How exactly does the inclination and direction (specifically retrograde) of orbit affect the velocity I need to attain orbit?

Question

this is my first question here; and a lot of what I've learnt so far has been pretty much self taught so I thought I best hear from professionals. :)

First let me set the scene as to what I am trying to achieve...

I am trying to calculate the fuel cost for a single stage to orbit spaceplane travelling from London Heathrow to Sydney Australia as part of my dissertation. For the purpose of avoiding supersonic overflight of populated areas (i.e. Europe), it is necessary to depart to the south-west in the climb to orbit. Since I am already facing that direction once I reach 7.84km/s at 100km above sea level, I thought I may as well just orbit in that direction (i.e between retrograde and polar) and follow a great circle that would take me over Sydney.

What I've calculated so far: is the change in net thrust, acceleration, mass, velocity, ect throughout the climb. What I've ended up with is a mass for my propellant and oxidiser used throughout the ascent under rocket thrust. Things like drag have been accounted for. I have assumed the atmosphere is under ISA conditions, with still and dry air. And I'm assuming a simplified circular orbit with a spherical earth.

Now, the point...

What I've NOT done, is take into account the direction I'm launching in. My calculations currently assume there's no spin at all. But I've read that a retrograde orbit is more costly to launch to than a prograde one. Is it a case of nothing more than increased drag? If so, are there altitudes at which this comes into effect? One would think that close to the surface (i.e. on a conventional airliner), the direction of the earths rotation is entirely negligible, so when is it not for spaceflight?

What I am assuming (just off the top of my head), that as acceleration due to gravity decreases in the ascent, the actual velocity of the aircraft in space and the velocity relative to the earths surface begins to differ, with the latter being the higher (if retrograde) and therefore with greater drag, if this is true than I suppose what I am asking is how to work it all out.

I understand that was a lot to read but I really appreciate your help! (Also, if my orbit passed over Sydney, is my inclination just the Latitude of Sydney? Thanks again!)

Answer 1

"Also, if my orbit passed over Sydney, is my inclination just the Latitude of Sydney?"

No, the inclination would be the inclination of the great circle passing through Sidney and London. A low-tech method of finding this great circle is to wrap a band of cardboard around a globe with one edge on the equator. Tape the band closed and you have a band of earth's circumference. Then tilt this band so the great circle edge passes through both Sidney and London.

The northern most or southern most latitude of this great circle is the inclination. When I do this, London is pretty close to the northern most part of the orbit. It's a fair approximation to call this orbit's inclination 52º.

"But I've read that a retrograde orbit is more costly to launch to than a prograde one. Is it a case of nothing more than increased drag?"

No. No matter where you are, you'd need to achieve an ~7.8 km/s wrt earth's center. If you're at the equator, you're already moving close to .5 km/s eastward. So you would only need another 7.3 km/s eastward velocity to achieve orbit (this is assuming you're already above earth's atmosphere). To achieve a 7.8 km/s westward velocity, you'd need to kill your .5 km/s velocity so the total westward burn would be 8.3 km/s.

Speed of a point on earth's surface is cos(latitude) * 2π * earth's radius / (24 hours). At 52º latitude, this comes to about .3 km/s. At London the earth's surface is moving .3 km/s eastward.

To enter a southwestward orbit that passes over Sidney You will need to achieve a ~7.8 km/s westward velocity vector from London. Since you're already moving .3 km/s eastward, your westward burn would have to be 8.1 km/s.

Answer 2

The speed you need to obtain is identical (Or close enough for most purposes) for all orbits. The velocity depends on the inclination of the orbit, basically your velocity vector needs to line up with the inclination.

What you are talking about, however, doesn't have to do with the orbital velocity at all, but rather with the amount of fuel required. The short version of this is, the Earth's rotation is a start velocity for the orbital velocity. Furthermore, the position has some influence, as without special effort, you can't have a lower inclination than your latitude (Actually, there is a few degrees difference by nature, but it isn't a lot...)