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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.
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  • $\begingroup$ Are you asking about reentry from Earth orbit? Or from interplanetary missions? Almost all Earth orbit missions are launched prograde, reentering retrograde would take far too much delta-v to be practical. $\endgroup$ – Hobbes Dec 31 '18 at 17:59
  • $\begingroup$ @hobbes I was wondering if anyone could explain why the re-entry retrograde would be more delta-v, and by how much. I only know basic reasons why it would likely be bad, but not to what scale. $\endgroup$ – Magic Octopus Urn Dec 31 '18 at 18:02
  • $\begingroup$ @Hobbes Not if it is a satellite already in a retrograde or polar orbit. $\endgroup$ – Muze the good Troll. Dec 31 '18 at 18:16
  • $\begingroup$ To go from a prograde orbit to retrograde reentry you need 16 km/s of delta-V, or twice what you needed to launch from Earth. $\endgroup$ – Hobbes Dec 31 '18 at 19:26
  • $\begingroup$ @hobbes that's true. This was a thought experiment mostly, I realize there's really no reason to launch like that. I kind of want to ask another question about what a retrograde orbit might be good for, probably not much it seems! $\endgroup$ – Magic Octopus Urn Dec 31 '18 at 19:32
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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.

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  • $\begingroup$ Cool! 12% is an impressive difference 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"? Great answer, exactly what I was looking for! $\endgroup$ – Magic Octopus Urn Dec 31 '18 at 18:31
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    $\begingroup$ Added some Mars info. $\endgroup$ – Russell Borogove Dec 31 '18 at 19:50
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I'd like to add to Russell Borogove's spot-on answer. You mentioned Coriolis forces on the retrograde object. The Coriolis "force" is actually a fictitious force required by a rotating frame of reference. Here's an example showing why this force is fictitious.

Imagine your head is at the center of an inertial (non-rotating, non-accelerating) reference frame with a small object, maybe a ball, placed some distance away—for our purposes, assume it's a couple of meters—and motionless in the inertial frame. Ignoring gravitational forces, which will be really small, the ball will just sit there: no applied force, no motion (Isaac Newton nods approvingly).

But now put yourself in rotation around your long axis (head-to-feet axis), with a rotating frame of reference fixed to you. Another of the rotating frame's axes has a direction from the origin (which is inside your head), and is aligned along your normal line of sight, so it emerges from your head directly between your eyes and at time zero continues straight at the ball. In the inertial frame you are rotating but the ball is still just sitting there, motionless, as you rotate. But in your rotating frame, the ball appears to be orbiting around your head with a uniform circular motion; some force (the $m(\Omega\times(\Omega\times R))$ force) appears to be making its path deviate from a straight line and instead describe a circle around your head, i.e. the origin of the rotating frame. It appears centrally directed, like the force due to gravity, except given that you're rotating at a constant rate, its magnitude increases proportionally with distance, not with the inverse square of distance as gravity would.

But if you allow yourself an out-of-body experience, and your astral projection looks at your body and the ball from a vantage point at rest in the inertial frame, you immediately know, "There's no such centrally-directed force. It's just an artifact of my rotation." And that is a perfect qualitative description of Coriolis forces: they are artifacts of the rotating frame of reference.

If you look at the prograde and retrograde entries from an inertial reference frame their trajectories are identical, in terms of speed and flight path angle vs altitude. As Russell Borogove says, the difference is that Earth's atmosphere is moving almost parallel to the prograde one, almost anti-parallel to the retrograde one. Notably, the curvature of the two trajectories is identical at the same altitude. They feel the same gravitational acceleration, and the curvature of their trajectories is identical, at the same altitude. There is no other significant force acting on the two objects. In the inertial frame!

But once you look at this situation in a frame of reference centered on and rotating with Earth, that simple situation becomes more complicated. In this frame, the velocity vector of the object entering on an eastward (prograde) trajectory has a magnitude smaller than its inertial-frame speed. The apparent centrifugal force (if treating this rotating reference frame as if it were inertial) is too small to account for the trajectory's curvature given the gravitational force and the velocity. The $-2(\Omega\times V_{pro})$ artificial force (outward) produces the proper curvature in the rotating frame.

The object entering westward has a velocity magnitude larger than its inertial-frame speed. The apparent centrifugal force is too large to account for the trajectory's observed (in the rotating frame) curvature. The $-2(\Omega\times V_{retro})$ artificial force (inward now, since the direction of $V_{retro}$ is close to opposite that of $V_{pro}$), acting somewhat like increased gravitational acceleration, produces the proper curvature in the rotating frame. Actually, it's not contributing to gravitational acceleration—it's correcting for too much centrifugal acceleration in the rotating frame! And in the prograde case, it's correcting for not enough centrifugal acceleration.

Once you return to viewing the situation from the inertial reference frame, those Coriolis forces go away. They are artifacts of the rotating reference frame.

As Russell says, the two cases have different atmosphere-relative entry speeds. They also have different atmosphere-relative entry flight path angles (EFPAs)! Relative to the atmosphere, they have identical radial (downward) speeds. But the atmosphere-relative tangential speed for the retrograde case is higher than for the prograde case, so its apparent EFPA is shallower than for the prograde case. EFPA is an important parameter for atmospheric entries: it has a large effect on peak heating rates, peak deceleration rates, and other quantities important to the design engineers.

