Why can't they just drop a solar winch down from a shuttle and have planes fly up and clip things on?
$begingroup$
Why can't they just drop a solar winch down from a shuttle and have planes fly up and clip things on?
I know of the idea to have a space lift but the cable necessary is too expensive.
Why can't we just have a small cable hanging down to the edge of space though?
spacecraft design space-elevator design-alternative
edited Jan 21 at 15:53
Machavity
2,32711037
asked Jan 20 at 6:59
SpaceMonkeySpaceMonkey
10013
$endgroup$
add a comment |
$begingroup$
Why can't they just drop a solar winch down from a shuttle and have planes fly up and clip things on?
I know of the idea to have a space lift but the cable necessary is too expensive.
Why can't we just have a small cable hanging down to the edge of space though?
spacecraft design space-elevator design-alternative
edited Jan 21 at 15:53
Machavity
2,32711037
asked Jan 20 at 6:59
SpaceMonkeySpaceMonkey
10013
$endgroup$
2
$begingroup$
In fact you can do this - in theory, anyway. The trick is that the cable has to be rotating, so that it dips into the atmosphere and is for a brief time moving slowly WRT the Earth's surface, so that an airplane could dock with it. See e.g. Momentum Exchange Tethers for more info:
$endgroup$
– jamesqf
Jan 20 at 17:58
6
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Worth reading Neal Stephenson's "Seveneves" for an excellent hard sci-fi description of a skyhook, centred at Geostationary orbit, and counter-rotating so the tether end reaches almost to ground and is briefly almost stationary.
$endgroup$
– Rory Alsop
Jan 21 at 14:32
1
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The first chapter of Soonish is about potentially cheap ways to get to space, including things like space elevators.
$endgroup$
– Barmar
Jan 22 at 1:31
$begingroup$
thank you for all your answers. i only just discovered them as i was expecting to get a notification but didn't. you are all very kind to take the time to explain it to me. thanks
$endgroup$
– SpaceMonkey
Jan 30 at 5:55
add a comment |
$begingroup$
Why can't they just drop a solar winch down from a shuttle and have planes fly up and clip things on?
I know of the idea to have a space lift but the cable necessary is too expensive.
Why can't we just have a small cable hanging down to the edge of space though?
spacecraft design space-elevator design-alternative
edited Jan 21 at 15:53
Machavity
2,32711037
asked Jan 20 at 6:59
SpaceMonkeySpaceMonkey
10013
$endgroup$
Why can't they just drop a solar winch down from a shuttle and have planes fly up and clip things on?
I know of the idea to have a space lift but the cable necessary is too expensive.
Why can't we just have a small cable hanging down to the edge of space though?
spacecraft design space-elevator design-alternative
spacecraft design space-elevator design-alternative
edited Jan 21 at 15:53
Machavity
2,32711037
asked Jan 20 at 6:59
SpaceMonkeySpaceMonkey
10013
edited Jan 21 at 15:53
Machavity
2,32711037
asked Jan 20 at 6:59
SpaceMonkeySpaceMonkey
10013
edited Jan 21 at 15:53
Machavity
2,32711037
edited Jan 21 at 15:53
Machavity
2,32711037
edited Jan 21 at 15:53
Machavity
2,32711037
2,32711037
asked Jan 20 at 6:59
SpaceMonkeySpaceMonkey
10013
asked Jan 20 at 6:59
SpaceMonkeySpaceMonkey
10013
asked Jan 20 at 6:59
SpaceMonkeySpaceMonkey
10013
10013
2
$begingroup$
In fact you can do this - in theory, anyway. The trick is that the cable has to be rotating, so that it dips into the atmosphere and is for a brief time moving slowly WRT the Earth's surface, so that an airplane could dock with it. See e.g. Momentum Exchange Tethers for more info:
$endgroup$
– jamesqf
Jan 20 at 17:58
6
$begingroup$
Worth reading Neal Stephenson's "Seveneves" for an excellent hard sci-fi description of a skyhook, centred at Geostationary orbit, and counter-rotating so the tether end reaches almost to ground and is briefly almost stationary.
$endgroup$
– Rory Alsop
Jan 21 at 14:32
1
$begingroup$
The first chapter of Soonish is about potentially cheap ways to get to space, including things like space elevators.
$endgroup$
– Barmar
Jan 22 at 1:31
$begingroup$
thank you for all your answers. i only just discovered them as i was expecting to get a notification but didn't. you are all very kind to take the time to explain it to me. thanks
$endgroup$
– SpaceMonkey
Jan 30 at 5:55
add a comment |
2
$begingroup$
In fact you can do this - in theory, anyway. The trick is that the cable has to be rotating, so that it dips into the atmosphere and is for a brief time moving slowly WRT the Earth's surface, so that an airplane could dock with it. See e.g. Momentum Exchange Tethers for more info:
$endgroup$
– jamesqf
Jan 20 at 17:58
6
$begingroup$
Worth reading Neal Stephenson's "Seveneves" for an excellent hard sci-fi description of a skyhook, centred at Geostationary orbit, and counter-rotating so the tether end reaches almost to ground and is briefly almost stationary.
$endgroup$
– Rory Alsop
Jan 21 at 14:32
1
$begingroup$
The first chapter of Soonish is about potentially cheap ways to get to space, including things like space elevators.
$endgroup$
– Barmar
Jan 22 at 1:31
$begingroup$
thank you for all your answers. i only just discovered them as i was expecting to get a notification but didn't. you are all very kind to take the time to explain it to me. thanks
$endgroup$
– SpaceMonkey
Jan 30 at 5:55
2
2
$begingroup$
In fact you can do this - in theory, anyway. The trick is that the cable has to be rotating, so that it dips into the atmosphere and is for a brief time moving slowly WRT the Earth's surface, so that an airplane could dock with it. See e.g. Momentum Exchange Tethers for more info:
$endgroup$
– jamesqf
Jan 20 at 17:58
$begingroup$
In fact you can do this - in theory, anyway. The trick is that the cable has to be rotating, so that it dips into the atmosphere and is for a brief time moving slowly WRT the Earth's surface, so that an airplane could dock with it. See e.g. Momentum Exchange Tethers for more info:
$endgroup$
– jamesqf
Jan 20 at 17:58
6
6
$begingroup$
Worth reading Neal Stephenson's "Seveneves" for an excellent hard sci-fi description of a skyhook, centred at Geostationary orbit, and counter-rotating so the tether end reaches almost to ground and is briefly almost stationary.
$endgroup$
– Rory Alsop
Jan 21 at 14:32
$begingroup$
Worth reading Neal Stephenson's "Seveneves" for an excellent hard sci-fi description of a skyhook, centred at Geostationary orbit, and counter-rotating so the tether end reaches almost to ground and is briefly almost stationary.
$endgroup$
– Rory Alsop
Jan 21 at 14:32
1
1
$begingroup$
The first chapter of Soonish is about potentially cheap ways to get to space, including things like space elevators.
$endgroup$
– Barmar
Jan 22 at 1:31
$begingroup$
The first chapter of Soonish is about potentially cheap ways to get to space, including things like space elevators.
