Reading this question: Is GOCE a satellite or aircraft?

I wondered what would happen if GOCE rolled 90° in either direction, so that it's solar panels become parallel to Earth's horizon, and then pitched up 1° relative to its velocity vector, so that it now has a positive non-zero angle of attack.

Let's ignore the sunlight not reaching the photovoltaïc panels anymore, and suppose the ion engine still can produce 20mN of thrust.

Drag would increase since cross-section is now bigger, but there would be one impact lift component, whose direction is pointing away from Earth's center, equivalent to one constant radial-out tiny burn.

Question is: would this configuration allow GOCE to orbit slightly slower than expected? (for similar apogee and perigee)

  • $\begingroup$ I thought the answer to that question said that any lift effects were negligible... $\endgroup$ – GdD May 8 '19 at 14:32
  • $\begingroup$ There is no lift in space, only drag. $\endgroup$ – AtmosphericPrisonEscape May 8 '19 at 16:11

Technically yes, practically: no.

In theory of course you can offset not being at orbital velocity by adding significant lift (all planes do this). And yes you can get that lift by pushing out the back and generating lift from deflecting incoming air. The question is can you do this efficiently (i.e. without burning too much fuel). In theory with a big enough dV and TWR: you can "orbit" at what ever speed you like. You can hover.

So could you glide in, and just re-boost the lost prograde velocity at a substantially lower velocity?

To achieve this with lift, the lift to drag ratio would need to be large. Enough so that the lowering of the orbit from the radial velocity lost is offset by the orbital height gained by burning upwards.

The thing to realise how large is how sensitive to orbital velocity apogee and perigee are. 1 m/s at 100km circular orbit boost apogee by more than 3 km. Continually fighting gravity with a the 'constant radial-out burn' to artificially maintain a significant off-neutral orbit in contrast uses a significant dV over an orbit.

In contrast the faster you are the lift coefficients get smaller roughly 1/4 those at low mach numbers. Specifically shaped planes for this task almost exclusively are not capable of this at low mach numbers. In practice there is no way rolling a satellite over is going to achieve this at mach 17.

That's not to say you can't orbit at a lower speed, but you're going to need the sort of thruster you could just point downwards to do the job.

Edit: Quick look around and: How could a 90 m/s delta-v be enough to commit the space shuttle to landing? Has some answers that help to get an intuition for this.

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  • $\begingroup$ "roughly 1/4 those at low mach numbers" - these approximations are only valid in reasonably dense atmosphere and are due to shock waves and compression happening. There is no such thing if you collide with individual molecules. $\endgroup$ – asdfex May 8 '19 at 17:05
  • $\begingroup$ @asdfex can it be generalized by saying : "individual particles (high knudsen number) having a mass will always be absorbed by the surface it hits, whatever angle of incidence and velocity relative to this surface, and therefore never produce lift"? (where massless photons can be reflected at an angle and can produce lift) so is this behaviour a matter of mass? $\endgroup$ – qq jkztd May 8 '19 at 19:01
  • $\begingroup$ @qq-jkztd not necessarily. Even if all particles are perfectly reflected, one can't use equations derived from fluid dynamics for this situation. Essentially everything boils down to simple momentum transfer between two bodies. $\endgroup$ – asdfex May 8 '19 at 20:27
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    $\begingroup$ What does this mean "In practice there is no way rolling a satellite over is going to cut to at mach 17." $\endgroup$ – Organic Marble May 9 '19 at 0:14
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    $\begingroup$ @OrganicMarble was supposed to say "cut it at" edited for clarity. $\endgroup$ – ANone May 9 '19 at 8:48

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