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Alright, this is an edit of my previous question, in response to a number of urges to clarify it. I'll do my best to be clear and concise.

How long would it take a spaceship, locked in geostationary orbit above a fixed point on the Earth's surface, to descend from orbit to the surface? I'm also curious as to how long it would take for said spaceship to reenter orbit, that is, to return to the exact geostationary position it had previously left.

I am, of course, speaking in terms of the technology we have today, but if this kind of maneuver is not possible with modern technology, please let me know why, and what would be required to pull it off.

So, once more: is it possible to move a spaceship orbiting the Earth, or an Earth-like planet, in and out of geostationary orbit, and if so, how long would it take?

-C

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    $\begingroup$ I'm not sure I understand, how can it be both in a geosynchronous orbit and being lowered onto the surface? Or do you mean it employs some sort of elevator system to descend part of it down? $\endgroup$ – Mormacil Apr 1 '17 at 19:01
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    $\begingroup$ can you clarify what you are asking? As it is now your question sounds like "can I move while standing still?", which is at best unclear. $\endgroup$ – L.Dutch Apr 1 '17 at 19:56
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    $\begingroup$ If it doesn't have any particular reason to be in a geosynchronous orbit (GEO) and you're concerned about the time it takes to get back to Earth, you could just fly the craft to a low earth orbit (LEO). It is much faster, as it doesn't have to travel as far (only 160~2000 km for LEO, versus nearly 36000 km for GEO). $\endgroup$ – Sazanami Apr 1 '17 at 21:31
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    $\begingroup$ Yes, at lower orbits, the atmosphere is denser, meaning there will be more friction. So, if you want to keep the altitude of the craft constant, you will have to apply corrections with, f.e., thrusters. However, if you have no particular mission, than you can just let the orbital decay take its course. The only thing you have to worry about is that the time between landings is less than the time it takes for the craft to reenter naturally (at 600km, we're talking years). Also, over the course of the mission, you'll safe fuel, as maintenance is less demanding than the planned reentry maneuvers. $\endgroup$ – Sazanami Apr 1 '17 at 22:27
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    $\begingroup$ Ok, un-downvoted because now it does look like you put a decent amount of thinking into it. $\endgroup$ – Mołot Apr 3 '17 at 8:36
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Yes, it would be possible, but prohibitively difficult (assuming an Earth-like planet) and it might not even be possible with only modern tech.

There is nothing about geostationary orbit that makes it any different from other orbits with regard to doing this (except for the greater delta-v requirements to get back than with a low orbit). Your satellite would just burn retrograde until its perigee is low enough, at which point it just falls back down to earth and lands.

From there, the satellite does whatever it does and returns to its orbit. If you care about being over the same spot while returning, then burn into a lower orbit and then time raising your orbit so that you end up over the right spot. Otherwise, you can launch into geostationary orbit directly.

The hard part about this is that you basically need to land something the size of a typical launch vehicle and keep it landed, which most such vehicles need flat ground and special infrastructure to do, as well as needing that much in fuel per launch (assuming modern or near-future tech), plus extra to be able to land on the surface again (if required). If you hand wave sci-fi tech of the sort used in Star Wars, Star Trek, or other similar works of sci-fi, then this is a lot more practical.

It would probably be possible to do, but it would most likely be not worth the effort (maybe leave the satellite in orbit and fly an airplane over the site it wanted to land at so that you can do what it was going to do).

As for the amount of time, https://space.stackexchange.com/questions/2914/how-long-does-it-take-for-a-satellite-to-reach-geo says that it would take at least 5 hours 15 minutes to launch to geostationary orbit, and doing the opposite would most likely be similar in timescale to do that, but it might need more time to do course correction so that it lands over the correct spot. In all, expect about an extra hour or so (approximate math, don't quote me on this) for landing, and up to another extra hour on take-off if you care about where you end up over.

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  • $\begingroup$ Thanks. I know this is a bit of a dodgy question, which is why I simply needed more information on it. Knowing the approximate time required to pull something like this off was particularly helpful. $\endgroup$ – C. S. Wright Apr 1 '17 at 20:42
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I assume you are thinking a geostationary orbit. (geosynchronous orbit above the equator, something on this always remain over the same spot.)

[I am unfamiliar with mathjax, edit by someone, who is, is appreciated.]

Geostationary orbits satisfy this equation:

(2*pi/T)2 R = G M / R2 (where R is the radius of the orbit, T is the sidereal period, G is the gravitational constant, and M is the mass of the planet.)

