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Assuming the spaceplane has no vertical (hover)engines, is there a possibility to land a spaceplane on a celestial body without atmosphere in a horizontal way as if it had one? One could lower the periapsis to (close to) 0 altitude and then use the front deorbit engines to further brake the spaceplane once it touches or gets very close to the surface. But wouldn't the craft still be too fast, killing or severly injuring the crew when touching down? Or is there a way one could decelerate it and follow the planet's curvature (by propulsion, beneath orbital velocity) to eventually touch down softly, assuming the craft has enough fuel?

In such case, would rather a high gravity (e.g. Mercury's) or a low gravity (like Ceres') be of advantage for the landing attempt?

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    $\begingroup$ Are you okay with landing backwards? $\endgroup$ – Charles Staats May 2 at 10:55
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    $\begingroup$ An airplane is used in air, a spacecraft in the vacuum of space, a spaceplane without air does not make sense. Do you ask about a spacecraft? $\endgroup$ – Uwe May 2 at 19:46
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    $\begingroup$ Spaceplanes are only used in starwars movies. Using both strong back engines and front engines does not make sense. $\endgroup$ – Uwe May 3 at 6:37
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    $\begingroup$ Turning around the craft for retrofire is easy, removing the front engines reduces the mass of the craft. $\endgroup$ – Uwe May 3 at 9:35
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    $\begingroup$ @Giovanni No, not the ones in the nose - those are only for attitude control. The OMS engines did the de-orbit. Those are the small AJ10 engines at the back (just above and outside the main 3 RS-25s). The shuttle de-orbited with its engines facing forward. Even then, you're talking about stopping the vehicle, not just a de-orbit. Stopping the vehicle requires about 7000x more energy than the de-orbit burn. $\endgroup$ – J... May 3 at 13:21
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Yes But...

This only works if your craft can generate enough thrust to reduce orbital velocity to zero quickly enough that acceleration due to gravity does not produce a crash landing, since you are effectively trying to reduce gravity losses to those that can be absorbed by the undercarriage.

For earth's moon, taking aircraft vertical touchdown speed at 2 m/s and lunar gravity at 1.6m/s this means completing the burn in at most 2 seconds. For a minimal starting orbital velocity of 1.6 kms this involves a burn at more than 800ms2 or 81 G.

Ceres is more reasonable, with surface gravity of 0.28 ms meaning we have 7 seconds to complete the burn, and an orbital velocity of around 300m/s for 42ms2 or 'only' 4G.

In practice, if you were in orbit around Ceres with a space plane capable of 4G you could probable make a conventional descent till close to the surface and land in whatever orientation was convenient on just RCS.

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    $\begingroup$ This answer could probably be improved by briefly explaining how "this means completing the burn in at most 2 seconds" follows from what is written before. $\endgroup$ – Thierry May 3 at 11:49
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    $\begingroup$ For reference, 80G is about what you would experience by crashing a car into a brick wall at about 90km/h (55mph). $\endgroup$ – J... May 3 at 13:28
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    $\begingroup$ It's also worth noting that a spaceplane is no different from any other shape of spacecraft in this instance - because there is no air, the aerodynamics are completely irrelevant. If the spacecraft was a cube or a torus or a snowflake fractal it would not change the equation for landing one bit. $\endgroup$ – J... May 3 at 18:27
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No, you can't. If you want to travel at below orbital velocity at a fixed distance from the centre of mass of some (approximately) spherical body you need a vertical component of thrust. Aeroplanes call this lift and they get it from their wings. In the absence of an atmosphere you must, therefore, have reaction engines which provide a vertical thrust component. From your comment ('I'm assuming the spaceplane has both back engines and front engines (like the shuttle's RCS but stronger), just no vertical engines.) you are ruling these out so what you want to do can't be achieved.

If you allow a vertical thrust component, well, that is how some people landed on an airless moon in the late 1960s and early 1970s, of course. They had only a single high-thrust engine (they also had attitude-control engines but these were relatively tiny) which was gradually turned from acting nearly purely horizontally to acting nearly purely vertically during the descent.

Lower gravity always helps, until it gets so low that tiny mistakes turn into vast leaps or put you back into orbit, or something.