$endgroup$
– Barmar
Jan 22 at 1:31
$begingroup$
thank you for all your answers. i only just discovered them as i was expecting to get a notification but didn't. you are all very kind to take the time to explain it to me. thanks
$endgroup$
– SpaceMonkey
Jan 30 at 5:55
$begingroup$
thank you for all your answers. i only just discovered them as i was expecting to get a notification but didn't. you are all very kind to take the time to explain it to me. thanks
$endgroup$
– SpaceMonkey
Jan 30 at 5:55
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Because space isn't about going high; it's about going fast! For example, in a 400 km orbit (like ISS) you need a speed of about 27,500 km/h or 7.66 km per second.
So if you would extend a pole, winch or anything else into the lower parts of the atmosphere, it would also move at about 27,500 km/h (if we ignore atmospheric drag and all other influences). Try to catch that hook! If you can, you might as well go straight into orbit yourself.
edited Jan 20 at 13:46
Peter Mortensen
20817
answered Jan 20 at 8:27
DarkDustDarkDust
7,47433056
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14
$begingroup$
Reminds old good what-if.xkcd.com/157
$endgroup$
– val
Jan 20 at 10:38
10
$begingroup$
The "Rotating Skyhoook" concept en.wikipedia.org/wiki/Skyhook_(structure) seems relevant here
$endgroup$
– Steve Linton
Jan 20 at 11:10
6
$begingroup$
@val this one what-if.xkcd.com/58 is even better :P
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– paul23
Jan 20 at 17:42
2
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Not to mention that you'd need a heck of a heat shield for that pole at 27,500 km/h ;-) ...
$endgroup$
– Peter A. Schneider
Jan 20 at 20:54
2
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@val please tag it as "Obligatory what if" ;)
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– Barranka
Jan 21 at 2:11
|
show 2 more comments
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As an alternative to DarkDust's answer, if you start higher, at the classic altitude for space elevators, the end of your cable is indeed stationary to the air. But your cable needs to reach from geostationary orbit to the upper atmosphere, something like 35,700 km. The clipping off the last 20-60 km does not make a big difference in the overall monumental cost and complexity.
It is also worth noting that as you climb the cable, you will also pull the cable and station down. So your elevator will need to burn similar amounts of fuel to keep in orbit as if you had flown there (though possibly in a more efficient engine), unless you can balance things with loads coming down and being dropped as you go up.
edited Jan 20 at 13:47
Peter Mortensen
20817
answered Jan 20 at 8:41
GremlinWrangerGremlinWranger
2,533218
$endgroup$
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The fuel need is a good observation. I assume that the fuel is not needed as much for the climb -- hey, it's ony 100 miles! -- as for the acceleration to orbital speed, as vividly illustrated in the xkcd strip. If not compensated, accelerating the additional load would slow down the shuttle so that it changes to a lower, elliptical orbit.
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– Peter A. Schneider
Jan 21 at 9:01
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@PeterA.Schneider No, you'd need to go the entire 35,700 km. When you get there you are in circular orbit. Leaving after only 100 km would require you to lift impractially large amount of fuel.
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– kubanczyk
Jan 21 at 9:16
6
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@GremlinWranger Classic space elevator design has a counterweight far above GEO which pulls it back up without any propulsion. It uses up a tiny bit of Earth's rotation though :)
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– kubanczyk
Jan 21 at 9:42
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@kubanczyk My remark was meant for low earth orbits when a shuttle hauls in a load starting at airplane altitude. For geostationary orbits I'm not sure what needs more energy, the climb or the acceleration (probably the climb).
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– Peter A. Schneider
Jan 21 at 9:57
1
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@PeterA.SChneider Of course, you don't need to carry that energy with you, and you need no propellant; even with tech we already have (excluding the magical tether itself), that'd mean orders of magnitude lower energy costs than rockets. And then you consider that as you go up, something else can go down, and it neatly balances out.
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– Luaan
Jan 21 at 16:41
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show 2 more comments
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Have you ever flown a kite with a tail? The tail flies almost horizontally, kept up by the wind. It would hang at a steeper angle if you made it heavy, but what would then happen to your kite?
Now imagine that wind to be Mach 20 or so. Even if you had a rope made of unobtainium and hence able to withstand the heat created by atmospheric compression it would always "skim" on top of the atmosphere, the end being horizontal.
The concept is unviable in almost all aspects.
answered Jan 21 at 6:49
Peter A. SchneiderPeter A. Schneider
26317
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add a comment |
$begingroup$
The end of the cable would be destroyed by the heat of reentry from a low orbit when reaching the height where planes may fly.
But if you try to drop a cable from a low orbit it would not drop, it would stay in orbit. There is no dropping of things in zero gravity.
Pulling up a load to the spacecraft in low orbit would slow down the spacecraft. It would lose height and fuel is needed to maintain orbit.
As written in the other answers, a plane is much too slow to catch the hook. Without a heatshield, the plane would be destroyed by the necessary speed. With a heatshield the plane would need a lot of fuel to maintain that speed.
edited Jan 20 at 17:56
Community♦
1
answered Jan 20 at 10:20
UweUwe
10.6k22956
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You could fire the cable in the retrograde direction out of a cannon. This would add complexity and cost, and does nothing to address the other challenges.
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– BobTheAverage
Jan 20 at 21:41
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@BobTheAverage A cannon is much too slow compared to orbital speed.
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– Uwe
Jan 20 at 22:00
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Use a railgun instead.
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– BobTheAverage
Jan 20 at 22:52
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I suppose a modest push towards the ground would suffice to get the cable going; the orbital speed is too low for lower orbits so that the cable would start to "fall" once it's left the current orbit. It would then experience drag fairly soon (which makes it clear that it would never reach the lower atmosphere unless it is propelled at the lower end and probably in between, too).
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– Peter A. Schneider
Jan 21 at 6:37
add a comment |
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There's plenty of answers discussing the problems with having a cable hanging into the Earth's atmosphere - the drag, the heat, the resistence on the orbiting craft, etc. However, I don't see much consideration of the orbital mechanics of the scenario, which would be an issue even if there were no atmosphere.
Simply put, once a craft is in orbit, all that keeps it there is its speed. If you had a craft that could somehow catch the bottom end of the cable at ground level (as it zips by at around twenty-something times the speed of sound), then you have a massive difference in speeds to resolve. If your cable was somehow strong enough (and/or stretchy enough) to bring the lower craft up to speed without breaking the cable, and the contents of each craft were somehow protected from accelerations that would turn cargo and crew into paste, you're now about to run into conservation of momentum. As the cable speeds you up, you're slowing the upper craft down... which means that it will stop orbiting and fall back down.
Even if you put your upper craft in geostationary orbit* (and lengthen the cable considerably to still reach the ground), as you climb up the cable, you're still pulling the upper craft down. Precisely how this works out depends on your relative masses; if you're about the same mass then you'd meet somewhere in the middle. Theoretically you could then push the upper craft down, perhaps by extending a massive pushrod, to continue on your way to orbit, at which point you've invented a rudimentary space elevator. Indeed, some space elevator concepts involve having something very massive (like a captured asteroid) in geostationary orbit*, with a huge tether to the ground - the huge mass means huge momentum and inertia, which mean that its position isn't affected as much by things climbing up to it; an effect that can be further minimised by sending something back down to balance out each thing that comes up.
