Space Elevator in 15 years?

[KrazyHorse]

Now, du is the extra mass density required to support the "weight" of the r to r+dr section.

Therefore du * s = u * dr * (GM/(r^2) - r*w^2)

or du/dr = (u/s) * (GM/(r^2) - r*w^2)

Now to solve this differential equation

u(r) = A * exp(-GM/(r*s) - (0.5s^-1) * (r^2) * (w^2))

with A the undetermined constant.

Set u(R0) = u0

Get A = m/s*(GM/R0^2 - R0*w^2) * exp(GM/(R0*s) + (0.5s^-1) * (R0^2) * (w^2))

EDIT: dropped a factor of 1/s

[KrazyHorse]

Originally posted by DanS
KH is a show-off.

Not at all. Trying to get some form of solution for total length of cable with anchoring mass m'.

[KrazyHorse]

Now to get the total force on the the cable we integrate from R0 to Rf (the radius of the anchoring mass) the expression:

u(r) * (GM/(r^2) - r*w^2)

or A * exp(v(r)) * dv/dr * s

So

F1 = A * s * (exp(v(Rf)) - exp(v(R0)))

Now add to this the force on the anchoring mass m'

F2 = m' * (GM/(r^2) - r*w^2)


F1 + F2 = 0 for minimal stable height.

[KrazyHorse]

If we have no anchoring mass F2 = 0, therefore F1 = 0

Therefore v(Rf) = v(R0)

Therefore -GM/R0 - 0.5(R0^2)*(w^2) = -GM/Rf - 0.5(Rf^2)*(w^2)

[Dis]

I'll take your word for it. I've been out of school too long.

[KrazyHorse]

or numerically

-6.260 * 10^7 = -3.986 * 10^14/Rf - (2.644 * 10^-9)Rf^2

Which (solving the cubic equation) gives a distance (from centre of the earth to end of cable) of

1.57 * 10^8 metres or 157 000 km. The length is this distance minus 6400 km (radius of the earth) or 150 600 km or so.

For comparative purposes the distance from the centre of the Earth to geosynchronous orbit is when

GM/(r^2) - r*w^2 = 0

or r = 42 200 km

So the height of geosynchronous orbit is about 35 800 km above the surface of the earth.

In other words the cable would be about 4 times longer than the height of geosynchronous orbit.

[DRoseDARs]

Freak.

[Dis]

the figure listed in the article in Discover is 62,000 miles long. Roughly 8 times the Earth's diameter.

That alone sounds far fetched. How would you coil that much ribbon. Even if it were thinner than paper, it would still take up a lot of space. How would you get it all up in space.

And what would happen if it got tangled up . Who would untangle all that?

[KrazyHorse]

62 000 miles (assuming I haven't dropped any factors here) would require an anchoring mass. Otherwise it would fall down...

[Lancer]

If this thing is ever built, all that came before it will be ancient history.

[Ted Striker]

KrazyHorse get the hell to work and make it happen.

[Urban Ranger]

Originally posted by KrazyHorse
To what? You definitely need to put your "anchor" further away than geosynchronous orbit (~40 000 km IIRC)

And the closer your anchor is to geosynchronous the more mass you need to put there. It's a simple question of requiring that the net force on the upper part balances out the net force on the lower end.

I don't think it's efficient to use a Space Elevator to put things into geosynchronous orbit. Wastes too much energy. Much better just move a rocket into low orbit and fire it from there.

[Lung]

We don't need a space elevator, we have Jerry Springer! Who needs a space elevator when we have the mundane to keep us eternally occupied?

Just imagine if we found alien artifacts that showed an ancient civilisation preoccupied with the worst of their species?

[Lung]

Originally posted by KrazyHorse
62 000 miles (assuming I haven't dropped any factors here) would require an anchoring mass. Otherwise it would fall down...

The whole point of the space elevator is that it won't. The centrifugal force would negate it's weight.

***Edit: Lazy Lung didn't read the rest of the thread! I hereby banish thyself to silence for the next 5 minutes!

[Impaler[WrG]]

KH - I dont belive you took into account the Thickness (mass per unit of length) of the Cable above Geosyncronus Orbit. Balancing the elevator is like a See-saw. If your using another person on the other end of the see-saw then the closer they are to the Fulcrum (Geosyncronus orbit) the heavier they need to be. Conversly the further away they are the lighter they can be. With a long enough plank you dont even need someone sitting on it at all. Also you must take into account that every part of the Cable is a differnt distance from the Earth and thus its weight is differnt.

A construction senario I read about involved bringing a 20 metric Ton spool of material up to GSO with conventional Rockets and then lowering the thread down to Earth. At 20,000 Kilos and aproimatly 20,000 Kilometers thats 1 Kilo per Kilometer or 1 gram per Meter. Thats obviusly to thin but if a large number of such filiments were spun together like silk from a spider they might be able to do the job.