The usual approach to reentry is fast and hot. There's a lot of energy to be lost, and doing it quickly has some advantages: You can dump energy into hypersonic air, and then leave that heat behind so that the hot part is over before the inside of the craft heats up.

As an alternative, what would it take to "fly" a slow reentry?

A typical LEO satellite, for example, is currently doing a stupendously slow reentry, dropping a few km at a time due to drag without significant heating. Eventually, however, when it hits denser atmosphere, it'll start losing energy and altitude fast, and convert to the fast & hot mode, usually burning up.

To avoid that, can a returning spacecraft could "fly" at the very top of the atmosphere for a long time? The craft would only slowly lose speed so keeping thermal issues small, without falling to lower altitude & denser air until the speed is far below orbital speed and normal flight can take over.

Normally, gliding flight is "balance lift and drag to fly at a descent rate that keeps speed constant". This would be more of a "balance lift and drag to fly at a slowly decreasing speed while maintaining altitude"

Is there any research available on what such a profile might look like, and what lift and drag performance would be needed to pull it off? (Even if we can't build such a craft, now or perhaps even ever)

Some related questions:

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    $\begingroup$ It's my understanding that if you have enough speed to get lift ≅ weight at high altitude, you're going to get significant heating. $\endgroup$ Commented Jul 4, 2018 at 20:59
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    $\begingroup$ So if you want to stay up as you slow down and the centrifugal force component drops off you have to increase lift somehow....and the easiest way to do that is to lose altitude. Which I think fights against the "stay high for a long time" goal. Good paper here: emits.sso.esa.int/emits-doc/ESTEC/… $\endgroup$ Commented Jul 4, 2018 at 21:18
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    $\begingroup$ @OrganicMarble " to increase lift somehow....and the easiest way to do that is to lose altitude": or use speed energy to overcome drag instead of gravitational. The F14 supersonic double-engine-out procedure was to start a 1k fpm climb at decreasing speed & work the problem, only transitioning to a more-traditional "max range" or "max duration" glide constant-speed glide once no longer supersonic. I was told that best managed total energy. $\endgroup$ Commented Jul 4, 2018 at 22:14
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    $\begingroup$ In the case of the Space Shuttle, the question is, "How do you deal with over 137 trillion joules?" You're "simply" moving energy around. If you're avoiding the engineering questions regarding materials, risk, etc., you can very easily plot a "Watts vs reentry time" curve. $\endgroup$ Commented Jul 4, 2018 at 22:16
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    $\begingroup$ I've done this in Kerbal once, it was a 5 stage rocket, the first 2 stages to reach the edge of the atmosphere and negate horizontal momentum, the last 3 to descend without a parachute. It's absolutely brutal to fight gravity for a lengthy period of time. I reentered the atmosphere going 0m/s and never sped up beyond 100m/s. I was completely starved of fuel by around ~10km up and had to bailout hah. $\endgroup$ Commented Jul 5, 2018 at 12:23

3 Answers 3


Not without propellant. Or at least not easily. However the "why" is a bit tricky.

This is far from the whole story but one of the problems is that generating enough lift to keep you from diving too fast into the atmosphere is tricky.

  • If you're going near orbital velocity you can scrub speed quite well as you don't need to generate much lift.

  • At low speed, you can survive going through the lower atmosphere so you can generate the lift aerodynamically

However, in-between is where things get difficult. You have to able to generate lift and deal with going fast. For example going say 70% of orbital-velocity you'd need to generate about half as much lift as you would at low speed. This is the bit that's hard to do without generating heat. The aerodynamics is a little tricky however the gist of the problem is that (drag -> heat) as the energy needs dissipating, and its hard to get a high enough lift-to-drag ratio to get the necessary lift, without turning too much of the air's energy into heat.

The first problem is that lift-to-drag ratios always get lower at high velocity. http://www.aerospaceweb.org/design/waverider/design.shtml has a bit on it but the important bit for us is lift to drag scales very roughly as $(M + 3)/M$, where $M$ is the mach number.

More significantly, the shape you need to keep the hot shockwave off the surface (a blunt curve) is not conducive to generating significant lift, but is very good at generating drag. There doesn't seem to be a way of doing this that either doesn't generate significant heat or keeps the heat off the body.

It might be possible to do better in future with shape altering materials that can always be 'optimal' but you can't generate significant lift from mach 20 airflow, without generating a lot of heat. Currently the plan is to get this bit over-with as fast as possible, not to prolong it, because peak temperature is less significant than how long you're exposed to 5000ish degrees.

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    $\begingroup$ "5000ish degrees" is that °C or °F? $\endgroup$
    – Calmarius
    Commented Jun 29, 2020 at 13:16
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    $\begingroup$ that would be F... even direct reentry from the Moon does not hit 5000C, not without also incurring triple-digit deceleration g-forces. "normal" reentry is about 3000K-3200K (2700C-2950~ish C), which is already hot enough to destroy any material if allowed to soak in over extended time. $\endgroup$ Commented Jun 25, 2021 at 8:17

It is possible to do a slow and cool reentry. But it would be extreamly expensive.

Just slow down from orbit speed to subsonic speed above the atmosphere and then fight gravity until it is possible to deploy parachutes. All that using rockets thrust to decelerate. No heat shield necessary. But you need a very large rocket with at least two stages in orbit attatched to the spacecraft. About that size necessary to lift the spacecraft from ground to orbit.

