5 ms·

Wait... most of the energy of going into orbit is in accelerating to orbital speed (lateral), not getting away from earth (vertical). An elevator's top, in a g

by hyp0 12y ago

Wait... most of the energy of going into orbit is in accelerating to orbital speed (lateral), not getting away from earth (vertical). An elevator's top, in a geosynchronous orbit) rotates much faster than the bottom (describing a larger circle, covered in the same time). Therefore, a payload moving up the elevator would either need to be accelerated laterally somehow, or it would bend the elevator over. There's no getting away from needing the energy for lateral acceleration - it must be supplied somehow. Piping fuel up (instead of blasting it up in a rocket) may be more efficient, but consider that its mass too will need lateral acceleration. \edit Fuel in the form of electricity plus reaction mass, to be sent at extremely high velocity, would probably have the least mass, and so minimize this effect (I think, high voltage electricity, to reduce the current, because the electron mass itself may have an effect at these scales!). Another alternative, for the reaction mass, is earth-encompassing rings at different altitudes. You can accelerate it without throwing it away. Of course, at lower altitudes, you can just use air... maybe there's even enough (for this purpose) at quite high altitudes... Finally, I think the basic solution has been a counter-weight: as one goes up (needs to speed up to orbital velocity), one comes down (needs to slow down from orbital velocity). If it was just one elevator each, the accelerations would be imparted at different points, twisting the evelator. But I guess a series of elevators - or even a continous elevator (like a conveyer-belt... or escalator...) would smoothen out this effect as much as you like, and only needs electricity to power it... and, any acceleration of the electron mass would also balance out, because they also go up and down in a circuit (or down and up for space solar panels). Sorry, nothing to see here. EDIT I see now a reply mentions this solution, upvoted.

