# Gliding into the atmosphere

The recent question about Cessna reentering from ISS got the answers that all imply a rapid drop.

But from what I know, air drag is proportional:

• to square of airspeed
• to air density
• to attack surface (with all the fancy caveats of aerodynamics).

Meanwhile, lift of a plane is proportional

• to square of airspeed
• to air density
• to wing area (with angle of attack and all the fancy caveats of aerodynamics).

That means if an airplane in air experiences X newtons of air drag, its wings are capable of producing Y newtons of lift, with Y being usually considerably higher than X, and mostly proportional. The actual ceiling limit comes from the engine performance which has not enough air to push against.

Is there, then, a specific reason why a plane couldn't enter from the sufficiently low, circular orbit, without need for heatshields? The moment it starts experiencing significant drag - that means considerable energy is being dissipated - wouldn't the same speed and air density that create that drag, contribute equally to its lift, allowing it to maintain altitude where the power dissipation would be kept in check? In other words, instead of dropping steeply into thick atmosphere like most reentry vehicles do, couldn't it just glide down, losing altitude very slowly, to keep both drag and lift to survivable/usable values? Or am I missing something here?

• That's a really good question. I'm guessing the limiting factor is going to be the total heat dissipation -- a high L/D "flying reentry" is going to spend an awful long time at high temperatures compared to STS or Apollo -- but I don't know for sure. – Russell Borogove Nov 9 '15 at 18:24
• X-15 and shuttle both have very poor lift/drag ratios. I assume SF is thinking about something more like a U-2 configuration with a 20+ L/D ratio. – Russell Borogove Nov 9 '15 at 19:03
• @called2voyage: considering the gravitational drag at orbital velocity is zero, you need actually very little atmospheric lift to remain "afloat" if you fly at speeds slightly slower than orbital. Of course aerodynamic control is just as "slight". But so is drag as well... – SF. Nov 9 '15 at 19:29
• @called2voyage Staying above the Kármán line is not reentry, pretty much by definition. – Russell Borogove Nov 9 '15 at 19:39
• IIRC there were total heat load problems associated with low-deceleration, high L/D entries. There's a short discussion of the different phases of STS entry traj at the end of this: spaceflight.nasa.gov/shuttle/reference/shutref/sts/profile.html – Organic Marble Nov 9 '15 at 23:53

It most certainly won't hurt anything. From FAA's document on returning from space, there is a very interesting chart, which I've included below.

So the maximum g load is almost always at around 4500 m, for this particular flight trajectory. From other charts, we find that a lower angle will spread out the re-entry forces, and specifically make the max deceleration happen at a higher altitude. Bottom line is, for our Cessna to have a higher change of landing, it would have to be capable of flight at high velocity around 10,000 m. The max flight ceiling is 13500 feet, roughly 4000 m, however, it should be capable of some lift at that high speeds.

I think you would run into a problem, however. The aircraft would effectively be going supersonic, and wouldn't have much steering capability. I don't believe the design of a Cessna would be capable of maintaining stability at such a high altitude and speed. A space plane would have it's wings further back, like the Space Shuttle, to allow it to have more stability. And if the stability can't be maintained, the lift can't be maintained.

Bottom line, this could assist, but I still think the Cessna wouldn't be around for long.

• I also think Cessna would be a poor choice - exactly due to poor stability; center of drag too far in front. But that doesn't mean a lightweight spaceplane without heat shielding would be unrealistic. The worst problems probably would be of political/geographic nature: this would need to complete a good few orbits before slowing down to speeds where turns become viable, and as result, would thoroughly randomize the landing site. (...and I'm still trying to imagine political results of Gagarin landing in Argentina...). – SF. Nov 9 '15 at 19:37
• The first graph is for a forty five degree entry slope. Manned vessels don't do 100-g reentry. OP's proposal would be a < 1º entry, I expect. – Russell Borogove Nov 9 '15 at 19:41
• @SF: It wouldn't be all that random. The key would be to remain in the realm of a, say, 2g acceleration for a lengthy period of time. – PearsonArtPhoto Nov 9 '15 at 19:46
• @PearsonArtPhoto: 2g is still a pretty high energy to dissipate. I was thinking of orders of 0.1g, near 1 degree angle and two and a half hour of deceleration. – SF. Nov 9 '15 at 19:59
• ${{1 \over 2} \cdot 1 ton \cdot (8000m/s)^2} \over{ 2.5 hours}$ still gives me over 3.5 megawatt to dissipate... – SF. Nov 9 '15 at 20:07

