20
$\begingroup$

The fiery re-entry of spacecraft has been a staple of spaceflight since the beginning, making ablative heat shielding a necessary component of any craft wishing to return to Earth intact. This is the result of the shockwave caused by the high-speed encounter with the upper atmosphere, in turn dictated by the angle of re-entry (~40 deg for the Space Shuttle). The literature on this topic always says that too shallow of an angle would result in the craft "skipping" off of the atmosphere. This seems to imply that the atmosphere has a "surface" similar to the ocean, which is certainly not intuitive for a gaseous fluid.

So my question is, why is it impossible to gradually, gently deorbit using a shallow glidepath? Is there no combination of aerodynamic surface & thrust profile that could make this happen?

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ If you're coming from a very high elliptical orbit, I think it's pretty clear that skipping is really just a matter of continuing on an orbit. However, that doesn't explain your shuttle fact. Is this 40 deg angle really for the angle at which it goes through that atmosphere, or is it the orientation angle of the craft? In that latter case, it wouldn't really have anything to do with skipping - just its aerodynamics. I don't see why getting lift should be a problem. $\endgroup$
    – AlanSE
    Commented Feb 20, 2014 at 15:46
  • $\begingroup$ I believe the two angles are almost the same, with the nose a little higher, but I'm getting this info from playing the (highly addictive) Space Shuttle Simulator SSM2007, not from real data, so I could very well be mistaken. $\endgroup$
    – Jerard Puckett
    Commented Feb 20, 2014 at 15:59
  • 1
    $\begingroup$ See my answer to a similar question. $\endgroup$
    – Mark Adler
    Commented Feb 21, 2014 at 16:41
  • 1
    $\begingroup$ This XKCD What if? answer may be relevant: what-if.xkcd.com/58 $\endgroup$
    – David Conrad
    Commented Feb 21, 2014 at 17:59
  • 1
    $\begingroup$ NASA has made very interesting educational movies detailing the requirements and actual phases of de-orbiting Apollo spacecraft: Apollo - Atmospheric Entry Phase (1968). Among many explanations with respect to entry angle, it reveals why a cone-shaped object can have lift. Some of these explanations provide answers to parts of your questions. $\endgroup$
    – Jens
    Commented Feb 22, 2015 at 12:07

4 Answers 4

Reset to default
9
$\begingroup$

There was a great little space simulator originally designed for simulating Orbital Tethers (the space tethers applet) that was expanded to do a variety of space based simulations, with readouts of various numbers relevant to the simulation:- distance, speed, temperature etc.

Running that software using a variety of the inbuilt scenarios, it was clear that a shallow re-entry, while lowering the maximum momentary heat the craft experiences, prolongs the duration of the heating phase significantly. As a result of that, a shallow re-entry will cause a craft to experience a greater overall heat load than a steep one.

Improve this answer
$\endgroup$
0
17
$\begingroup$

When you reenter at a shallow angle your velocity is almost horizontal and you have orbital velocity. A small amount of lift will lift your velocity above the vertical and send you back into orbit. This is what is meant by skipping off the atmosphere. As long as you have enough oxygen, you will survive this. After a few skips (losing velocity each time due to drag) you will be under orbital velocity, so if you are going to reenter shallow you need to generate lift. At hypersonic velocity the lift to drag ratio tends not to be very good, so generating that lift will generate a lot of drag, slowing you even more. Then you drop lower into the atmosphere because you aren't generating enough lift, generate more heat, and burn up. Theoretically if you had thrust at your disposal you could stay high while you lose the orbital velocity. The amount of fuel required would be enormous, though.

