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Date: Sat, 11 Nov 89 03:21:27 -0500 (EST)
Subject: SPACE Digest V10 #235

SPACE Digest                                     Volume 10 : Issue 235

Today's Topics:
			    Space Elevator
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Date: 11/10/89 15:34:13
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Subject: Space Elevator

In a mooving coordinate system whoose origin is at Earth's center and
turning with Earth's daily revolution, the acceleration of any static
point in the equator's plane is:

   g  = -K.M/r^2 + w^2.r, where
      g (gamma : m.s-2) is the accleration along the radius,
      K is the gravitational constant (m3.s-2.kg-1)
      M is the mass of the Earth (kg)
      r is the distance from that point to Earth's center (m),
      w (omega : s-2) is Earth's rotation speed.

The ground acceleration 'g0' at radius 'r0' is given by:

   g0  = K.M/r0^2 (the other term is negligible), so that:
   K.M = g0.r0^2

At some point r1 above the equator line, the two terms cancel out and
provide a geosynchroneous trajectory:

  r1 = (g0.r0^2/w^-2)^(1/3)   <=>   K.M/r1^2 = w^2.r1

Now, seen from a geosynchroneous position, any point closer from
Earth will be accelerated downward, and any point above that would
be accelerated toward space. This is used to stabilize satellites in
such trajectories: two poles (or cables) bearing masses are directed
in opposite directions, and the force differential is such that
these poles tend to keep aligned toward Earth's center.  Of course,
the longer the pole, the better the stabilisation.  And if one of
the pole is really long, you can either make the other long as well,
or put more mass at its end. Just balance the corresponding forces.

Let's see, now, what happens when the downward one is so long as to
reach the Earth? The answer is obvious: you can then anchor it at
some place, and use it as an elevator cable to send various things
into geosynchroneous orbit. As a matter of fact, one need not
restrict oneself to that orbit: any object droped from under the
orbit would revolve around Earth faster, and the ones dropped from
the outer part of the cable would be thrown further out in space.

The first problem is that such a cable might collapse under its own
weight. The solution is to build it in such a way that its section
is proportional to the force it has to withstand, that is, the section
must follow the following differential equation:

  s.dS = g.p.S.dr, where
      s (sigma : N.m-2=kg.m-1.s-2) is the traction a given area can
        bear without splitting,
      S is the cross-area of the cable at point r, (m2)
        and dS its variation (m2 as well),
      p (rho : kg.m-3) is the density of the material used for the cable.

Given the value of 'g' from the first equation, one obtains:

    Delta( Log S )  = p/s.Delta( K.M/r + w^2.r/r ),

the variation beeing taken between 'r1' (geostationary) and 'r0'
(ground).  It turns out that this quantity can be expressed simply
as:

    Delta( Log S ) = p/s.g0.r0.( 1 + x/2 - 3/2.x^(1/3) ),

where  'x' = w^2.r0/g0 is the ratio between the centrifuge force on
the equator and the gravitational force if rotation is not taken
into account. (Please check - I may be wrong.)

The second problem, then, is that the 'g0.r0' factor is quite large.
Since its influence on the maximal cross-section is exponential, one
need to find materials where s will be large enough to cancel our
gravity. In our case, we have:

   g0.r0 = 62.5E6 m2.s-2 (or Joules per kg)
   p     = 5E3 for most materials, so that 's' needs to be:
   s = 300E9 kg.m-1.s-2 .

This is quite large: it corresponds to a cable capable of sustaining
30 tons with a cross-section of one square milimeter, under Earth's
gravity. (Non-International Units users may need to do the
conversion:  1inch is 25mm, one ton (metric, of course...) is about
the same in Her Majesty's unit.) If 's' is only one-tenth of that
value (3tons/mm2), the diameter variation between ground and orbit
will jump to 150: begin with 1cm and end with 1.5m...  Of course,
the 'x' factor that takes into account Earth's rotation is not
negligible, and reduces the strength needed by about one third.
Still... Who wants to compute the corresponding volume?

So this is why I regularly check the Guiness Book of Records to see
what kind of material holds the resistance/density record. Up to now,
it seems to be Ruby. Imagine that: 36_000 km of jewel in space...
What I'd like to know is wether any compound material can do better
than that. I know this is a space discussion list, not a material
one, but if someone has more up-to-date information please say so.

Now for the advange: I gave the 'g0.r0' figure in a strange
'Joule/kg' unit, and this is precisely what that figure represents:
the number of joules to send one kilogram up there. In other words,
given that a kilowatt-hour is 3.6e6J, to send 1kg in space via such
elevator requires as much energy as one electric bulb burning for
200 hours.  Furthermore, you get the energy back as soon as the load
comes back to Earth.  I don't have the figures at hand, but I think
it is cheaper than air mail. (Well at least for the marginal cost,
the investment being *much* larger.)

Now, the main problem here is that Earth is too big and turns too
slowly. If no material can withstand such stress, an alternative would
be to accelerate Earth's rotation, or decrease its gravitation or
radius. Well, why not change planet and see where that 'g0.r0'
product is lower? Mars is too big and Venus too slow, but such a
cable on the Moon might be realistic, and on other satellites as well.
Right in the middle of the far side of the Moon, of course.

I guess this idea was first presented by A.C.Clarke in one of
his books (I don't remember which one), and I think it's a realistic
one, but I haven't seen it discussed for some time. Any ideas?

                             Bertrand MICHELET

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End of SPACE Digest V10 #235
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