* Actually, it would have to be a little above geostationary, so that when the mass of the cable is taken into account, the centre of mass is at geostationary altitude. There's also challenges to overcome with regards to the tendency of the whole system to want to rotate relative to the Earth, so that the tether breaks away and drifts out into space, but that's tangential to the question.
answered Jan 22 at 11:27
anaximanderanaximander
1405
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add a comment |
$begingroup$
If the shuttle is at 100 km, and planes fly at say 20 km, that would mean the cable would have to be 80 km in length. Such a long cable would produce so much atmospheric drag that the shuttle would need to constantly burn to counter it, otherwise it would de-orbit pretty quickly.
answered Jan 21 at 14:42
stackzebrastackzebra
2063
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1
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A shuttle at 100 km would deorbit very quickly anyway, the final orbit height of the Space Shuttle was about 380 km.
$endgroup$
– Uwe
Jan 21 at 15:22
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because space isn't about going high; it's about going fast! For example, in a 400 km orbit (like ISS) you need a speed of about 27,500 km/h or 7.66 km per second.
So if you would extend a pole, winch or anything else into the lower parts of the atmosphere, it would also move at about 27,500 km/h (if we ignore atmospheric drag and all other influences). Try to catch that hook! If you can, you might as well go straight into orbit yourself.
edited Jan 20 at 13:46
Peter Mortensen
20817
answered Jan 20 at 8:27
DarkDustDarkDust
7,47433056
$endgroup$
14
$begingroup$
Reminds old good what-if.xkcd.com/157
$endgroup$
– val
Jan 20 at 10:38
10
$begingroup$
The "Rotating Skyhoook" concept en.wikipedia.org/wiki/Skyhook_(structure) seems relevant here
$endgroup$
– Steve Linton
Jan 20 at 11:10
6
$begingroup$
@val this one what-if.xkcd.com/58 is even better :P
$endgroup$
– paul23
Jan 20 at 17:42
2
$begingroup$
Not to mention that you'd need a heck of a heat shield for that pole at 27,500 km/h ;-) ...
$endgroup$
– Peter A. Schneider
Jan 20 at 20:54
2
$begingroup$
@val please tag it as "Obligatory what if" ;)
$endgroup$
– Barranka
Jan 21 at 2:11
|
show 2 more comments
$begingroup$
Because space isn't about going high; it's about going fast! For example, in a 400 km orbit (like ISS) you need a speed of about 27,500 km/h or 7.66 km per second.
So if you would extend a pole, winch or anything else into the lower parts of the atmosphere, it would also move at about 27,500 km/h (if we ignore atmospheric drag and all other influences). Try to catch that hook! If you can, you might as well go straight into orbit yourself.
edited Jan 20 at 13:46
Peter Mortensen
20817
answered Jan 20 at 8:27
DarkDustDarkDust
7,47433056
$endgroup$
14
$begingroup$
Reminds old good what-if.xkcd.com/157
$endgroup$
– val
Jan 20 at 10:38
10
$begingroup$
The "Rotating Skyhoook" concept en.wikipedia.org/wiki/Skyhook_(structure) seems relevant here
$endgroup$
– Steve Linton
Jan 20 at 11:10
6
$begingroup$
@val this one what-if.xkcd.com/58 is even better :P
$endgroup$
– paul23
Jan 20 at 17:42
2
$begingroup$
Not to mention that you'd need a heck of a heat shield for that pole at 27,500 km/h ;-) ...
$endgroup$
– Peter A. Schneider
Jan 20 at 20:54
2
$begingroup$
@val please tag it as "Obligatory what if" ;)
$endgroup$
– Barranka
Jan 21 at 2:11
|
show 2 more comments
$begingroup$
Because space isn't about going high; it's about going fast! For example, in a 400 km orbit (like ISS) you need a speed of about 27,500 km/h or 7.66 km per second.
So if you would extend a pole, winch or anything else into the lower parts of the atmosphere, it would also move at about 27,500 km/h (if we ignore atmospheric drag and all other influences). Try to catch that hook! If you can, you might as well go straight into orbit yourself.
edited Jan 20 at 13:46
Peter Mortensen
20817
answered Jan 20 at 8:27
DarkDustDarkDust
7,47433056
$endgroup$
Because space isn't about going high; it's about going fast! For example, in a 400 km orbit (like ISS) you need a speed of about 27,500 km/h or 7.66 km per second.
So if you would extend a pole, winch or anything else into the lower parts of the atmosphere, it would also move at about 27,500 km/h (if we ignore atmospheric drag and all other influences). Try to catch that hook! If you can, you might as well go straight into orbit yourself.
edited Jan 20 at 13:46
Peter Mortensen
20817
answered Jan 20 at 8:27
DarkDustDarkDust
7,47433056
edited Jan 20 at 13:46
Peter Mortensen
20817
edited Jan 20 at 13:46
Peter Mortensen
20817
edited Jan 20 at 13:46
Peter Mortensen
20817
20817
answered Jan 20 at 8:27
DarkDustDarkDust
7,47433056
answered Jan 20 at 8:27
DarkDustDarkDust
7,47433056
answered Jan 20 at 8:27
DarkDustDarkDust
7,47433056
7,47433056
14
$begingroup$
Reminds old good what-if.xkcd.com/157
$endgroup$
– val
Jan 20 at 10:38
10
$begingroup$
The "Rotating Skyhoook" concept en.wikipedia.org/wiki/Skyhook_(structure) seems relevant here
$endgroup$
– Steve Linton
Jan 20 at 11:10
6
$begingroup$
@val this one what-if.xkcd.com/58 is even better :P
$endgroup$
– paul23
Jan 20 at 17:42
2
$begingroup$
Not to mention that you'd need a heck of a heat shield for that pole at 27,500 km/h ;-) ...
$endgroup$
– Peter A. Schneider
Jan 20 at 20:54
2
$begingroup$
@val please tag it as "Obligatory what if" ;)
$endgroup$
– Barranka
Jan 21 at 2:11
|
show 2 more comments
14
$begingroup$
Reminds old good what-if.xkcd.com/157
$endgroup$
– val
Jan 20 at 10:38
10
$begingroup$
The "Rotating Skyhoook" concept en.wikipedia.org/wiki/Skyhook_(structure) seems relevant here
$endgroup$
– Steve Linton
Jan 20 at 11:10
6
$begingroup$
@val this one what-if.xkcd.com/58 is even better :P
$endgroup$
– paul23
Jan 20 at 17:42
2
$begingroup$
Not to mention that you'd need a heck of a heat shield for that pole at 27,500 km/h ;-) ...
$endgroup$
– Peter A. Schneider
Jan 20 at 20:54
2
$begingroup$
@val please tag it as "Obligatory what if" ;)
$endgroup$
– Barranka
Jan 21 at 2:11
14
14
$begingroup$
Reminds old good what-if.xkcd.com/157
$endgroup$
– val
Jan 20 at 10:38
$begingroup$
Reminds old good what-if.xkcd.com/157
$endgroup$
– val
Jan 20 at 10:38
10
10
$begingroup$
The "Rotating Skyhoook" concept en.wikipedia.org/wiki/Skyhook_(structure) seems relevant here
$endgroup$
– Steve Linton
Jan 20 at 11:10
$begingroup$
The "Rotating Skyhoook" concept en.wikipedia.org/wiki/Skyhook_(structure) seems relevant here
$endgroup$
– Steve Linton
Jan 20 at 11:10
6
6
$begingroup$
@val this one what-if.xkcd.com/58 is even better :P
$endgroup$
– paul23
Jan 20 at 17:42
$begingroup$
@val this one what-if.xkcd.com/58 is even better :P
$endgroup$
– paul23
Jan 20 at 17:42
2
2
$begingroup$
Not to mention that you'd need a heck of a heat shield for that pole at 27,500 km/h ;-) ...
