph equatorial velocity then, it just fades away.

 

[Urwumpe]

So what happens to the ~1000mph equatorial velocity then, it just fades away.

No, it is just already included in your inertial velocity, when you launch from Earths surface. In space, this velocity is of course uninteresting. Earth just rotates below your orbit, and the rotation of it has no impact on you.

 

[Zatnikitelman]

Sort of, like tblaxland said, you have to cancel it out by pointing slightly westward. Once you're free of Earth's surface canceling out the lateral velocity is easy, just point a few degrees west, then continue with the orbit. That's why vessels usually launch eastward unless a special orbit calls for it. You get about 1000 MPH added at the equator and I think about 900 MPH added at KSC's inclination.

P.S. Woot! 300th post!

 

[darian]

Okay I got it now, thanks.

 

[ar81]

Orbit will not rotate with Earth after you are in space, therefore you would not pass over the cape every single time.

 

[tomek]

In short, 90° inclination already implies you canceled out the velocity you got from Earth's rotation. You can't have exactly polar orbit if you have any velocity component along the equator.

 

[darian]

Orbit will not rotate with Earth after you are in space, therefore you would not pass over the cape every single time.

Ahhh I see I guess I wanted to know at which point during Orbit insertion would the velocity be neglegent. Right after Orbit insertion the velocity would be neglegent or perhaps when you are nearing this velocity. Looks like a Calculus equation.

In short, 90° inclination already implies you canceled out the velocity you got from Earth's rotation. You can't have exactly polar orbit if you have any velocity component along the equator.

Ok I see this too, thanks