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!)

  • $\begingroup$ Suborbital "kangaroo route" (UK to Australia) is also discussed in Would a point-to-point suborbital spaceflight have a “negative” perigee? $\endgroup$
    – TildalWave
    Commented Apr 2, 2014 at 15:04
  • $\begingroup$ Thanks for pointing to AlanSe's interesting question. I posted my own answer with a diagram of the minimum energy suborbital ellipse. As the angle between cities approaches 180º, the 2nd focus approaches the 1st focus and a approaches r. At 180º the minimum energy ellipse becomes a circular orbit with a=r. Caprik's westward path would be 208º. The semi major axis of this ellipse would be greater than earth's radius and it's energy would be greater than a low earth orbit. It would have a perigee below earth's surface, but I'm not sure "suborbital" is a good label. $\endgroup$
    – HopDavid
    Commented Apr 4, 2014 at 18:18

2 Answers 2


"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.

  • $\begingroup$ Yes! Thank you, that was exactly what I was looking for. Much appreciated. $\endgroup$
    – caprlk
    Commented Apr 10, 2014 at 17:52

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...)

  • $\begingroup$ I see, so to clarify: If I for example launch on the equator, in the opposite direction to rotation and the rotational speed at the equator is 465m/s. To be in orbit at 100km I need to be travelling about 7844m/s, does that mean that for a retrograde orbit I should consider my final "earth relative speed" 8309m/s, with my "inertial speed" as 7844m/s? And if so, at what point (be it altitude or speed) does my speed stop being earth relative and become inertial? (I'm still not 100% if I've even gotten the terminology right here). $\endgroup$
    – caprlk
    Commented Apr 2, 2014 at 18:33
  • $\begingroup$ True orbital speed ignores the speed of the ground below it, it's measured in ECI coordinates, which are fixed even with a rotating Earth. But yes, you have the idea. $\endgroup$
    – PearsonArtPhoto
    Commented Apr 2, 2014 at 18:36

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