$endgroup$
– Peter A. Schneider
Jan 20 at 20:54
$begingroup$
Not to mention that you'd need a heck of a heat shield for that pole at 27,500 km/h ;-) ...
$endgroup$
– Peter A. Schneider
Jan 20 at 20:54
2
2
$begingroup$
@val please tag it as "Obligatory what if" ;)
$endgroup$
– Barranka
Jan 21 at 2:11
$begingroup$
@val please tag it as "Obligatory what if" ;)
$endgroup$
– Barranka
Jan 21 at 2:11
|
show 2 more comments
$begingroup$
As an alternative to DarkDust's answer, if you start higher, at the classic altitude for space elevators, the end of your cable is indeed stationary to the air. But your cable needs to reach from geostationary orbit to the upper atmosphere, something like 35,700 km. The clipping off the last 20-60 km does not make a big difference in the overall monumental cost and complexity.
It is also worth noting that as you climb the cable, you will also pull the cable and station down. So your elevator will need to burn similar amounts of fuel to keep in orbit as if you had flown there (though possibly in a more efficient engine), unless you can balance things with loads coming down and being dropped as you go up.
edited Jan 20 at 13:47
Peter Mortensen
20817
answered Jan 20 at 8:41
GremlinWrangerGremlinWranger
2,533218
$endgroup$
$begingroup$
The fuel need is a good observation. I assume that the fuel is not needed as much for the climb -- hey, it's ony 100 miles! -- as for the acceleration to orbital speed, as vividly illustrated in the xkcd strip. If not compensated, accelerating the additional load would slow down the shuttle so that it changes to a lower, elliptical orbit.
$endgroup$
– Peter A. Schneider
Jan 21 at 9:01
$begingroup$
@PeterA.Schneider No, you'd need to go the entire 35,700 km. When you get there you are in circular orbit. Leaving after only 100 km would require you to lift impractially large amount of fuel.
$endgroup$
– kubanczyk
Jan 21 at 9:16
6
$begingroup$
@GremlinWranger Classic space elevator design has a counterweight far above GEO which pulls it back up without any propulsion. It uses up a tiny bit of Earth's rotation though :)
$endgroup$
– kubanczyk
Jan 21 at 9:42
$begingroup$
@kubanczyk My remark was meant for low earth orbits when a shuttle hauls in a load starting at airplane altitude. For geostationary orbits I'm not sure what needs more energy, the climb or the acceleration (probably the climb).
$endgroup$
– Peter A. Schneider
Jan 21 at 9:57
1
$begingroup$
@PeterA.SChneider Of course, you don't need to carry that energy with you, and you need no propellant; even with tech we already have (excluding the magical tether itself), that'd mean orders of magnitude lower energy costs than rockets. And then you consider that as you go up, something else can go down, and it neatly balances out.
$endgroup$
– Luaan
Jan 21 at 16:41
|
show 2 more comments
$begingroup$
As an alternative to DarkDust's answer, if you start higher, at the classic altitude for space elevators, the end of your cable is indeed stationary to the air. But your cable needs to reach from geostationary orbit to the upper atmosphere, something like 35,700 km. The clipping off the last 20-60 km does not make a big difference in the overall monumental cost and complexity.
It is also worth noting that as you climb the cable, you will also pull the cable and station down. So your elevator will need to burn similar amounts of fuel to keep in orbit as if you had flown there (though possibly in a more efficient engine), unless you can balance things with loads coming down and being dropped as you go up.
edited Jan 20 at 13:47
Peter Mortensen
20817
answered Jan 20 at 8:41
GremlinWrangerGremlinWranger
2,533218
$endgroup$
$begingroup$
The fuel need is a good observation. I assume that the fuel is not needed as much for the climb -- hey, it's ony 100 miles! -- as for the acceleration to orbital speed, as vividly illustrated in the xkcd strip. If not compensated, accelerating the additional load would slow down the shuttle so that it changes to a lower, elliptical orbit.
$endgroup$
– Peter A. Schneider
Jan 21 at 9:01
$begingroup$
@PeterA.Schneider No, you'd need to go the entire 35,700 km. When you get there you are in circular orbit. Leaving after only 100 km would require you to lift impractially large amount of fuel.
$endgroup$
– kubanczyk
Jan 21 at 9:16
6
$begingroup$
@GremlinWranger Classic space elevator design has a counterweight far above GEO which pulls it back up without any propulsion. It uses up a tiny bit of Earth's rotation though :)
$endgroup$
– kubanczyk
Jan 21 at 9:42
$begingroup$
@kubanczyk My remark was meant for low earth orbits when a shuttle hauls in a load starting at airplane altitude. For geostationary orbits I'm not sure what needs more energy, the climb or the acceleration (probably the climb).
$endgroup$
– Peter A. Schneider
Jan 21 at 9:57
1
$begingroup$
@PeterA.SChneider Of course, you don't need to carry that energy with you, and you need no propellant; even with tech we already have (excluding the magical tether itself), that'd mean orders of magnitude lower energy costs than rockets. And then you consider that as you go up, something else can go down, and it neatly balances out.
$endgroup$
– Luaan
Jan 21 at 16:41
|
show 2 more comments
$begingroup$
As an alternative to DarkDust's answer, if you start higher, at the classic altitude for space elevators, the end of your cable is indeed stationary to the air. But your cable needs to reach from geostationary orbit to the upper atmosphere, something like 35,700 km. The clipping off the last 20-60 km does not make a big difference in the overall monumental cost and complexity.
It is also worth noting that as you climb the cable, you will also pull the cable and station down. So your elevator will need to burn similar amounts of fuel to keep in orbit as if you had flown there (though possibly in a more efficient engine), unless you can balance things with loads coming down and being dropped as you go up.
edited Jan 20 at 13:47
Peter Mortensen
20817
answered Jan 20 at 8:41
GremlinWrangerGremlinWranger
2,533218
$endgroup$
As an alternative to DarkDust's answer, if you start higher, at the classic altitude for space elevators, the end of your cable is indeed stationary to the air. But your cable needs to reach from geostationary orbit to the upper atmosphere, something like 35,700 km. The clipping off the last 20-60 km does not make a big difference in the overall monumental cost and complexity.
