# Why do spacecraft enter the atmosphere violently instead of a smooth spiral?

Why do spaceships almost make a straight line in the atmosphere when coming back to earth? This makes the ship undergo high stress and temperature. Why don't they make a spiral trajectory so that they can slowly brake?

• What you mean with "spiral". Did you mean the so-called "S-shaped banking turns"? Because that was and is actually used with lifting-body designs like Space Shuttle Orbiter, SNC's Dream Chaser, or ESA's IXV that flew on its inaugural test mission today. Please edit to clarify. – TildalWave Feb 11 '15 at 18:57
• Worryingly, neither of the answers so far mention the S-shaped banking turns – Qsigma Feb 11 '15 at 20:44
• Possible dupe of Why is it not possible to deorbit in a shallow glidepath? – Deer Hunter Feb 12 '15 at 3:08
• @DeerHunter I mulled over saying "duplicate," but on reflection, OP is suggesting a different alternative to a (mostly straight) shallow glidepath. – Jerard Puckett Feb 12 '15 at 4:45
• Please describe how a spiral entry could possibly be less violent than common controlled atmospheric entry is in nearly all cases now. – user2338816 Feb 12 '15 at 6:04

Spiraling down in the sense you mean is not possible, the reason is that when a spaceship is orbiting Earth, it is travelling extremely fast relative to the surface, it is not that space is so high up, but that a spaceship needs to travel very fast in order to orbit. So in order to reenter, it is not the velocity of falling that needs to be shed, but the velocity of orbiting. Think of the spaceship skimming very fast over the surface of the Earth but not that high up - that is orbiting. Now think about what it would need to do in order to spiral, it would need to shed all of that orbital velocity - all of it - and start moving in the opposite direction, in other words it would have to have already shed all that orbital velocity, before it could spiral! That is why it is simply not possible to spiral down from orbit.

When they reenter, spaceships enter at a very shallow angle so they travel a long way through the atmosphere, with density gradually increasing. If the angle is too shallow it would skip off back into space, if the angle is too steep it will be destroyed by heat and gforces. So spaceships already enter in a way which is as gentle as possible.

As you can see from the image, in a sense a spaceship does spiral down when it re-enters, but it spirals down around the globe of the Earth. But that is because the Earth is a globe, considered from the frame of reference of the surface of the Earth, the reentry trajectory is rather like a straight line.

• Nice answer, I understand that it is the orbiting velocity that needs to be shed, but what is the problem of overshooting since you will still loose velocity, then you can do it again and again until there is no velocity left. – agemO Feb 11 '15 at 15:46
• @agemO Re-entry cannot be retried over and over again because it takes fuel. Given how fast they are going, it takes a LOT of fuel to retry a second re-entry. – Nelson Feb 11 '15 at 17:23
• @agemO in theory, for certain designs, there may not be a "window": obviously this is accounted for during the design stage. If you overshoot, you risk bouncing off the atmosphere, or being pushed into a re-entry which involves too much deceleration. If you don't have enough control to get into the entry corridor on the first attempt, you sure as hell don't have enough control to mess up the first attempt in such a way as to allow the second to be successful – Jon Story Feb 11 '15 at 17:34
• @agemO, the problem with repeatedly overshooting to gradually shed velocity is that once you reach minimum orbital velocity, there's no longer a "drag too low" region in the re-entry corridor. At this point, you either miss the atmosphere entirely, or are forced to re-enter, and you've still got a great deal of velocity to shed (for Earth, about 7.9 km/s). – Mark Feb 11 '15 at 22:06
• @NelsonChan No. Skipping on atmosphere does not magically add velocity. You energy will be lower when you leave the atmosphere than what it was when you enered the atmosphere - this is where the heat comes from. That energy will mainly come from your orbital velocity. The problem is heat management. – Taemyr Feb 12 '15 at 12:47

There are three major constraints that have to be taken into account:

• maximum deceleration (equipment and structural elements can withstand much higher g's than the crew, so it's about medical limits);
• peak heat flux which allows one to determine the worst case temperatures that the spacecraft structure is heated to during re-entry (heat transfer and dissipation are not instantaneous);
• total heat load - the spacecraft accumulates heat faster than it radiates/convects it away, raising interior temperatures, possibly beyond what the crew and avionics can tolerate. Heat sinks are designed to alleviate this problem.

The total heat load to be absorbed/re-emitted is lower during a normal re-entry than that in a slow "spiral" trajectory.

Please also note that to maintain a spiral trajectory you need some source of ever increasing lift, since the slower you go, the less lift your craft generates. Adding engines and fuel to maintain lift increases initial mass to be put into orbit several orders of magnitude, thus making the whole affair much more costly.

