The shuttle had a empty gross weight of 172,000 lb (78,000 kg), and the only feasible way shedding the 27,724 kilometres (17,227 mi) per hour, of the relative speed of the ISS is Aerobraking. A four passenger Cessna 172 has an empty weight of under 2,500 lb (1,136 kg).

While actually using Cessna as a reentry vehicle would not be practical, the image allows for exploring the concept of an lite reentry vehicle (as opposed to How possible are 'space jumps'? ). Assuming we swap out the gas engine of the Cessna, and replace the mass with fuel, then strap some rockets (clearly an oversimplification) to the plane to decrease the orbital velocity, we have a nice glider with 2% of the mass and twice or more of the glide capacity.

If your pilot and passengers are in space suits for the trip, you won't need to add mass for keeping the vehicle pressurized.

So could a lite winged reentry vehicle use rockets to decelerate, and allow reentry with out having drag/heat issues? Would the fuel requirements to decelerate safely outweigh the benefits of creating vehicle that does not need to stand up to high speed Aerobraking?

Edit This answer to this question comes down what what is more economical; The extra mass for aerobraking heat shielding or the mass for fuel to decelerate. There was a comment addressing this but it has been deleted.

  • $\begingroup$ Only if you have some kind of crazy fusion rockets, in which case this proposal seems increasingly kerbal. $\endgroup$
    – ikrase
    Commented Feb 22, 2020 at 23:21

3 Answers 3


We can use the rocket equation to get an estimate of the amount of fuel we need to slow down from orbital speed to something survivable by a Cessna.
Orbital speed is 7890 m/s. We also need to dissipate the potential energy that was stored as altitude: if we fire the rocket at orbital altitude until the Cessna's speed is zero, it'll start dropping in freefall and gaining speed again. Here's the speed gained in a drop:
$\sqrt(2 * g * h)$
Starting from 200 km and ignoring atmospheric influences, I get 1980 m/s. If the rocket has to brake away that speed as well, the total $\Delta V$ is 9870 m/s. Let's use an Isp of 300 seconds (e.g. the SpaceX Merlin rocket engine) and an empty mass of 1200 kg. Solving the rocket equation to give us the initial mass:
$M_0 = M_1 * e^{\frac{9870}{300*9.81}}$
So we end up with an initial mass of 34.3 tons.

We'll have to work out a flight profile where we save some of the rocket fuel until the Cessna hits the atmosphere so we can prevent overspeeding and overheating during the descent.

We also need a structure to carry 34 tons of fuel in, and the rocket engine is going to have lots more thrust than the original Cessna engine so we may have to beef up the Cessna's structure to take the loads. It starts piling up quickly, and it gets worse when you consider that those 34 tons have to be launched as well. To get the original 1200 kg up, a Soyuz launcher would suffice. At 34 tons, you're beyond the heaviest launch vehicles we currently have.
Compare this with a Soyuz descent module which weighs about 3000 kg. Clearly aerobraking and parachutes carries much less of a weight penalty than removing all that delta-V via thrust alone.

  • 2
    $\begingroup$ IIRC Cessna's maximum structural stress load is roughly around 1g, and that's for the fancy Cessnas like 441. By the time you reduce orbital velocity to subsonic speeds, gravity takes over reducing your altitude and you hit denser atmosphere head on at velocities it couldn't possibly tolerate. Deploying a sufficient sized parachute would only help tearing it apart. $\endgroup$
    – TildalWave
    Commented Feb 17, 2014 at 14:25
  • $\begingroup$ That's why I added a factor to brake away any speed gains from freefall. $\endgroup$
    – Hobbes
    Commented Feb 17, 2014 at 14:31
  • $\begingroup$ OK but that's already close to 1g, so the burn in the deceleration vector would only end up being longer, requiring even more reaction mass because only a small fraction of that is actually reducing your orbit. $\endgroup$
    – TildalWave
    Commented Feb 17, 2014 at 14:36
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    $\begingroup$ You'd be surprised. Light aircraft are stressed to +3.8g and -1.5g. $\endgroup$
    – Hobbes
    Commented Feb 18, 2014 at 11:27
  • 2
    $\begingroup$ Light aircraft in the normal category are stressed to +3.8g, but if you picked an aerobatic category aircraft you'd get +6g and -3g. $\endgroup$
    – GdD
    Commented Nov 10, 2015 at 13:08

Cessna is at the edge of the atmosphere where its potential energy is very high which must be dissipated in some form (which re-entry vehicles dissipate as heat) since energy in this case is conserved. It is not about gliding through the atmosphere.

