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12 Mar 15
3 edits
Originally posted by sonhouseWhat forces are we talking about here? If we for example are thinking about a x-y-z coordinate system. Note, there was a long time since my university course in elementary mechanics.
If you are at geo altitude, about 36K up, you will by definition have a velocity around 11,000 Km/hr. Isn't that already above escape velocity at that altitude? Isn't escape velocity at that altitude about 7000 km/hr?
a (N)= mass of 'you'.
a (N) * e = weight of 'you'. e is earths gravitational pull.
This gives us one dimention. A line between earths core and core of the 'you' object. According to Newton (1 2 or 3, I keep mixing them up). Then bodies 'earth' and 'you' are falling towards each other. When the sky is sufficiently dence for friction to work on object 'you' the power of friction are increasing with the cube when body 'you' are accelerating.
Perhaps. I said perhaps. We also need to accout for time change since there will be relativistic time. According to Einstein. We could do 2 equation systems. One for relativistic time and one for constant time.
12 Mar 15
Relativity is not going to produce significant corrections to the Newtonian picture for this problem. Escape velocity at a given distance from the earth's centre is the radial velocity required to just escape the earth's gravitational pull. The orbital velocity is tangential to the orbit. In other words at right angles to the direction of the force, assuming a circular orbit which it has to be for a geostationary object. So the two velocities you are considering are at right angles to one another.
12 Mar 15
1 edit
Originally posted by DeepThoughtI just did a back of the envelope calculation of the escape velocity at geo orbit height, inverse square law and all that, 9.8 M/S^2 at the surface and using (2GM/r)^0.5 come up with 11,117 and change meters per second at the surface and 4687 M/Sec at geo.
Relativity is not going to produce significant corrections to the Newtonian picture for this problem. Escape velocity at a given distance from the earth's centre is the radial velocity required to just escape the earth's gravitational pull. The orbital velocity is tangential to the orbit. In other words at right angles to the direction of the f ...[text shortened]... tationary object. So the two velocities you are considering are at right angles to one another.
I see in that case at Geo you would be going 2606 M/S and escape being 4687, you need to add about 2000 meters per second to say bye bye to Earth.
Can someone verify my numbers?
I used 6.667 E-11 for G, 5.9E24 Kg for mass of Earth, 6,371,000 meters for r. That should give 11,117 meters per second as escape velocity of Earth or about 6.9 miles per second.
12 Mar 15
http://en.wikipedia.org/wiki/Geostationary_orbit
No need to do any calculations. Its all on Wikipedia. Geostationary orbit is at 35,786 kilometres. If you are geostationary higher than that, you are above escape velocity. If you are at geostationary orbit lower than that, you will crash into the earth.
A space elevator however has a significantly massive tether and thus needs to extend beyond geostationary orbit in order to counterbalance.
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12 Mar 15
Originally posted by twhiteheadNo, that's not correct, although it's an enticing assumption.
http://en.wikipedia.org/wiki/Geostationary_orbit
No need to do any calculations. Its all on Wikipedia. Geostationary orbit is at 35,786 kilometres. If you are geostationary higher than that, you are above escape velocity.
If you are higher than the geostationary orbit, but going at
geostationary orbit speed, you are above the orbital speed
for that altitude. You are not necessarily above escape velocity
for that altitude.
If you are orbiting at above the local orbital velocity you will
gain altitude. As you gain altitude you slow down as you convert KE*
into GPE**.
Only if your excess KE is greater than the GPE gain at infinity are
you going faster than escape velocity.
If your excess KE is less than the GPE gain at infinity then you are
simply in an elliptical orbit.
*Kinetic Energy
**Gravitational Potential Energy
12 Mar 15
Originally posted by googlefudgeI see. So essentially you will get transferred to a higher orbit that is no longer geostationary.
No, that's not correct, although it's an enticing assumption.
Although with a space elevator it doesn't matter that much whether or not you have obtained escape velocity unless your aim is to be able to send objects off its end into interplanetary space. For getting objects into orbit, which will most likely be its main use, all you need it to get high enough to ensure it holds itself up. Interplanetary vehicles can carry extra boosters to get them selves to escape velocity.
