Tukukkk

I was thinking about how many things are affected by the Earth's rotation, and was wondering how much more difficult a retrograde re-entry would be (where the Earth's rotation direction is assumed prograde). I know about the Coriolis force, which states:

The Coriolis effect is the behavior added by the Coriolis
acceleration. The formula implies that the Coriolis acceleration is
perpendicular both to the direction of the velocity of the moving mass
and to the frame's rotation axis. So in particular:

[...]

• if the velocity is against the direction of local rotation, the acceleration is inward to the axis. (On Earth, this situation occurs
  for a body on the equator moving west, which would deflect downward as
  seen by an observer.)

Meaning that we'd essentially be seeing the Coriolis effect added to the gravity vector (right?). My guess is that the craft would basically accelerate towards Earth at a greater rate as opposed to being "ejected"? I've not really done calculations on exactly how much the Coriolis effect would be in a retrograde re-entry (because I don't really know how to).

Getting back to the point, I'd like the answer to the question to focus on:

• What are adjustments we'd need to make for a retrograde (against rotation) re-entry on Earth?


• If none, would there be a planet in our solar system that would require adjustments?


• Am I misunderstanding any key concepts here?

Things I'd like to ignore:

• In-feasibility of retrograde launches, orbits, or anything else to do with getting there.
  
  
  
  
  • The craft came back on the wrong side of the Earth from a Mars voyage or something.
  
  
  
  

I was wondering if anyone could explain why the re-entry retrograde would be more delta-v, and by how much.

The Earth rotates eastward, carrying the atmosphere with it. For prograde reentry, you're moving "with the wind" so to speak, so the relative speed between you and the atmosphere is lessened; for retrograde reentry you're going into a headwind; the relative speed is increased. Earth's rotation speed at the equator is around 460 m/s eastward, so retrograde reentry adds around 920 m/s to the entry speed.

LEO orbital speed is about 7800 m/s, so for retrograde you're coming in $frac {7800 + 460} {7800 - 460} $ or 12.5% faster than prograde. Kinetic energy is proportional to the speed squared, so you have to dump 27% more energy.

Coming from the moon, where reentry speed is about 11 km/s, your speed would be 9% higher and energy 19% greater if you entered retrograde.

Some reentry-heating considerations go up as higher powers than the square of speed, though I don't fully understand them, and you'd want to take those into consideration.

What are adjustments we'd need to make for a retrograde (against rotation) re-entry on Earth?

You'd need a slightly more robust thermal protection system, primarily, and you'd suffer a little more g-force on re-entry -- the exact amounts would depend on the exact trajectory, and there's a lot of room for tradeoff - less g-force means a longer time being heated; a high-g reentry gives a relatively short heat pulse but higher peak temperatures.

If I remember rightly, the heat shield of the Apollo command module was substantially stronger than was actually needed for lunar reentry, and off the top of my head I would guess that it could survive a retrograde reentry without modification. (I might think about using any remaining propellant in the SPS to decelerate before entry, though!)

I'd assume these numbers are lessened by the density of the atmosphere, so it really wouldn't be as big a deal landing on Mars "the wrong way"?

For Mars, the surface speed is a lot lower (~240 m/s); low Mars orbit speed is ~3360 m/s, so it's 15% faster to enter retrograde from LMO.

For Earth-Mars transit to direct entry, this WP page says Pathfinder hit at 7.3km/s, so ~7% speed difference prograde vs retrograde.

The lower density of the atmosphere doesn't affect how much total energy needs to be dissipated; it just constrains how you do so.