## You can't just slow it down over many orbits

I think the question is suggesting letting a little bit of drag slow the Cessna down until it's at a normal speed before gliding through the atmosphere. That's nice, but it won't work. Orbiting is ballistic flight. Ballistic flight without lateral speed (i.e. once you've started to really slow the Cessna down) is just falling - your ballistic trajectory intersects the Earth's surface. There is no low-speed way to miss the Earth without lots of thrusters, which is not gliding.

## Orbit isn't high, it's fast

Orbit is moving sideways so fast you miss the Earth. Something like 7.8 km/s, which is 17,500 mph. A Cessna flies at rounding error speeds in comparison.

So instead of thinking that it would be possible if the Cessna were dropped low enough, think of it being possible if the Cessna is dropped at a slow enough air speed.

## Kinetic Energy is your enemy

We have a Cessna strapped to an orbital craft doing 7.8 km/s (or it's not orbiting), and by the time it gets into the atmosphere to avoid burning up it needs to be going about 130 mph, or 58 m/s, which is zero by comparison to the initial speed. Our 1,000 kg Cessna thus has a relative KE of 0.5*1000*(7.8e3^2) = 30.4 Gigajoules which we need to lose.

## Decelerockets!

There is only one way at present to transfer 30 GJ of energy to a Cessna before it reaches the atmosphere (i.e. before needing heatshields) and that is a rocket.

Essentially you are going to drop the Cessna off the orbital vehicle and fire a large booster rocket with a delta-v of 7.8 km/s. That is energetically similar to launching a LEO rocket from Earth (LEO has a delta-v of 9 or so from Earth), because again it's about the speed, not the height, and you're just accelerating backwards instead of forwards.

An appropriately sized rocket might be the Japanese Epsilon rocket; it has a payload of 1200 kg for LEO which should get the Cessna and a pilot from +7.8 km/s to 0.06 km/s over the course of a couple of minutes.

## The Tyranny

There's one snag; the Epsilon is 91,000 kg because of the Tyranny Of The Rocket Equation - to accelerate a thing you have to stick it on a rocket, which you also have to accelerate, which takes more fuel, which also has to be accelerated, etc. So 1,200 kg payload requires a 91,000 kg rocket.

That rocket is now a payload we have to put in orbit in the first place, on the top of a much bigger rocket.

But wait, we've lifted more before; Saturn V managed 140,000 kg. The smallest previous launch capable of that scale is the Russian Energia rocket, a follow-on from the unsuccessful N1 project (Russia's Saturn V equivalent). It is about 2,400,000 kg (2.4 kt), about the size of Big Ben's tower and would cost something over \$1.5bn.

## Or just put a coat on

I've let it get a bit silly. It should hopefully be clear that a heatshield is rather cheaper than a rocket to decelerate with, which is essentially why Shuttle was designed the way it was - decelerating by airbrake only costs you the weight of a heatshield.

• You seem to be treating the atmosphere as binary. He's talking about using the very outer edge of the atmosphere. – Loren Pechtel Sep 14 '17 at 15:20
• +n! This is a great answer! I think you should move You can't just slow it down over many orbits to the top and call it a tl;dr. As soon as you loose a little velocity you start dropping. Gravity is not scaling, it's always close to 1g so until your lift force exceeds 1g, you are going to keep falling faster and faster. You burn up long before you reach sufficient density for 1g of lift. The question is misleading in a very clever way and forces us to examine our preconceived notions about what an orbit is and what it isn't. – uhoh Sep 14 '17 at 15:59
• @uhoh: Good plan. – Phil H Sep 18 '17 at 8:08

First of all, the heat shields aren't there just for the dense atmosphere with high deceleration, the upper atmosphere part of reentry also generates heat. If you had a craft that could fly in the upper atmosphere it would still heat up just from its forward speed. Hypersonic vehicles also heat up considerably even though they are just normal aircraft. The SR71 Blackbird reached speeds of 'only' mach 3.3 and had heated up to 300°C, so it had to be made from titanium instead of regular aluminium. Heating also increases with velocity squared.