Improve this answer
$\endgroup$
8
  • $\begingroup$ Upvoted. It would be ideal to add a diagram with Mach number vs. L/D for an Apollo-type capsule. $\endgroup$
    – Deer Hunter
    Commented Feb 20, 2014 at 18:53
  • $\begingroup$ @DeerHunter: I think the idea behind the question is not for capsules, but for things that look like aircraft. I don't have a nice L/D plot for hypersonic aircraft. $\endgroup$
    – Ross Millikan
    Commented Feb 20, 2014 at 18:55
  • $\begingroup$ And the reason this is not done would be .... what? Repeated heating? Difficulty in predicting or controlling where you actually land? Historical inertia? Something else entirely? $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Feb 20, 2014 at 22:31
  • 1
    $\begingroup$ Sorry, I didn't mean why don't they maintain height with thrust to slow down, but "Why don't they plan on a few skips to spread the heating over more time?" $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Feb 20, 2014 at 22:34
  • 4
    $\begingroup$ Um, no, that's completely wrong. You don't need any lift up to "skip". "Skip" is a horrible term for this, and has always caused this confusion about "surfaces" and lift due to the familiarity with a stone skipping off a pond, which this is not like at all. You need a steep enough entry angle to not re-exit the atmosphere, for a completely ballistic entry with no lift, because the planet and the atmosphere is curved. If your drag is not high enough, then your descent rate will be slower than the planet curving away from you, and you will re-exit. $\endgroup$
    – Mark Adler
    Commented Feb 21, 2014 at 16:36
9
$\begingroup$

You actually can deorbit using a shallow glidepath that "skips" off the atmosphere, slowing down each time, called a "skip trajectory." This helps to dissipate the heat the craft experiences over a longer time frame, and allows for a greater range of landing sites.

But why does a craft skip when there is no well defined surface to the atmosphere? It happens because of the lift-to-drag ratio of the craft. Most craft have angled surfaces, and as it deorbits into thicker parts of the atmosphere, they will create more lift. If the object reentering the atmosphere is a sphere, there will be almost no lift, but it will decelerate too quickly (for humans, anyway). Having flat surfaces is important for a softer deceleration, which creates a higher lift-to-drag ratio, and the potential to "fly back out" of the atmosphere.

Check out this website for more information: http://www.aerospaceweb.org/question/spacecraft/q0218.shtml

(sorry I don't have specific examples, my physics background is not very strong)

Improve this answer
$\endgroup$
3
  • $\begingroup$ No, it has nothing to do with lift. See comment on other answer and comment on question pointing to an answer. $\endgroup$
    – Mark Adler
    Commented Feb 22, 2014 at 3:53
  • $\begingroup$ Hmm, gotcha. I understand what you are saying about the atmosphere being curved. I thought that lift and drag went hand-in-hand. $\endgroup$
    – Stu
    Commented Feb 24, 2014 at 13:38
  • $\begingroup$ You can have drag and no lift. I.e. L/D = 0. For example, the Mars Pathfinder and Mars Exploration Rover entry vehicles had zero lift. In fact, they entered spinning at 2 rpm in order to average out any inadvertent residual lift to assure it was zero. They would still quite easily "skip" out if entered at too shallow of a flight path angle. $\endgroup$
    – Mark Adler
    Commented Feb 24, 2014 at 20:01
3
$\begingroup$

It actually is possible (in a sense). In such cases, the trajectory is a slow spiral downward, but since the majority of the velocity will be lateral, the craft will tend to slow down faster than falling, therefore making the trajectory progressively steeper. When this happens, the lifting effects of the other answers apply.

This actually happens frequently, just usually not on purpose. The ISS, for example, is very slowly deorbiting (due to air resistance). This is why it has had to be boosted several times to a higher orbit.

Also note that this needs to happen very gradually. If you're in a circular orbit and then burn retrograde, you'll be in an elliptical orbit, which will result in far and close passes to the planet (which could be mistaken for skipping off its atmosphere).

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ There is huge difference between the ISS losing altitude and spacecraft reentry. Very rarefied monoatomic oxygen vs. dense air. Without considering heating, your answer is, I'm afraid, meaningless. $\endgroup$
    – Deer Hunter
    Commented Feb 20, 2014 at 22:08
  • 2
    $\begingroup$ @DeerHunter the original question asks "So my question is, why is it impossible to gradually, gently deorbit using a shallow glidepath?" I'm answering exactly what is being asked. Furthermore, the ISS losing altitude is a form of reentry/deorbiting. The fact that appreciable heating is not generated is irrelevant. Plus, that example was showing how (complete) reentry is prevented. $\endgroup$
    – geometrian
    Commented Feb 20, 2014 at 22:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.