You get the radius as:

R = 3th root( G M / (2*pi/T)2 ) This yields 42,267 kilometers in Earth's case, the sum of the famous 35,786 and Earth's radius.

The orbital velocity: v = (2*pi/T)*R, about 3074 m/s by Earth.

The lowest deltaV orbit from here, witch touches the planetary surface has periapsis at sea level, and apoapsis at 42,267km. (Sort of Hohmann transfer to surface orbit.)

Such orbit would have an orbital velocity at apoapsis of v1 = sqrt(G M ( 2/R - 2/(r+R) ) )

(where r is the radius of the planet, and R is the geostationary radius as above.) For Earth it is 1571 m/s. To deorbit, you have to lower your velocity by v-v1

So first of all you need 1500 m/s deltaV.

The time to periapsis will be t = pi*sqrt( ( (r+R)/2)3/ (G M) )

It would take 18874 s to touch down on Earth.

But on periapsis you would have an orbital velocity of v2 = sqrt(G M ( 2/r - 2/(r+R) ) )

The ground only moves with (2*pi/T)*r, so you have to decelerate the difference to land. (instead of hypervelocity crash.)

In Earths case you would need 9940 m/s additional deltaV for landing.

This is quite a lot. But if the planet has a suitable atmosphere, you could use aerobraking and parachutes to do the most of the deceleration.

Using these formulas, one can calculate deorbiting dV for any celestial body.

The ascent would ideally take the same time and dV. But since atmospheric and gravity drag work against you in liftoff case, a margin of some thousand m/s is needed. (Depending on planet size and atmosphere thickness.)

We need 11 hour for minimal dV landing and getting back on GEO.

A spacecraft with chemical engines to accomplish this on Earth would have a mass ratio of roughly 10. This is impossibly high for a compact, entry capable, non-staging spacecraft.

If we would have engines with better specific impulse and ALSO decent thrust-to -weight ratio, and capability to work efficiently in atmosphere,(Say 50,000m/s exhaust velocity and 20,000N thrust.) we could do it. But such a thing needs 0.5 GW thrust power, making it into a bulky fusion drive with large heat radiators.

My best idea (for futuristic but physically plausible solution) would be a solid core antimatter ramjet/rocket. It would use air as reaction mass during liftoff, (and also as coolant) and once orbit is achieved, make GEO transfer by low thrust - high efficiency plasma electromagnetic drive.

If the drive is overkill enough, you can make transfer with continuos thrust too, reducing needed time, but raising fuel consumption.

Of course, there is no way you could maintain 'syncroisedness' during ascent/descent, but if you have plenty of time to wait, or powerful continuos thrust drive, you can manage landing on the same spot you have orbited above before.

11 hour is the basic time, with no waiting (for landing on the same spot or getting back on the same GEO position) but minimal dV elliptical trajectory. If your engine is very efficient, it can be greatly reduced.

All distances in meter and all times in second. If I have miscalculted something (it is late) let me know.

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  • $\begingroup$ It seems sound to me, b.Lorenz (possibly because most of it is beyond me). One question, though; would the 11-hour timescale for ascent and descent remain the same, even if the ship were using your solid core antimatter ramjet/rocket model, or would it take less time/more time? The timing you mentioned lines up with one of the previous answers so I'm sure that's how it would work under normal circumstances, I'm just curious if there's any way to shorten the amount of time necessary to move the satellite in and out of GEO/LEO (specifically out of GEO/LEO and down to the planet's surface). $\endgroup$ – C. S. Wright Apr 1 '17 at 21:49
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    $\begingroup$ @C. S. Wright Yes, if the drive is strong and efficient enough, you can reduce the travel time greatly, but by the cost of using up more dV. For example, a Gravitic Handwavium drive, which cancels out outside gravitational effects, and provides constant 1g acceleration, ca do it in 4100seconds $\endgroup$ – b.Lorenz Apr 2 '17 at 5:21
  • $\begingroup$ Is that 4100 seconds each way (i. e., out of orbit/into orbit), or roundtrip? $\endgroup$ – C. S. Wright Apr 3 '17 at 16:22
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    $\begingroup$ @C. S. Wright Each way. But this is just a very simple approximation for a very unrealistic drive. Accelerating upwards for 1800s with 1g, then decelerating with 1g, and 300s for building orbital velocity of 3000m/s $\endgroup$ – b.Lorenz Apr 3 '17 at 16:31

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