Florin_Andrei 12y ago
What if there's constant up/down traffic on two opposite sides of the ribbon? Wouldn't ascending cars try to pull the ribbon "back" (from orbital direction), whereas descending cars would try to push the ribbon "forward"? I'm pretty tired now and I can't think very well, but my gut feeling tells me the two effects would compensate each other to a pretty large fraction of 100%.
- desdiv 12y ago
  But where will that extra downward traffic come from? The total weight of all space debris is only around 6000 tons[0] so that'll get exhausted within a matter of years (assuming we're even capable of collecting said debris). The only other alternative seems to be pulling near-Earth asteroids into geosynchronous orbit and using that. [0] http://www.dnaindia.com/analysis/comment-space-debris-constant-threat-to-life-on-earth-1632965 http://www.dnaindia.com/analysis/comment-space-debris-consta...
  - hyp0 12y ago
    There could be bounty hunters, who wrangle asteriods and other debris to serve the ever-hungry fleet of elevators. Also, there's legit asteriod/moon mining possibilities to at least partially counter the up-traffic.
mrfusion 12y ago
I thought centrifugal force held it straight up and you're taking a tiny amount of energy from the earths rotation when sending up payloads. Please correct me if I'm wrong.
- deleted 12y ago
  [deleted]
- idlewords 12y ago
  This is the correct answer. Angular momentum in the system is conserved; as you move mass up the space elevator, the earth spins (imperceptibly) slower. Note that the tether only moves at orbital velocity at one point. Below that point, if you want to get into orbit, you must accelerate laterally after releasing the tether.
  - hyp0 12y ago
    So the cable pulls the payload across, to (lateral) orbital speed: this was my point about the payload bending the cable over. 1. The cable has to be strong enough to do this (which may be reasonable, given how strong it needs to be anyway - plus it's in tension) 2. The cable will still be bent over, by a force imparted at the payload. If the bottom is attached and the top is "fixed" in geostationary orbit, the force will move the payload laterally (lagging the orbit - west), looking like a kind of arrow, or "V" on its side. The force on the cable itself might be OK, as the force is in tension; and the curve of the bend might be gradual enough. (Or maybe it would end up straight, but tilted westward?) It will also pull on both earth and satellite, slowing both.
gliese1337 12y ago
Therefore, a payload moving up the elevator would either need to be accelerated laterally somehow, or it would bend the elevator over. Or both. A payload moving up the elevator does pull on the cable, tending to slow it down. The cable, however, being massive and under very high tension, pulls back, accelerating the payload. If the cable is not anchored to the Earth, this causes the cable to slow down, and wobble a bit since the force is applied sequentially along the cable's length, rather than always on the center of mass, though tidal effects will tend to damp the wobble and keep it oriented practically straight up and down. For any one payload, the effect is negligible, as a space elevator cable would be freakin' massive, but without some sort of active stationkeeping it would eventually come down. If the cable is anchored to the Earth, however, as soon as it begins to be pulled over even a little bit by the rising payload, the lateral force is transmitted to the ground. The cable pulls on the Earth and the Earth pulls back, re-accelerating the cable with no need for active stationkeeping and ever so slightly slowing the rotation of the Earth. Since the Earth is Gigantic and has ridiculously large quantities of angular momentum and rotational kinetic energy, no one will ever notice this loss in practice.
- lotsofmangos 12y ago
  I like the extension to this which is to make a cable that stretches much further out than geostationary, then you clip payloads to the cable above geostationary and let the earth catapult them up the cable and slingshot them into the outer solar system.
Derbasti 12y ago
the math actually works, given a strong enough cable. You would lower a tether from geosynchronous orbit, and simultaneously extend a tether into space. Thus, the total mass stays centered in geosynchronous orbit, and the lower tether gets pulled towards earth because of gravity, whole the upper tether gets pulled away because of centrifugal forces. Thus, in the end, you have a tether that lightly touches the ground without force and stays in the same place, and an outer tether that extends about a third of the way out to the moon. The tensile forces would be strongest in geosynchronous orbit and weakest at the ends of the tether. Another benefit of this construction is that you can use the outer tether to launch spaceships into space at escape velocity without any need for propulsion. You can even use tethers on other planets to catch such payloads again. This construction does not eliminate the need for propulsion to reach orbit. But, you now can push against the tether instead of slippery and ever-thinning air, and you don't need to worry about lateral acceleration.
DavidSJ 12y ago
Wait... most of the energy of going into orbit is in accelerating to orbital speed (lateral), not getting away from earth (vertical). That's true for Low Earth Orbit, but as the radius of a circular orbit increases the kinetic energy decreases and the potential energy increases. On the surface of the Earth, we have: -(398600 km^3/s^2 [1]) / (6378 km [2])) = -62.5 km^2/s^2 of potential energy (0.465 km/s [3])^2/2 = 0.1 km^2/s^2 of kinetic energy At LEO, we have: -(398600 km^3/s^2 [1]) / (6678 km [4])) = -59.7 km^2/s^2 of potential energy (7.8 km/s [5])^2/2 = 30.4 km^2/s^2 of kinetic energy At GEO, we have: -(398600 km^3/s^2 [1]) / (42164 km [6])) = -9.5 km^2/s^2 of potential energy (3.07 km/s [7])^2/2 = 4.7 km^2/s^2 of kinetic energy Notice how LEO has only 2.8 km^2/s^2 of additional potential energy compared to the surface, but 30.3 km^2/s^2 of additional kinetic energy. However, at GEO there is 53 km^2/s^2 additional potential energy compared to the surface and only 4.6 km^2/s^2 additional kinetic energy. [1] Earth's gravitational parameter [2] Earth's equatorial radius [3] Earth's equatorial rotation speed [4] A 300 km altitude orbit [5] Velocity at LEO [6] GEO radius [7] Velocity at GEO

by hyp0 12y ago

Florin_Andrei 12y ago

What if there's constant up/down traffic on two opposite sides of the ribbon? Wouldn't ascending cars try to pull the ribbon "back" (from orbital direction), whereas descending cars would try to push the ribbon "forward"? I'm pretty tired now and I can't think very well, but my gut feeling tells me the two effects would compensate each other to a pretty large fraction of 100%.

desdiv 12y ago
But where will that extra downward traffic come from? The total weight of all space debris is only around 6000 tons[0] so that'll get exhausted within a matter of years (assuming we're even capable of collecting said debris). The only other alternative seems to be pulling near-Earth asteroids into geosynchronous orbit and using that. [0] http://www.dnaindia.com/analysis/comment-space-debris-constant-threat-to-life-on-earth-1632965 http://www.dnaindia.com/analysis/comment-space-debris-consta...
- hyp0 12y ago
  There could be bounty hunters, who wrangle asteriods and other debris to serve the ever-hungry fleet of elevators. Also, there's legit asteriod/moon mining possibilities to at least partially counter the up-traffic.

mrfusion 12y ago

I thought centrifugal force held it straight up and you're taking a tiny amount of energy from the earths rotation when sending up payloads. Please correct me if I'm wrong.