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  • $\begingroup$ And what if you kept a near-perfectly circular orbit that you'd lower more and more until touching the ground softly, is this possible, assuming there aren't strong mascons like on the airless moon you mentioned? $\endgroup$ – Giovanni May 2 at 11:46
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    $\begingroup$ You'd end up touching the ground softly at multiple km/s, dissipating the heat usually shed in reentry in your landing gear instead. Wheels made of any real-world material will fly apart at the rotation rates needed. Maybe some kind of maglev/gas cushion system could deal with it together with a very long landing strip, but it'd be pretty exotic and specialized, and not something you'd ever put on a spaceplane...this is a job for a specialized vehicle. $\endgroup$ – Christopher James Huff May 2 at 12:54
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    $\begingroup$ @Giovanni: you'd contact the ground at orbital velocity. Which, generally, would be extremely bad for you indeed. $\endgroup$ – user21103 May 2 at 13:15
  • $\begingroup$ I wonder why I hadn't got a notification in the infobox for tfb's answer. $\endgroup$ – Giovanni May 2 at 13:56
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    $\begingroup$ Something which is probably obvious to the people answering this question, but perhaps not to those asking this question, is the notion that "a planet's surface does not orbit that planet". Perhaps this answer can be improved by incorporating that notion. If the surface were orbiting, then the act of landing would be akin to orbit matching, and it that case the effective vertical component would reach (in practice: nearly) 0m/s at the instant the horizontal ('overland') speed difference reaches 0m/s and the height above terrain/target-orbit reaches 0m; i.e. the moment you touch down. $\endgroup$ – Thierry May 3 at 11:57
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If you ever looked at the photos of the surfaces of Earth's deserts, Mars' plains, Venus' Venera landing site, Jupiter's moons, or some of the recently visited asteroids, you will notice there are few if any areas smooth enough for such landings. Aircraft and spaceplane landing gear are not designed for more than minimal roughness in the landing site. This is partly to save mass, partly because landing gear can't perform miracles of strength, even if they are just skids. Higher speeds mean higher energy collisions with small rocks (whatever their substance), sand drifts, and cavities or craters. All designated rough (unpaved) landing fields on Earth try to minimize these. (Small rocks can bounce and damage or puncture the vehicle skin.) On a distant planet or planetoid, there is no such advance party to clean up the landing site unless a "highway repair" robot lands first. The previous responder is correct, retro thrusting is the only sensible way and the mass allotment is better spent on engines and propellant than the now useless wings. (Mass ratios are unforgiving.) Bear in mind also that on Earth we nearly always conduct long distance travel by changing from one type of vehicle to another, each suited to the type of travel involved. Example: Travel from one's home in NorAm or Eurasia to a research station in Anarctica. If you start in Emporium, Kansas, you won't leave in a Boeing 747 and won't arrive in one either.

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  • $\begingroup$ I never wrote anything about tire wheel gear. Except that some fighter planes landed in dry lake beds (including the spaceplane X-15) a spaceplane might use a ski-like gear like the shuttles in the Armageddon movie. But the question assumes the gear is strong enough to withstand a rough terrain (but not friction heating as described by Mr. Huff). $\endgroup$ – Giovanni May 2 at 14:56
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It can be done but only if you have a very long, very good runway and your wheel bearings are very, very good. You put your apoapsis low and then put your periapsis at the elevation of your runway. Note that you will need to immediately retrofire when you touch the runway or you'll just take off again. Brakes will not work at all until you have slowed enough that your apoapsis is at the runway and your periapsis is underground.

I'm not aware of any wheels that would work on any substantial world. However, there's a better way to handle this. What's the usual way to do this on Earth? Magnetic levitation. Build a doubled track--instead of simply riding on the rails it rides between two rails. The spacecraft comes in and sets it's periapsis very low as it passes over the track. A vehicle zips down the track at the same speed and releases a cable. The plane grabs the cable, then the cable is reeled back in to bring the plane down to the capture vehicle, then the capture vehicle slows down. Note that you can launch the same way.

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  • $\begingroup$ I used to think that cable arrestor landings were really scary. Now I can't stop thinking about trying to capture a wire at orbital speed. :-) $\endgroup$ – TooTea May 3 at 12:47
  • $\begingroup$ @TooTea This is safer than a carrier landing because you have more time to get perfect alignment and you don't have the atmosphere and waves messing with you. Like with a carrier, if you miss the cable you just go around. $\endgroup$ – Loren Pechtel May 3 at 15:24

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