It is also worth noting that as you climb the cable, you will also pull the cable and station down. So your elevator will need to burn similar amounts of fuel to keep in orbit as if you had flown there (though possibly in a more efficient engine), unless you can balance things with loads coming down and being dropped as you go up.
edited Jan 20 at 13:47
Peter Mortensen
20817
answered Jan 20 at 8:41
GremlinWrangerGremlinWranger
2,533218
edited Jan 20 at 13:47
Peter Mortensen
20817
edited Jan 20 at 13:47
Peter Mortensen
20817
edited Jan 20 at 13:47
Peter Mortensen
20817
20817
answered Jan 20 at 8:41
GremlinWrangerGremlinWranger
2,533218
answered Jan 20 at 8:41
GremlinWrangerGremlinWranger
2,533218
answered Jan 20 at 8:41
GremlinWrangerGremlinWranger
2,533218
2,533218
$begingroup$
The fuel need is a good observation. I assume that the fuel is not needed as much for the climb -- hey, it's ony 100 miles! -- as for the acceleration to orbital speed, as vividly illustrated in the xkcd strip. If not compensated, accelerating the additional load would slow down the shuttle so that it changes to a lower, elliptical orbit.
$endgroup$
– Peter A. Schneider
Jan 21 at 9:01
$begingroup$
@PeterA.Schneider No, you'd need to go the entire 35,700 km. When you get there you are in circular orbit. Leaving after only 100 km would require you to lift impractially large amount of fuel.
$endgroup$
– kubanczyk
Jan 21 at 9:16
6
$begingroup$
@GremlinWranger Classic space elevator design has a counterweight far above GEO which pulls it back up without any propulsion. It uses up a tiny bit of Earth's rotation though :)
$endgroup$
– kubanczyk
Jan 21 at 9:42
$begingroup$
@kubanczyk My remark was meant for low earth orbits when a shuttle hauls in a load starting at airplane altitude. For geostationary orbits I'm not sure what needs more energy, the climb or the acceleration (probably the climb).
$endgroup$
– Peter A. Schneider
Jan 21 at 9:57
1
$begingroup$
@PeterA.SChneider Of course, you don't need to carry that energy with you, and you need no propellant; even with tech we already have (excluding the magical tether itself), that'd mean orders of magnitude lower energy costs than rockets. And then you consider that as you go up, something else can go down, and it neatly balances out.
$endgroup$
– Luaan
Jan 21 at 16:41
|
show 2 more comments
$begingroup$
The fuel need is a good observation. I assume that the fuel is not needed as much for the climb -- hey, it's ony 100 miles! -- as for the acceleration to orbital speed, as vividly illustrated in the xkcd strip. If not compensated, accelerating the additional load would slow down the shuttle so that it changes to a lower, elliptical orbit.
$endgroup$
– Peter A. Schneider
Jan 21 at 9:01
$begingroup$
@PeterA.Schneider No, you'd need to go the entire 35,700 km. When you get there you are in circular orbit. Leaving after only 100 km would require you to lift impractially large amount of fuel.
$endgroup$
– kubanczyk
Jan 21 at 9:16
6
$begingroup$
@GremlinWranger Classic space elevator design has a counterweight far above GEO which pulls it back up without any propulsion. It uses up a tiny bit of Earth's rotation though :)
$endgroup$
– kubanczyk
Jan 21 at 9:42
$begingroup$
@kubanczyk My remark was meant for low earth orbits when a shuttle hauls in a load starting at airplane altitude. For geostationary orbits I'm not sure what needs more energy, the climb or the acceleration (probably the climb).
$endgroup$
– Peter A. Schneider
Jan 21 at 9:57
1
$begingroup$
@PeterA.SChneider Of course, you don't need to carry that energy with you, and you need no propellant; even with tech we already have (excluding the magical tether itself), that'd mean orders of magnitude lower energy costs than rockets. And then you consider that as you go up, something else can go down, and it neatly balances out.
$endgroup$
– Luaan
Jan 21 at 16:41
$begingroup$
The fuel need is a good observation. I assume that the fuel is not needed as much for the climb -- hey, it's ony 100 miles! -- as for the acceleration to orbital speed, as vividly illustrated in the xkcd strip. If not compensated, accelerating the additional load would slow down the shuttle so that it changes to a lower, elliptical orbit.
$endgroup$
– Peter A. Schneider
Jan 21 at 9:01
$begingroup$
The fuel need is a good observation. I assume that the fuel is not needed as much for the climb -- hey, it's ony 100 miles! -- as for the acceleration to orbital speed, as vividly illustrated in the xkcd strip. If not compensated, accelerating the additional load would slow down the shuttle so that it changes to a lower, elliptical orbit.
$endgroup$
– Peter A. Schneider
Jan 21 at 9:01
$begingroup$
@PeterA.Schneider No, you'd need to go the entire 35,700 km. When you get there you are in circular orbit. Leaving after only 100 km would require you to lift impractially large amount of fuel.
$endgroup$
– kubanczyk
Jan 21 at 9:16
$begingroup$
@PeterA.Schneider No, you'd need to go the entire 35,700 km. When you get there you are in circular orbit. Leaving after only 100 km would require you to lift impractially large amount of fuel.
$endgroup$
– kubanczyk
Jan 21 at 9:16
6
6
$begingroup$
@GremlinWranger Classic space elevator design has a counterweight far above GEO which pulls it back up without any propulsion. It uses up a tiny bit of Earth's rotation though :)
$endgroup$
– kubanczyk
Jan 21 at 9:42
$begingroup$
@GremlinWranger Classic space elevator design has a counterweight far above GEO which pulls it back up without any propulsion. It uses up a tiny bit of Earth's rotation though :)
$endgroup$
– kubanczyk
Jan 21 at 9:42
$begingroup$
@kubanczyk My remark was meant for low earth orbits when a shuttle hauls in a load starting at airplane altitude. For geostationary orbits I'm not sure what needs more energy, the climb or the acceleration (probably the climb).
$endgroup$
– Peter A. Schneider
Jan 21 at 9:57
$begingroup$
@kubanczyk My remark was meant for low earth orbits when a shuttle hauls in a load starting at airplane altitude. For geostationary orbits I'm not sure what needs more energy, the climb or the acceleration (probably the climb).
$endgroup$
– Peter A. Schneider
Jan 21 at 9:57
1
1
$begingroup$
@PeterA.SChneider Of course, you don't need to carry that energy with you, and you need no propellant; even with tech we already have (excluding the magical tether itself), that'd mean orders of magnitude lower energy costs than rockets. And then you consider that as you go up, something else can go down, and it neatly balances out.
$endgroup$
– Luaan
Jan 21 at 16:41
$begingroup$
@PeterA.SChneider Of course, you don't need to carry that energy with you, and you need no propellant; even with tech we already have (excluding the magical tether itself), that'd mean orders of magnitude lower energy costs than rockets. And then you consider that as you go up, something else can go down, and it neatly balances out.
$endgroup$
– Luaan
Jan 21 at 16:41
|
show 2 more comments
$begingroup$
Have you ever flown a kite with a tail? The tail flies almost horizontally, kept up by the wind. It would hang at a steeper angle if you made it heavy, but what would then happen to your kite?
Now imagine that wind to be Mach 20 or so. Even if you had a rope made of unobtainium and hence able to withstand the heat created by atmospheric compression it would always "skim" on top of the atmosphere, the end being horizontal.
The concept is unviable in almost all aspects.
answered Jan 21 at 6:49
Peter A. SchneiderPeter A. Schneider
26317
$endgroup$
add a comment |
$begingroup$
Have you ever flown a kite with a tail? The tail flies almost horizontally, kept up by the wind. It would hang at a steeper angle if you made it heavy, but what would then happen to your kite?