deleted 12y ago
[deleted]
idlewords 12y ago
This is the correct answer. Angular momentum in the system is conserved; as you move mass up the space elevator, the earth spins (imperceptibly) slower. Note that the tether only moves at orbital velocity at one point. Below that point, if you want to get into orbit, you must accelerate laterally after releasing the tether.
- hyp0 12y ago
  So the cable pulls the payload across, to (lateral) orbital speed: this was my point about the payload bending the cable over. 1. The cable has to be strong enough to do this (which may be reasonable, given how strong it needs to be anyway - plus it's in tension) 2. The cable will still be bent over, by a force imparted at the payload. If the bottom is attached and the top is "fixed" in geostationary orbit, the force will move the payload laterally (lagging the orbit - west), looking like a kind of arrow, or "V" on its side. The force on the cable itself might be OK, as the force is in tension; and the curve of the bend might be gradual enough. (Or maybe it would end up straight, but tilted westward?) It will also pull on both earth and satellite, slowing both.

gliese1337 12y ago

Therefore, a payload moving up the elevator would either need to be accelerated laterally somehow, or it would bend the elevator over. Or both. A payload moving up the elevator does pull on the cable, tending to slow it down. The cable, however, being massive and under very high tension, pulls back, accelerating the payload. If the cable is not anchored to the Earth, this causes the cable to slow down, and wobble a bit since the force is applied sequentially along the cable's length, rather than always on the center of mass, though tidal effects will tend to damp the wobble and keep it oriented practically straight up and down. For any one payload, the effect is negligible, as a space elevator cable would be freakin' massive, but without some sort of active stationkeeping it would eventually come down. If the cable is anchored to the Earth, however, as soon as it begins to be pulled over even a little bit by the rising payload, the lateral force is transmitted to the ground. The cable pulls on the Earth and the Earth pulls back, re-accelerating the cable with no need for active stationkeeping and ever so slightly slowing the rotation of the Earth. Since the Earth is Gigantic and has ridiculously large quantities of angular momentum and rotational kinetic energy, no one will ever notice this loss in practice.

lotsofmangos 12y ago
I like the extension to this which is to make a cable that stretches much further out than geostationary, then you clip payloads to the cable above geostationary and let the earth catapult them up the cable and slingshot them into the outer solar system.

Derbasti 12y ago

the math actually works, given a strong enough cable. You would lower a tether from geosynchronous orbit, and simultaneously extend a tether into space. Thus, the total mass stays centered in geosynchronous orbit, and the lower tether gets pulled towards earth because of gravity, whole the upper tether gets pulled away because of centrifugal forces. Thus, in the end, you have a tether that lightly touches the ground without force and stays in the same place, and an outer tether that extends about a third of the way out to the moon. The tensile forces would be strongest in geosynchronous orbit and weakest at the ends of the tether. Another benefit of this construction is that you can use the outer tether to launch spaceships into space at escape velocity without any need for propulsion. You can even use tethers on other planets to catch such payloads again. This construction does not eliminate the need for propulsion to reach orbit. But, you now can push against the tether instead of slippery and ever-thinning air, and you don't need to worry about lateral acceleration.

DavidSJ 12y ago

Wait... most of the energy of going into orbit is in accelerating to orbital speed (lateral), not getting away from earth (vertical). That's true for Low Earth Orbit, but as the radius of a circular orbit increases the kinetic energy decreases and the potential energy increases. On the surface of the Earth, we have: -(398600 km^3/s^2 [1]) / (6378 km [2])) = -62.5 km^2/s^2 of potential energy (0.465 km/s [3])^2/2 = 0.1 km^2/s^2 of kinetic energy At LEO, we have: -(398600 km^3/s^2 [1]) / (6678 km [4])) = -59.7 km^2/s^2 of potential energy (7.8 km/s [5])^2/2 = 30.4 km^2/s^2 of kinetic energy At GEO, we have: -(398600 km^3/s^2 [1]) / (42164 km [6])) = -9.5 km^2/s^2 of potential energy (3.07 km/s [7])^2/2 = 4.7 km^2/s^2 of kinetic energy Notice how LEO has only 2.8 km^2/s^2 of additional potential energy compared to the surface, but 30.3 km^2/s^2 of additional kinetic energy. However, at GEO there is 53 km^2/s^2 additional potential energy compared to the surface and only 4.6 km^2/s^2 additional kinetic energy. [1] Earth's gravitational parameter [2] Earth's equatorial radius [3] Earth's equatorial rotation speed [4] A 300 km altitude orbit [5] Velocity at LEO [6] GEO radius [7] Velocity at GEO