Russell mentioned that some entry conditions are not proportional to velocity squared. One example of that is the radiative component of heating. Heat shields ("Thermal Protection Systems", TPS, in the lingo) are heated primarily by two mechanisms: convection and radiation. Air coming through the shock wave in front of the entering vehicle is compressed and heated tremendously, and when that heated air moves into contact with the TPS it heats the TPS. The air heated at the shock wave also radiates electromagnetic energy, and when that energy impinges on the TPS it heats the TPS. Typically, in a given atmosphere (composition is important!) the higher the velocity, the larger the fraction of total heating is due to radiation. I can't point to any specific references, but I participated in a workshop at NASA Ames Research Center discussing entry conditions at the gas giant planets, and one of the presenters commented that for prograde entries at Saturn the radiative heating rate goes as velocity to the 15th power!

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  • $\begingroup$ I wasnt the one who downvoted this! I love the more indepth view of this same question which includes explanations of my gaps in thinking! Thanks- Im on mobile for new years but will return home soon and get a chance to read it I may have more questions. Thanks Tom. $\endgroup$ – Magic Octopus Urn Jan 1 at 6:25
  • $\begingroup$ @MagicOctopusUrn Thank you for your comment. Your questions and comments over the past several months tell me that you're here to learn, not just to be entertained, and I appreciate that very much. Let me know when you have more questions! $\endgroup$ – Tom Spilker Jan 1 at 18:38
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While @RussellBorogove's answer mentions 12% faster and 25% more energy, and @TomSpilker's answer points out that instantaneous heating is highly relevant separate from the total energy, I think the slide below makes it clear that a little bit more velocity can mean an incredible increasing in the potential heating rate and the necessity to rethink the function of the heat shield.

The title of @KimHolden's question What aspects of reentry heating 'scale as the 8th power'? sets the stage. Increase velocity by 12% and then raise it to the 8th power and you get a factor of 2.5!

But according to the slide below from my unnecessarily down-voted answer the behavior is more like the 5th power below about 10 km/sec, but a whopping 30th power above that! So going from 11-0.46 km/sec (return from the Moon) to 11 + 0.46 km is more than an order of magnitude (factor of 10) increase in heating!

That doesn't mean making the heat shield a little thicker, that means re-engineering the heat shield to block the radiative heating from the plasma much more effectively.

Returning from the Moon going the wrong way won't work, unless you can come up with almost 1 km/s of "extra" delta-v from some where.


In these randomly selected NASA slides I found a plot of heating rate as a function of reentry velocity. I'm pulling it out of the context of a much more complex problem, but it serves the purpose.

To get the value for the power of a power law behavior, you use the ratio of the logs of the ratios:

If $y=x^p$, then

$$p = \frac{\log\left(\frac{y2}{y1}\right)} {\log\left(\frac{x2}{x1}\right)}. $$

After counting pixels, the power of the very steep, thiner red line is about 30, and the power of the upper, ending part of the thick red line at the top drops to about 5.3.

Radiative heating is a complex process, and organic molecules emitted from an ablative heat shield (if one is used) can go a very long way to reduce heat transport to the spacecraf spacecraft due to their very strong absorption of infrared light. But the plasma can get really, really hot, and so in addition to CFD one must do detailed modeling of the plasma as well. You can't just use the familliar $T^4$ power law alone when the properties of the plasma are also temperature dependent and there is a lot of self-absorption.

enter image description here


When Musk says "certain aspects" depend on the 8th power, he may be remembering an equation he once read in one of the rocketry books he has borrowed and read over the years:

enter image description here

See also Quora Did Elon Musk return the books he borrowed from Jim Cantrell? Hold your cursor over the spoiler alert box to reveal the answer. Hint - it's what you think.

No he never did

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    $\begingroup$ I suspect the sudden onset of significant radiative heating (as velocity increases) is due to the per-molecule energy of the heaviest molecules in the incoming flow reaching some threshold in the air mixture, something like an ionization energy or a large rotational- or vibrational-state transition energy. One problem in talking about a power law when the phenomenon being described is a combination of two mechanisms with different power-law radices is that no single-radix power law can accurately describe the actual behavior. The behavior is always the sum of two terms of different radices... $\endgroup$ – Tom Spilker Jan 1 at 21:44
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    $\begingroup$ When the independent variable takes on small values, the term with the smaller radix dominates; at larger values, the term with the larger radix dominates. In between (in the chart you show, where the two curves cross) neither dominate. Looking at that crossover point, if the blue curve has a slope indicating a power-law radix of 2.5, and the red curve has a slope indicating a radix of 15, I calculate the slope of the sum of the two curves as indicating a power-law radix of 8.75, which is of course useless except in the immediate vicinity of the crossover. $\endgroup$ – Tom Spilker Jan 1 at 21:53
  • $\begingroup$ @TomSpilker I like it when I'm wrong and I learn something. In this case though I'm not sure how I got the power law value from a log-lin plot to begin with, a straight line on a log-lin would be an exponential. A quick check before coffee shows 11.5 to 13 km/s results in a factor of 10, so it's ~exp( +v / 0.65 km/s) and not a power law at all. Will have to come back to this one. Learned the word radix today too! $\endgroup$ – uhoh Jan 2 at 2:31

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