Now imagine that wind to be Mach 20 or so. Even if you had a rope made of unobtainium and hence able to withstand the heat created by atmospheric compression it would always "skim" on top of the atmosphere, the end being horizontal.
The concept is unviable in almost all aspects.
answered Jan 21 at 6:49
Peter A. SchneiderPeter A. Schneider
26317
$endgroup$
add a comment |
$begingroup$
Have you ever flown a kite with a tail? The tail flies almost horizontally, kept up by the wind. It would hang at a steeper angle if you made it heavy, but what would then happen to your kite?
Now imagine that wind to be Mach 20 or so. Even if you had a rope made of unobtainium and hence able to withstand the heat created by atmospheric compression it would always "skim" on top of the atmosphere, the end being horizontal.
The concept is unviable in almost all aspects.
answered Jan 21 at 6:49
Peter A. SchneiderPeter A. Schneider
26317
$endgroup$
Have you ever flown a kite with a tail? The tail flies almost horizontally, kept up by the wind. It would hang at a steeper angle if you made it heavy, but what would then happen to your kite?
Now imagine that wind to be Mach 20 or so. Even if you had a rope made of unobtainium and hence able to withstand the heat created by atmospheric compression it would always "skim" on top of the atmosphere, the end being horizontal.
The concept is unviable in almost all aspects.
answered Jan 21 at 6:49
Peter A. SchneiderPeter A. Schneider
26317
answered Jan 21 at 6:49
Peter A. SchneiderPeter A. Schneider
26317
answered Jan 21 at 6:49
Peter A. SchneiderPeter A. Schneider
26317
answered Jan 21 at 6:49
Peter A. SchneiderPeter A. Schneider
26317
26317
add a comment |
add a comment |
$begingroup$
The end of the cable would be destroyed by the heat of reentry from a low orbit when reaching the height where planes may fly.
But if you try to drop a cable from a low orbit it would not drop, it would stay in orbit. There is no dropping of things in zero gravity.
Pulling up a load to the spacecraft in low orbit would slow down the spacecraft. It would lose height and fuel is needed to maintain orbit.
As written in the other answers, a plane is much too slow to catch the hook. Without a heatshield, the plane would be destroyed by the necessary speed. With a heatshield the plane would need a lot of fuel to maintain that speed.
edited Jan 20 at 17:56
Community♦
1
answered Jan 20 at 10:20
UweUwe
10.6k22956
$endgroup$
$begingroup$
You could fire the cable in the retrograde direction out of a cannon. This would add complexity and cost, and does nothing to address the other challenges.
$endgroup$
– BobTheAverage
Jan 20 at 21:41
$begingroup$
@BobTheAverage A cannon is much too slow compared to orbital speed.
$endgroup$
– Uwe
Jan 20 at 22:00
$begingroup$
Use a railgun instead.
$endgroup$
– BobTheAverage
Jan 20 at 22:52
$begingroup$
I suppose a modest push towards the ground would suffice to get the cable going; the orbital speed is too low for lower orbits so that the cable would start to "fall" once it's left the current orbit. It would then experience drag fairly soon (which makes it clear that it would never reach the lower atmosphere unless it is propelled at the lower end and probably in between, too).
$endgroup$
– Peter A. Schneider
Jan 21 at 6:37
add a comment |
$begingroup$
The end of the cable would be destroyed by the heat of reentry from a low orbit when reaching the height where planes may fly.
But if you try to drop a cable from a low orbit it would not drop, it would stay in orbit. There is no dropping of things in zero gravity.
Pulling up a load to the spacecraft in low orbit would slow down the spacecraft. It would lose height and fuel is needed to maintain orbit.
As written in the other answers, a plane is much too slow to catch the hook. Without a heatshield, the plane would be destroyed by the necessary speed. With a heatshield the plane would need a lot of fuel to maintain that speed.
edited Jan 20 at 17:56
Community♦
1
answered Jan 20 at 10:20
UweUwe
10.6k22956
$endgroup$
$begingroup$
You could fire the cable in the retrograde direction out of a cannon. This would add complexity and cost, and does nothing to address the other challenges.
$endgroup$
– BobTheAverage
Jan 20 at 21:41
$begingroup$
@BobTheAverage A cannon is much too slow compared to orbital speed.
$endgroup$
– Uwe
Jan 20 at 22:00
$begingroup$
Use a railgun instead.
$endgroup$
– BobTheAverage
Jan 20 at 22:52
$begingroup$
I suppose a modest push towards the ground would suffice to get the cable going; the orbital speed is too low for lower orbits so that the cable would start to "fall" once it's left the current orbit. It would then experience drag fairly soon (which makes it clear that it would never reach the lower atmosphere unless it is propelled at the lower end and probably in between, too).
$endgroup$
– Peter A. Schneider
Jan 21 at 6:37
add a comment |
$begingroup$
The end of the cable would be destroyed by the heat of reentry from a low orbit when reaching the height where planes may fly.
But if you try to drop a cable from a low orbit it would not drop, it would stay in orbit. There is no dropping of things in zero gravity.
Pulling up a load to the spacecraft in low orbit would slow down the spacecraft. It would lose height and fuel is needed to maintain orbit.
As written in the other answers, a plane is much too slow to catch the hook. Without a heatshield, the plane would be destroyed by the necessary speed. With a heatshield the plane would need a lot of fuel to maintain that speed.
edited Jan 20 at 17:56
Community♦
1
answered Jan 20 at 10:20
UweUwe
10.6k22956
$endgroup$
The end of the cable would be destroyed by the heat of reentry from a low orbit when reaching the height where planes may fly.
But if you try to drop a cable from a low orbit it would not drop, it would stay in orbit. There is no dropping of things in zero gravity.
Pulling up a load to the spacecraft in low orbit would slow down the spacecraft. It would lose height and fuel is needed to maintain orbit.
As written in the other answers, a plane is much too slow to catch the hook. Without a heatshield, the plane would be destroyed by the necessary speed. With a heatshield the plane would need a lot of fuel to maintain that speed.
edited Jan 20 at 17:56
Community♦
1
answered Jan 20 at 10:20
UweUwe
10.6k22956
edited Jan 20 at 17:56
Community♦
1
edited Jan 20 at 17:56
Community♦
1
edited Jan 20 at 17:56
Community♦
1
1
answered Jan 20 at 10:20
UweUwe
10.6k22956
answered Jan 20 at 10:20
UweUwe
10.6k22956
answered Jan 20 at 10:20
UweUwe
10.6k22956
10.6k22956
$begingroup$
You could fire the cable in the retrograde direction out of a cannon. This would add complexity and cost, and does nothing to address the other challenges.
$endgroup$
– BobTheAverage
Jan 20 at 21:41
$begingroup$
@BobTheAverage A cannon is much too slow compared to orbital speed.
$endgroup$
– Uwe
Jan 20 at 22:00
$begingroup$
Use a railgun instead.
$endgroup$
– BobTheAverage
Jan 20 at 22:52
$begingroup$
I suppose a modest push towards the ground would suffice to get the cable going; the orbital speed is too low for lower orbits so that the cable would start to "fall" once it's left the current orbit. It would then experience drag fairly soon (which makes it clear that it would never reach the lower atmosphere unless it is propelled at the lower end and probably in between, too).