When entering the atmosphere while returning from the Moon (or other planets), it is indeed possible to split the heat load into at least two phases (with some radiative cooling in between), viz. Skip re-entry, however the resulting trajectory is not a spiral.

• Spacecraft can generate lift from aerodynamic forces, it doesn't really need engines or fuel for that. Consider a glider that start from top of a mountain. In a stable flight it will lose X meters of altitude per km flown, without any external source of energy. And in orbital coordinate system glider trajectory will be actually a spiral. – Sergei Ozerov May 20 '20 at 15:09

It's a precarious balance between heating and heat dissipation - it seems that heat dissipation - heat loss - is significantly lower in thin atmosphere, than in much denser atmosphere. Meanwhile, heating from air resistance grows much sooner.

If you "plunge" into deep atmosphere, it can remove produced heat very fast. How fast? Well, the ballpark figure is 0.2 gigawatt during the communicational blackout. This is the energy that is being dissipated as the glowing plasma surrounding the plunging capsule.

Now, if you want to descend slowly, you'll need to mostly depend on radiative cooling. And that takes time. How much time?

Let's be generous, and let our craft dissipate 20 kilowatt of energy during the descent. And let's make it pretty light, a mere 5 tons.

$$E_k = 0.5 \cdot 5000kg \cdot (7800m/s)^2 = 304 GJ = 3.04\cdot10^{11} J$$

This is what we have to dissipate at 20 KW.

$$3\cdot 10^{11}J / t = 20000W$$

$$t= 3\cdot 10^{11} /20000 [s]$$

$$t = 1.5 \cdot 10^7 s$$

That comes up as almost half a year.

Of course over time convective cooling will increase so the heat dissipation will grow and rate of descent could be increased, so the figure wouldn't be that bad, but we're still looking at weeks of descent time if we don't "dunk" the craft into atmosphere dense enough to remove the heat as fast as it's created.

• Can you clarify the source of the 20kW ballpark? – Nathan Tuggy Dec 25 '15 at 20:07
• @NathanTuggy: Definitely a ballpark - a 3kW heater in a closed small room will make it oppressively hot but survivable; now let's surround our craft with a heaters that add up to said 3kW emitting in each of 6 directions, 1/6th directed at the craft (down), the remainder away (back, front, left, right, up) - that would be 18kW out of which 1/6 is directed at the craft. Of course the mechanism of cooling of a small room on Earth is quite different, so a rather large error is possible, but - that's a ballpark of what would be survivable without extreme cooling mechanisms. – SF. Dec 25 '15 at 20:39
• You did not multiply by 0.5 in "Ek=0.5⋅5000kg⋅(7800m/s)2". The result is 1.52E11 J, not 3.04E11 J. So 7.6E6 s or 88 days. Almost a quarter of a year. – Uwe May 17 '20 at 15:54
• @Uwe Thanks, I keep forgetting that 0.5 – SF. May 17 '20 at 16:30

Getting to space is hard and expensive, so naturally spaceship designers try to minimize weights everywhere. This leads to spaceships being mostly in form of capsules using parachutes, rather than winged bodies, as for spacecraft with better aerodynamic forms you need to lift greater weight to orbit - you carry all those wings or even just having higher surface-to-volume ratio for spacecraft.

Capsules have modest aerodynamic abilities, they can't maintain horizontal flight for long, as they lose speed. So they mostly land fast, and that leads to higher deceleration along the way. Winged spacecrafts can lose comparatively little speed - and therefore height of flight - per unit of time, they decelerate gentler, take longer to land and can even jump out of atmosphere, say, to dissipate some heat.

Capsules can also make some similar maneuvers (say, double dip into atmosphere), they're just more limited in that, because they have to lose more horizontal speed to get the same vertical speed. You may, for example, come back from the Moon in a capsule spacecraft, gently skip upper atmosphere so that you just lower the apogee of your orbit, and get back to space until the next perigee. If your goal is to land soon, you don't want to reduce speed too slowly, but theoretically you can.

Since most spacecrafts are capsules, they are limited to high acceleration returns, while more expensive spacecrafts with good aerodynamics can - and do - more gentle, longer, in time and distance covered, returns. Those returns, following curvature of the Earth, can be viewed as spirals.

There are some projects to equip a spacecraft having modest aerodynamics with inflatable heat shields, which would provide much more aerodynamic control during landing, see e.g. https://space.skyrocket.de/doc_sdat/irdt-fregat.htm . There are some promising results, but I'd say those projects haven't left R&D phase yet.