Secondly, the density of the atmosphere decreases exponentially with height. So the speed an 'aircraft' needs to fly also increases exponentially with height. On the other hand orbital speed (though not total orbital energy) decreases the higher your orbit is. So the faster/higher you go the less lift you need as the closer you are to orbiting. At the Karman line at 100 km altitude both effects are roughly equal, and thus above the Karman line the speed you need to fly is higher than your orbital speed, so 'flying' there would mean shooting away from the earth. Or another way around, a stable altitude above the Karman line would imply an almost-orbit with only a very little bit contributed by lift. However there is still enough atmosphere that you need an engine to overcome the drag, so you would need to continually spend rocket fuel (no airbreathing engine is going to work at that altitude) to not drop down/descend slowly.

I guess you could just use bigger wings, increasing the lift, but with this exponential thinning of the atmosphere your wing size needs to be exponential with height, so that gives problems with total drag and with weight and getting up there in the first place. Weight is also the main reason the Space Shuttle did not have bigger wings than it did, with its rather poor lift to drag ratio.

Third, a shallower entry actually produces more total heat for an entry vehicle to absorb. A very steep entry produces higher peak heating and higher G forces, but lower total heating on the vehicle. For manned reentry vehicles shallowest possible angle of entry is limited by the total heating and the steepest angle by the maximum allowable G forces on the crew.

Despite all of the above, your idea works (but in the lower atmosphere) and is used. Though not to eliminate heat shields (as mentioned before shallower angle = more total heating), but to limit G forces experienced by crews. The Soyuz in normal entry mode actively controls its attitude to generate lift, giving a relatively shallow trajectory with a maximum of 5G deceleration or so. It also has an emergency ballistic mode in which the reentry vehicle is spin stabilised. In that case it does not generate lift, its descent is much steeper, and the crew experiences up to 9G deceleration. The space shuttle also made use of lift, with a maximum deceleration of 3G.

In short, heating is caused by high speed, not (only) by low altitude. Entry can be done without a heat shield, but then you need to make sure you slow down enough before hitting the dense atmosphere. So that means using rockets. Reentering from orbit using rockets to break requires as much fuel as launching in the first place, so that is not practical. However the SpaceX Falcon 9 booster reaches much lower top speeds and uses this approach (the 'entry burn') to slow down and limit reentry heating before it lands.

I like the idea you're using, I've thought about it myself as well. Unfortunately, it doesn't work out, for the reason that drag is not the limiting factor, at least not directly.

# Amount of drag <> amount of generated heat

Air drag is the amount of force that is generated by air acting on your vehicle. It is indeed proportional to airspeed squared
It imposes some limits (to prevent your wings from being torn off, for example), and can be the limiting factor for some portions of the flight. But it's not the hardest limit.

Reentry heat is also caused by air hitting your aircraft, but is proportional to the amount of energy lost, and proportional to airspeed cubed.
And when coming from orbit, going 8 km/s, that's a far more serious problem.

So while a cessna probably wouldn't handle very well when flying supersonic, there are a lot of airplanes that have the structural strength and handling characteristics that could make your approach viable, as long as they were heat-resistant.

The only reason that most spacecraft encounter high-g loads when re-entering is because the most efficient approach is to lose your energy in a short time, because that means only having to withstand the heat for a short time. And for the space shuttle, aerodynamic stresses (= drag, or force), are actually maximal during the launch, when it goes faster through the denser parts of the atmosphere (the maximum is around 11 km altitude)