  • During re-entry the shuttle experience enormous stress which Cessna would collapse .
  • Eventhough Cessna glides through it would eventually reach hypersonic speed (Apollo capsule reached Mach number nearly equal to 36) which leads to a bow shock wave which inturn causes increase in temperature near its front end which Cessna would not survive (would melt or it will become weak leads to failure). But re-entry vehicles are shielded from this intense heat (shuttles having tiles in the nose).
  • If the Cessna is not pressurised it will leads to more force (due to high pressure difference inside and outside the Cessna where the pressure would be very high) being exerted on the Cessna body which is not designed to withstand it .
  • $\begingroup$ Rockets/fuel would be used to lower the orbital speed and provided a controled decent through 418 km (260 mi) so you don't overheat. $\endgroup$ Commented Aug 23, 2013 at 16:06
  • $\begingroup$ @JamesJenkins - ISS speed on orbit. And the Cessna wouldn't mind falling apart by either acceleration or deceleration. As far as its structural stability goes, it's the same thing. $\endgroup$
    – TildalWave
    Commented Aug 23, 2013 at 17:00
  • $\begingroup$ @TildalWave, if you keep your deceleration to say half a gee, it should not cause undue stress on the Cessna. I don't know what the design criteria is but I would imagine it could easly take a 1.5 gees. $\endgroup$ Commented Aug 23, 2013 at 19:04
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    $\begingroup$ @JamesJenkins - Those deceleration rockets better be something we've yet to see, otherwise you'll be flying through their exhaust in a tin can of soon to be extremely charred baked beans. Which wouldn't be any different without the rockets anyway, you're just needlessly adding on weight and its potential energy you need to decelerate. And I haven't a clue where to attach the rockets on a Cessna, or expect it to support their acceleration force, or indeed their weight and drag they'd produce, once you're approaching lower atmosphere with increased air resistance and gravitational pull. $\endgroup$
    – TildalWave
    Commented Aug 23, 2013 at 22:45
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    $\begingroup$ @TildalWave I think you're right on the mark. By the time you complete the design you're discussing, it would be a bullet-proof reentry vehicle carrying a foam-encased Cessna inside. That might work. $\endgroup$ Commented Sep 6, 2013 at 15:44

While a Cessna is extremely implausible, due to a variety of stress and temperature factors, Burt Rutan developed a means for non-heat-shielded descent.

The mode is called "shuttlecock mode" and involves a high bypass configuration on a transsonic airframe; SpaceShip One wasn't shedding orbital velocity. SpaceShip One was at very low orbital speed, insufficient to even qualify as an orbit in all but the most technical terms. The reason for shuttlecock mode is that it generates a huge amount of drag on a very low amount of material.

The proposed Cessna can't do it, but a SpaceShipOne style shuttlecock mode could, in theory, make use of the principles used for removing vertical velocity in order to remove orbital velocity via near-vacuum drag; put it into shuttlecock mode, maintain the correct angle, and have the angle continuously change as it decays into an Earth-impacting ellipse. (Rutan has spoken of this in interviews.) Just as a badminton shuttlecock slammed due forward at 80 miles per hour rapidly drops to under 1 MPH, and as it does so, angles down, so also could an orbital craft.

The trick will be maintaining a drag coefficient that doesn't result in heating too quickly - and a proper shuttlecock mode theoretically can do just that... provided the angles are correct on initial entry. Further, since the primary drag is behind the center of mass, it should, as with a badminton shuttlecock, maintain the correct angle by dynamic stability.

  • $\begingroup$ Are there any tests or simulations from orbital speed? $\endgroup$ Commented Feb 18, 2014 at 11:07
  • $\begingroup$ Rutan's said he's done some simulations... but it's not the goal of the program at SC from what can be seen. Virgin Galactic is all about getting people on non-oribting flights. $\endgroup$
    – aramis
    Commented Feb 19, 2014 at 7:13
  • $\begingroup$ To the person who edited in all the links - you added some totally irrelevant ones. Don't do so. Bad form, and bad editing. $\endgroup$
    – aramis
    Commented Feb 19, 2014 at 7:17
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    $\begingroup$ Minimum Earth orbital velocity is ~ 17,500 MPH. Space Ship One's max velocity was ~ 2,100 MPH. To say that SpaceShip One "was at a very low orbital speed" is not correct as it was never anywhere near orbital velocity. My edits to this effect were rolled back by the OP; is there something I'm missing? $\endgroup$ Commented Feb 19, 2014 at 17:41
  • $\begingroup$ Orbital speed in this regard refers to direction, not magnitude. That's what you missed, plus you mangled the wordings elsewhere by use of search and replace. $\endgroup$
    – aramis
    Commented Feb 19, 2014 at 21:00

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