I asked What does account for a high coffin corner of a plane? in Aviation SE. Now I wonder what factors would contribute to a spacecraft being able to maintain low circular orbits (e.g. within 200 km / 125 mi from Earth's sea level) or orbits with very low perigees (e.g. within 120 km / 75 mi or so) without decaying too rapidly nor re-entering the atmosphere.

To survive low orbits, would it help if the spacecraft in question was smaller? Larger? More massive?

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    $\begingroup$ Technically, could you describe a non-moving blimp as being in geo-stationary VLEO? If that blimp starts circling the Earth (possibly requiring some mid-air refueling), that could be described as a very low circular orbit... $\endgroup$ Sep 16 at 19:26
  • $\begingroup$ Hi Giovanni I saw your edit to the very-low-earth-orbit tag but the number you've assigned seems arbitrary and in disagreement with the tag's usage., for example How low is VLEO? (FCC's newest approval for SpaceX) Unless you can cite a formal definition it's better to leave the tag blank than to just put a number that seems okay. Thanks! $\endgroup$
    – uhoh
    Sep 17 at 0:52
  • $\begingroup$ @uhoh It is convention that LEO is defined at 160-2000 km and VLEO would be beneath. Look on WP or Quora, it's not hard to find out. Whether we agree with these demarcations is a different story. $\endgroup$
    – user43968
    Sep 17 at 5:25
  • $\begingroup$ @DarrelHoffman Would you say a sailplane or helium balloon that circumnavigated the Earth has orbited it? I don't think so, you should reach orbital velocity for that. $\endgroup$
    – user43968
    Sep 17 at 5:28
  • $\begingroup$ @Giovanni 2000 km seems standard, but now that the FCC recognizes very-low-earth orbit as distinct from LEO I think Wikipedia and Quora need to be updated. Usage guidance should be up-to-date and consistent with its use on the site. Since SpaceX and the FCC are using VLEO in official actions, I think that it is here to stay. So LEO may have to nudged up a bit, or maybe brought down to 80 km and VLEO become a subset. Either way the very-low-earth-orbit and LEO tags should be updated to acknowledge each other. $\endgroup$
    – uhoh
    Sep 17 at 14:37

The lower the orbit, the higher the drag. Since the drag force is roughly proportional to area, but the resulting acceleration is force divided by mass, then the denser the spacecraft, the less it is affected by drag.

The simplest way to fight drag is to carry a lot of fuel, and use it often. This will make the spacecraft less dense as the tank empties, but also lighter and thus have less mass to accelerate by the same fuel, in a way that cancels out. The main thing to worry about is just that you have finite fuel, and thus will eventually run out and have no remaining way to prevent eventual decay.

Drag models of spacecraft are almost always spheres, but you could try making a long, narrow vehicle -- that is, rocket-shaped -- and keeping the nose oriented into the direction of travel. This will reduce drag, but if your vehicle has one end heavier than the other, then the gravity gradient will tend to pull it upright, so that its long side faces into the wind, thus increasing drag, and pointing the tail engine in the wrong direction. You will have to monitor attitude, and have additional thrusters oriented properly to correct it.

You could do very interesting things by designing the body so that it generates lift as well as drag, and maintaining attitude like an aircraft. This is tricky to get right and requires a lot more fluid dynamics for proper modeling than is usually done, so you will need custom software to control it.

Also, drag is friction, which heats the spacecraft. The lower you go, and the more drag you encounter, the faster things heat up. You need ways to control that heat, and get rid of it, or your very low altitude spacecraft may melt long before it runs out of fuel.

  • $\begingroup$ But there can't be generated enough heat at say 110 km (361K ft) altitude that would be of any danger for the craft at orbital velocity, can it? $\endgroup$
    – user43968
    Sep 16 at 13:45
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    $\begingroup$ "will make the spacecraft less dense as the tank empties, so you will need gradually more fuel to achieve the same effect against drag." is incorrect. Yes the spaceship will mass less for the same drag profile, thus slow down faster. equally (exactly equally!!) the lesser mass will require less fuel to make up the lost speed. $\endgroup$
    – PcMan
    Sep 16 at 13:48
  • $\begingroup$ I'm just speculating about a polar orbit vs. the Earth's rotation. Orbital velocity is about 17 times faster than the velocity of the air at the equator and they are perpendicular to each other. So if I can design a spacecraft with a lift/drag ratio more than 17 then I should be able to get a net positive force component along the direction of travel. Modern aircraft reach about 20 but I gather the numbers are much lower at supersonic speeds. Any thoughts ... ? $\endgroup$
    – Roger Wood
    Sep 16 at 17:09
  • $\begingroup$ @PcMan thank you, fixed now. $\endgroup$
    – Ryan C
    Sep 17 at 1:02
  • $\begingroup$ just to continue this earlier line of thought, I see the lift/drag ratio for a flat plate in a rarified atmosphere and with specular reflection is just 1/tan(theta) which is arbitrarily large for very small angles of attack. I suppose the problem at these velocities it getting molecules to bounce nicely of the surface in the correct direction and without losing energy. Even if you are able to find a surface smooth enough to give specular reflection at an atomic level, the effective temperature of an O2 molecule at orbital speed is about 2000 K which is surely enough to disrupt most surfaces. $\endgroup$
    – Roger Wood
    Sep 17 at 4:21

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