But you would need multiple rockets of Saturn V size to lift the spacecraft and the rocket for return into orbit.

It is much cheaper to use the atmosphere for braking using a heat shield. Entering the atmosphere with hypersonic speed will transform energy from speed into heat. Much slower airplanes faster than sonic heat up too. You can't get atmospheric drag to slow down and lift to keep height without the heat caused by compression of air. Thats physics.

No manned orbital missions like Soyuz, Mercury, Gemini and Shuttle would have been possible using this slow and cool reentry.

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    $\begingroup$ xkcd worked this out in a What-If article a while ago. He basically says the same thing as this answer, talking about the tyranny of the rocket equation and the high velocities involved. Check out the article if you want a more in-depth (and humorous) explanation. $\endgroup$
    – Cody
    Commented Jul 5, 2018 at 20:37
  • $\begingroup$ Good answer. I'm pondering the difference in heat between the heat of compression vs. the heat of descending through one's rocket exhaust. Cooler, but "cool"? Not sure. $\endgroup$ Commented Jul 5, 2018 at 20:55
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    $\begingroup$ Above the atmosphere, the rocket exhaust will spread so fast that the resulting heat load will be very small. $\endgroup$
    – Uwe
    Commented Jul 8, 2018 at 15:07
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    $\begingroup$ More exactly, you need a launch rocket that is proportional to the size of the return rocket at the same proportion the return rocket is to the craft itself. For some figures, if you have a 1 megagram craft and 150 megagram launch rocket, then to turn that latter rocket into a slow-return rocket means you now have to have a $\left(\frac{150}{1}\right)^2$, i.e. 150x the 150x craft mass, or 22 500 megagram launch rocket for this slow-return spacecraft. That's roughly 7 times larger than the Saturn V (2970 Mg). $\endgroup$ Commented May 4, 2019 at 23:13
  • $\begingroup$ Hence, far cheaper to just design the craft to either be expendable (satellites) or able to take a hot and fast return (all returning craft to date). $\endgroup$ Commented May 4, 2019 at 23:14

Let's try to work out what a slow reentry would be.

A spacecraft on orbit balances the Earth's gravity with the centrifugal force in the oribital reference frame to keep altitude. This orbital lift is proportional to $v^2/r$, where $v$ is the spacecraft's speed r is the radius of the orbit.

Once the spacecraft slows down below the orbital speed it will need some additional lift to maintain altitude and balance it with Earth's gravity, so:

$$ g = \frac{v^2}{r} + L $$

Where $L$ is the lift needed. And $g$ is the standard gravity. The best lift-to-drag ratio you can get based on what I found is:

$$ \frac{4(M+3)}{M} $$

Where $M$ is the mach number. or to turn this into metric in meters per second it is approximately:

$$ \frac{4(v+1000)}{v} $$

Speed of sound is temperature dependent but so for simplicity I assumed room temperature. In real life it would vary as temperature of air changes during descent.

This is an empirical maximum, Concorde had the lift to drag ratio of 7 at 2 Mach. So really that factor is more like 3 than 4 if we consider all design compromises. So this means we can get a lift of $3k(v+1000)$ and the corresponding drag of $kv$. Where $k$ is a scale factor that can be controlled by changing altitude (eg. to increase it you'll need to descend into a denser atmosphere). $k$ is measured in units of 1/s so the lift is given in terms of acceleration (force per mass if you like).

So if we plug in this lift for $L$, and solve for $k$ we will get:

$$ k = \frac{g - \frac{v^2}{r}}{3(v+1000)} $$

The $kv$ is the drag that is the rate of deceleration during reentry. And then the heat produced in an unit time is given by the product of the drag force times the speed. That is $kv^2$.

Then we can plug in the $g$ which is approximated to be 10m/s$^2$ and the r which is approximated to be 6371000m.

So the plot of deceleration ($kv$) is:

Plot of deceleration

Which shows the deceleration would be very gentle, just 0.2g at peak.

Now plot $kv^2$ for the heating:


Which at the peak shows about 8kW/kg heat production. So for an 1 ton spacecraft it would be 8MW heat production at peak. Probably the majority of this energy is dissipated into the surrounding air rather than absorbed by the craft, but it's still a lot of energy that needs to be handled.

Moreover a high lift design is likely a waverider that purposely rides the shock wave created by it. So it's in contact with the plasma. And since the peak deceleration is just 0.2g it would take like 2-3 hours to slow down and land, so the vehicle would need to endure the hot plasma that long.

If we can solve this problem and find a way to harness this energy to eg. genenrate fuel, then waveriders will make sense. But until that it's easier to just plunge through the atmosphere as quickly and possible and hope the heat shield of spacecraft doesn't burn through in the process.

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    $\begingroup$ Note that the conversion from the Mach number to meters per second that you have used is a really rough approximation. The speed of sound varies a whole lot with altitude. $\endgroup$
    – TooTea
    Commented Jul 12, 2020 at 9:31
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    $\begingroup$ @TooTea It depends on temperature. I don't know if it's the temperature of the ambient gas, or the hot plasma though... $\endgroup$
    – Calmarius
    Commented Jul 12, 2020 at 11:53

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