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12 Mar 15
Originally posted by twhitehead
I see. So essentially you will get transferred to a higher orbit that is no longer geostationary.
Although with a space elevator it doesn't matter that much whether or not you have obtained escape velocity unless your aim is to be able to send objects off its end into interplanetary space. For getting objects into orbit, which will most likely be its m ...[text shortened]... self up. Interplanetary vehicles can carry extra boosters to get them selves to escape velocity.
Interplanetary vehicles can carry extra boosters to get them selves to escape velocity
Interplanetary vehicles with boosters do not need to get to "escape velocity". "escape velocity" is the velocity which is needed at any given altitude to escape earths gravitational field at fuel burnout. If they have the fuel they can escape at relatively low velocity.
13 Mar 15
Originally posted by googlefudgeYour link went to a comic strip. Was that intentional? Kebal is not free, costs $30 US.
Basically everyone here should go spend a few [hundred 😉 ] hours playing Kerbal Space Program.
After which they will all intuitively understand orbital dynamics...
Or will have gone insane.
http://xkcd.com/1356/
13 Mar 15
2 edits
Originally posted by googlefudgeYes, a nice ad for Kebal! BTW, I found the formula for figuring geo synch orbits:
Yes it was intentional... Did you read the comic in question?
r= cube root of (T^2*G*M)/4*PI^2
4Pi^2= 39.4784176....
G=6.67E-11
M=mass in Kilograms.
T=length of day for whatever body you are working with.
So I did Mars and it came out there to 16,687 km above the surface
and Jupiter, 475,100 km above surface.
13 Mar 15
1 edit
Originally posted by sonhouseDo the moon.
So I did Mars and it came out there to 16,687 km above the surface
and Jupiter, 475,100 km above surface.
Is a space elevator even viable on the moon given its slow rotation?
If we put a railway all the way around the equator, and ran a train with the space elevator attached to it such that it went twice as fast as the earth spins, could we have a shorter tether?
How fast would the train have to go?
13 Mar 15
Originally posted by twhiteheadI already did the train thing, about ten years ago. I envisioned a train going round the lunar equator with the idea it would be about a mile long and 50 odd feet wide where the top would be plastered with solar cells. The idea there would be to generate energy, feed it to the tracks, and colonies would be able so suck off the energy given to the tracks anywhere round the system. The train would want to maintain itself in local noon all the time and would therefore need to be traveling about 10 miles per hour, 16 km per hour or so. That puts it at local noon forever.
Do the moon.
Is a space elevator even viable on the moon given its slow rotation?
If we put a railway all the way around the equator, and ran a train with the space elevator attached to it such that it went twice as fast as the earth spins, could we have a shorter tether?
How fast would the train have to go?
I envisioned circular platforms used to allow people to exit and enter the train which never stops, I guess it could stop for a few minutes so that might not be needed.
Anyway, it would be getting the full 1300 odd watts per square meter for a total energy gathering area of about 80,000 square meters for a possible 100 megawatts, of course times say, 25% cells, 25 megawatts on the ground.
I thought that was a pretty slick system🙂
I think geo synch orbit for the moon would be about 1000 miles below the surface🙂
Escape velocity is about 1.8 km/sec for the moon and orbital velocity right at the surface would be about 1.3 km/sec. Since the moon is already only spinning at a rate of about 4 meters per second, there is a big disconnect there for a geo orbit.
13 Mar 15
1 edit
Originally posted by twhiteheadYour space elevator would have to cope with the fairly crazy orbit the moon has. Supermoons are caused when the moon is closer to the earth than normal. It's average orbital distance increases by a centimetre per year. The orbit is perturbed by the sun, and all the other planets, although obviously Jupiter is the most important. A space elevator to the moon isn't a practical proposition. Although a space elevator from the moon to lunar orbit has a lot fewer practical difficulties than one from the earth to earth orbit.
Do the moon.
Is a space elevator even viable on the moon given its slow rotation?
If we put a railway all the way around the equator, and ran a train with the space elevator attached to it such that it went twice as fast as the earth spins, could we have a shorter tether?
How fast would the train have to go?
Edit: A second look at your post indicates this (elevator to lunar orbit) is probably what you meant. As long as the rotation is non-zero you just need the counterweight far enough up.