$endgroup$
– Peter A. Schneider
Jan 21 at 6:37
add a comment |
$begingroup$
You could fire the cable in the retrograde direction out of a cannon. This would add complexity and cost, and does nothing to address the other challenges.
$endgroup$
– BobTheAverage
Jan 20 at 21:41
$begingroup$
@BobTheAverage A cannon is much too slow compared to orbital speed.
$endgroup$
– Uwe
Jan 20 at 22:00
$begingroup$
Use a railgun instead.
$endgroup$
– BobTheAverage
Jan 20 at 22:52
$begingroup$
I suppose a modest push towards the ground would suffice to get the cable going; the orbital speed is too low for lower orbits so that the cable would start to "fall" once it's left the current orbit. It would then experience drag fairly soon (which makes it clear that it would never reach the lower atmosphere unless it is propelled at the lower end and probably in between, too).
$endgroup$
– Peter A. Schneider
Jan 21 at 6:37
$begingroup$
You could fire the cable in the retrograde direction out of a cannon. This would add complexity and cost, and does nothing to address the other challenges.
$endgroup$
– BobTheAverage
Jan 20 at 21:41
$begingroup$
You could fire the cable in the retrograde direction out of a cannon. This would add complexity and cost, and does nothing to address the other challenges.
$endgroup$
– BobTheAverage
Jan 20 at 21:41
$begingroup$
@BobTheAverage A cannon is much too slow compared to orbital speed.
$endgroup$
– Uwe
Jan 20 at 22:00
$begingroup$
@BobTheAverage A cannon is much too slow compared to orbital speed.
$endgroup$
– Uwe
Jan 20 at 22:00
$begingroup$
Use a railgun instead.
$endgroup$
– BobTheAverage
Jan 20 at 22:52
$begingroup$
Use a railgun instead.
$endgroup$
– BobTheAverage
Jan 20 at 22:52
$begingroup$
I suppose a modest push towards the ground would suffice to get the cable going; the orbital speed is too low for lower orbits so that the cable would start to "fall" once it's left the current orbit. It would then experience drag fairly soon (which makes it clear that it would never reach the lower atmosphere unless it is propelled at the lower end and probably in between, too).
$endgroup$
– Peter A. Schneider
Jan 21 at 6:37
$begingroup$
I suppose a modest push towards the ground would suffice to get the cable going; the orbital speed is too low for lower orbits so that the cable would start to "fall" once it's left the current orbit. It would then experience drag fairly soon (which makes it clear that it would never reach the lower atmosphere unless it is propelled at the lower end and probably in between, too).
$endgroup$
– Peter A. Schneider
Jan 21 at 6:37
add a comment |
$begingroup$
There's plenty of answers discussing the problems with having a cable hanging into the Earth's atmosphere - the drag, the heat, the resistence on the orbiting craft, etc. However, I don't see much consideration of the orbital mechanics of the scenario, which would be an issue even if there were no atmosphere.
Simply put, once a craft is in orbit, all that keeps it there is its speed. If you had a craft that could somehow catch the bottom end of the cable at ground level (as it zips by at around twenty-something times the speed of sound), then you have a massive difference in speeds to resolve. If your cable was somehow strong enough (and/or stretchy enough) to bring the lower craft up to speed without breaking the cable, and the contents of each craft were somehow protected from accelerations that would turn cargo and crew into paste, you're now about to run into conservation of momentum. As the cable speeds you up, you're slowing the upper craft down... which means that it will stop orbiting and fall back down.
Even if you put your upper craft in geostationary orbit* (and lengthen the cable considerably to still reach the ground), as you climb up the cable, you're still pulling the upper craft down. Precisely how this works out depends on your relative masses; if you're about the same mass then you'd meet somewhere in the middle. Theoretically you could then push the upper craft down, perhaps by extending a massive pushrod, to continue on your way to orbit, at which point you've invented a rudimentary space elevator. Indeed, some space elevator concepts involve having something very massive (like a captured asteroid) in geostationary orbit*, with a huge tether to the ground - the huge mass means huge momentum and inertia, which mean that its position isn't affected as much by things climbing up to it; an effect that can be further minimised by sending something back down to balance out each thing that comes up.
* Actually, it would have to be a little above geostationary, so that when the mass of the cable is taken into account, the centre of mass is at geostationary altitude. There's also challenges to overcome with regards to the tendency of the whole system to want to rotate relative to the Earth, so that the tether breaks away and drifts out into space, but that's tangential to the question.
answered Jan 22 at 11:27
anaximanderanaximander
1405
$endgroup$
add a comment |
$begingroup$
There's plenty of answers discussing the problems with having a cable hanging into the Earth's atmosphere - the drag, the heat, the resistence on the orbiting craft, etc. However, I don't see much consideration of the orbital mechanics of the scenario, which would be an issue even if there were no atmosphere.
Simply put, once a craft is in orbit, all that keeps it there is its speed. If you had a craft that could somehow catch the bottom end of the cable at ground level (as it zips by at around twenty-something times the speed of sound), then you have a massive difference in speeds to resolve. If your cable was somehow strong enough (and/or stretchy enough) to bring the lower craft up to speed without breaking the cable, and the contents of each craft were somehow protected from accelerations that would turn cargo and crew into paste, you're now about to run into conservation of momentum. As the cable speeds you up, you're slowing the upper craft down... which means that it will stop orbiting and fall back down.
Even if you put your upper craft in geostationary orbit* (and lengthen the cable considerably to still reach the ground), as you climb up the cable, you're still pulling the upper craft down. Precisely how this works out depends on your relative masses; if you're about the same mass then you'd meet somewhere in the middle. Theoretically you could then push the upper craft down, perhaps by extending a massive pushrod, to continue on your way to orbit, at which point you've invented a rudimentary space elevator. Indeed, some space elevator concepts involve having something very massive (like a captured asteroid) in geostationary orbit*, with a huge tether to the ground - the huge mass means huge momentum and inertia, which mean that its position isn't affected as much by things climbing up to it; an effect that can be further minimised by sending something back down to balance out each thing that comes up.
* Actually, it would have to be a little above geostationary, so that when the mass of the cable is taken into account, the centre of mass is at geostationary altitude. There's also challenges to overcome with regards to the tendency of the whole system to want to rotate relative to the Earth, so that the tether breaks away and drifts out into space, but that's tangential to the question.
answered Jan 22 at 11:27
anaximanderanaximander
1405
$endgroup$
add a comment |
$begingroup$
There's plenty of answers discussing the problems with having a cable hanging into the Earth's atmosphere - the drag, the heat, the resistence on the orbiting craft, etc. However, I don't see much consideration of the orbital mechanics of the scenario, which would be an issue even if there were no atmosphere.
Simply put, once a craft is in orbit, all that keeps it there is its speed. If you had a craft that could somehow catch the bottom end of the cable at ground level (as it zips by at around twenty-something times the speed of sound), then you have a massive difference in speeds to resolve. If your cable was somehow strong enough (and/or stretchy enough) to bring the lower craft up to speed without breaking the cable, and the contents of each craft were somehow protected from accelerations that would turn cargo and crew into paste, you're now about to run into conservation of momentum. As the cable speeds you up, you're slowing the upper craft down... which means that it will stop orbiting and fall back down.
Even if you put your upper craft in geostationary orbit* (and lengthen the cable considerably to still reach the ground), as you climb up the cable, you're still pulling the upper craft down. Precisely how this works out depends on your relative masses; if you're about the same mass then you'd meet somewhere in the middle. Theoretically you could then push the upper craft down, perhaps by extending a massive pushrod, to continue on your way to orbit, at which point you've invented a rudimentary space elevator. Indeed, some space elevator concepts involve having something very massive (like a captured asteroid) in geostationary orbit*, with a huge tether to the ground - the huge mass means huge momentum and inertia, which mean that its position isn't affected as much by things climbing up to it; an effect that can be further minimised by sending something back down to balance out each thing that comes up.
* Actually, it would have to be a little above geostationary, so that when the mass of the cable is taken into account, the centre of mass is at geostationary altitude. There's also challenges to overcome with regards to the tendency of the whole system to want to rotate relative to the Earth, so that the tether breaks away and drifts out into space, but that's tangential to the question.
answered Jan 22 at 11:27
anaximanderanaximander
1405
$endgroup$
There's plenty of answers discussing the problems with having a cable hanging into the Earth's atmosphere - the drag, the heat, the resistence on the orbiting craft, etc. However, I don't see much consideration of the orbital mechanics of the scenario, which would be an issue even if there were no atmosphere.
Simply put, once a craft is in orbit, all that keeps it there is its speed. If you had a craft that could somehow catch the bottom end of the cable at ground level (as it zips by at around twenty-something times the speed of sound), then you have a massive difference in speeds to resolve. If your cable was somehow strong enough (and/or stretchy enough) to bring the lower craft up to speed without breaking the cable, and the contents of each craft were somehow protected from accelerations that would turn cargo and crew into paste, you're now about to run into conservation of momentum. As the cable speeds you up, you're slowing the upper craft down... which means that it will stop orbiting and fall back down.
Even if you put your upper craft in geostationary orbit* (and lengthen the cable considerably to still reach the ground), as you climb up the cable, you're still pulling the upper craft down. Precisely how this works out depends on your relative masses; if you're about the same mass then you'd meet somewhere in the middle. Theoretically you could then push the upper craft down, perhaps by extending a massive pushrod, to continue on your way to orbit, at which point you've invented a rudimentary space elevator. Indeed, some space elevator concepts involve having something very massive (like a captured asteroid) in geostationary orbit*, with a huge tether to the ground - the huge mass means huge momentum and inertia, which mean that its position isn't affected as much by things climbing up to it; an effect that can be further minimised by sending something back down to balance out each thing that comes up.
* Actually, it would have to be a little above geostationary, so that when the mass of the cable is taken into account, the centre of mass is at geostationary altitude. There's also challenges to overcome with regards to the tendency of the whole system to want to rotate relative to the Earth, so that the tether breaks away and drifts out into space, but that's tangential to the question.
answered Jan 22 at 11:27
anaximanderanaximander
1405
answered Jan 22 at 11:27
anaximanderanaximander
1405
answered Jan 22 at 11:27
anaximanderanaximander
1405
answered Jan 22 at 11:27
anaximanderanaximander
1405
1405
add a comment |
add a comment |
$begingroup$
If the shuttle is at 100 km, and planes fly at say 20 km, that would mean the cable would have to be 80 km in length. Such a long cable would produce so much atmospheric drag that the shuttle would need to constantly burn to counter it, otherwise it would de-orbit pretty quickly.
answered Jan 21 at 14:42
stackzebrastackzebra
2063
$endgroup$
1
$begingroup$
A shuttle at 100 km would deorbit very quickly anyway, the final orbit height of the Space Shuttle was about 380 km.
$endgroup$
– Uwe
Jan 21 at 15:22
add a comment |
$begingroup$
If the shuttle is at 100 km, and planes fly at say 20 km, that would mean the cable would have to be 80 km in length. Such a long cable would produce so much atmospheric drag that the shuttle would need to constantly burn to counter it, otherwise it would de-orbit pretty quickly.
answered Jan 21 at 14:42
stackzebrastackzebra
2063
$endgroup$
1
$begingroup$
A shuttle at 100 km would deorbit very quickly anyway, the final orbit height of the Space Shuttle was about 380 km.
$endgroup$
– Uwe
Jan 21 at 15:22
add a comment |
$begingroup$
If the shuttle is at 100 km, and planes fly at say 20 km, that would mean the cable would have to be 80 km in length. Such a long cable would produce so much atmospheric drag that the shuttle would need to constantly burn to counter it, otherwise it would de-orbit pretty quickly.
answered Jan 21 at 14:42
stackzebrastackzebra
2063
$endgroup$
If the shuttle is at 100 km, and planes fly at say 20 km, that would mean the cable would have to be 80 km in length. Such a long cable would produce so much atmospheric drag that the shuttle would need to constantly burn to counter it, otherwise it would de-orbit pretty quickly.
answered Jan 21 at 14:42
stackzebrastackzebra
2063
answered Jan 21 at 14:42
stackzebrastackzebra
2063
answered Jan 21 at 14:42
stackzebrastackzebra
2063
answered Jan 21 at 14:42
stackzebrastackzebra
2063
2063
1
$begingroup$
A shuttle at 100 km would deorbit very quickly anyway, the final orbit height of the Space Shuttle was about 380 km.
$endgroup$
– Uwe
Jan 21 at 15:22
add a comment |
1
$begingroup$
A shuttle at 100 km would deorbit very quickly anyway, the final orbit height of the Space Shuttle was about 380 km.
$endgroup$
– Uwe
Jan 21 at 15:22
1
1
$begingroup$
A shuttle at 100 km would deorbit very quickly anyway, the final orbit height of the Space Shuttle was about 380 km.
$endgroup$
– Uwe
Jan 21 at 15:22
$begingroup$
A shuttle at 100 km would deorbit very quickly anyway, the final orbit height of the Space Shuttle was about 380 km.
$endgroup$
– Uwe
Jan 21 at 15:22
add a comment |
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In fact you can do this - in theory, anyway. The trick is that the cable has to be rotating, so that it dips into the atmosphere and is for a brief time moving slowly WRT the Earth's surface, so that an airplane could dock with it. See e.g. Momentum Exchange Tethers for more info:
$endgroup$
– jamesqf
Jan 20 at 17:58
6
$begingroup$
Worth reading Neal Stephenson's "Seveneves" for an excellent hard sci-fi description of a skyhook, centred at Geostationary orbit, and counter-rotating so the tether end reaches almost to ground and is briefly almost stationary.
$endgroup$
– Rory Alsop
Jan 21 at 14:32
1
$begingroup$
The first chapter of Soonish is about potentially cheap ways to get to space, including things like space elevators.
$endgroup$
– Barmar
Jan 22 at 1:31
$begingroup$
thank you for all your answers. i only just discovered them as i was expecting to get a notification but didn't. you are all very kind to take the time to explain it to me. thanks
$endgroup$
– SpaceMonkey
Jan 30 at 5:55