Space Grain Elevator

Really Wild Stuff - just for fun!

---

The classical "power of the exponential" problem.

A grateful king asks a mathematician to choose a reward. The mathematican asks for a chessboard, with a grain of wheat on the first square, two grains on the second square, four on the third, etc. For 64 squares, the last square has 2⁶³ grains on it, and the total is 2⁶⁴-1 grains, or 18,446,744,073,709,551,615 grains, 1.845e19 grains. At 30 grains of dry wheat per cubic centimeter (3e7/m³), and a density of 0.7, that is 6.15e11 cubic meters of wheat, and 4.3e14 kilograms. 615 cubic kilometers, 430 billion metric tonnes. With annual global wheat production about 730 Mmt, this is 590 years of production, 40% of the volume of Mount Everest. Most leave this problem with a vexed king beheading the smarmy mathematician.

But the problem describes half this mass and volume (3.07e11 m³, 2.15e14 kg) stacked on the last square of the chessboard, not spread out in a vast mountain. What would happen if we stacked all that wheat above one single chessboard square?

It would not hold its shape very long.

Unless we put it in a container. Presume an incredibly tall container made of unobtainium. Infinitely strong and rigid, zero weight and transparent. Regulation tournament chessboard squares are 5 to 6.5 centimeters on a side; let's choose 6.325 centimeters because it makes the area an even 40 square centimeters, 0.004 square meters. Assume the chessboard is at the equator, also made of unobtainium, and well anchored to the entire planet earth. As you will see, full planet anchoring is important; we don't want to yank any continents loose.

If the grain maintained a density of 0.7, the last square stack (half the total wheat) would be 3.07e11 m³ / 0.004 m² meters high; 7.7e10 kilometers. For comparison:

At a density of 0.7, our stack would extend ten times as far as Pluto's apehelion. But as you already guessed, the stack will get compressed. A lot. whole lot. Wheat is not unobtanium. Hold on - this is going to get weird.

How Much Grain below Geosynchronous Orbit?

A good practice question. \mu = 3.9864e14 is the gravitational parameter, \omega = 7.271e-5 radians/second is the angular velocity of the rotating earth. The perceived gee force at height r is:

gee(r) ~ = ~ \mu / r^2 - {\omega}^2 r

The additional pressure of a mass element height dr is

dP ~ = ~ - \rho(P) gee(r) dr ~ = ~ - \rho(P) \left( \mu / r^2 - {\omega}^2 r \right) dr

If the grain was uniformly dense, with ρ(P) = ρ0= 1000 kg/m², then we can integrate from the earth surface to GEO and get

P(surface) ~ = ~ \rho \left( \mu ( 1/r_e - 1/r_{geo} ) + {\omega}^2 ( {r_e}^2 - {r_{geo}}^2 ) \right)

P(surface) ~ = ~ 1000 × ( 3.9864e14 × ( 1/6.378e6 - 1/4.216e7 ) + (7.271e-5)^{2 × ( 6.378e6}2 ^{- 4.216e7}2^ ) )

P(surface) ~ = ~ 4.39e10 Pascals

That seems like a lot of pressure, 44 GPa. It is enough to increase the density of "incompressable" water to about 1300 kg/m³. Assuming that is averaged over the distance near the earth, that might increase the relevant column density (near the earth) by 15%, increasing P(surface) to ≈ 50 GPa. That does not increase the density significantly.

How much does this column weigh? Assuming an average density of 1000 kg/m³, with lower density in the low gravity near GEO and higher density near the surface averaging out, the volume of the column weighs 4 kg/m × ( 4.216e7 - 6.378e6 ) meters, which computes to 140 metric tons - a tiny fraction of the total 430 trillion metric tons. Essentially all of the mass is above GEO, where centrifugal force dominates and the earth's gravity is insignificant.

Rotational Speed and Acceleration, Take One

The Earth turns^{[citation needed]}. Because it goes around the sun, it makes one more turn per year relative to the fixed stars than there are days in a year, and consequently a "sidereal day" is 86164.09 seconds. The angular velocity ω is 2 π/86164.09 or 7.2921e-5 radians per second. The end of a massless, empty 7.7e10 kilometer silo would be whipping around the solar system once a day at a speed of ω×R or 5.6 million kilometers per second, 19 times faster than the speed of light at a mere 300,000 km/s , almost as fast as Einstein would turn in his grave at the thought of this.

Worse, the centrifugal acceleration at the end would be ω²×R or 42 thousand gees; the wheat would get compressed at the far end. The pressure at the end of the stack would be the mass times the acceleration divided by the cross sectional area: 4.09e5*2.15e14/0.004 = 2.2e22 Pascals, much larger than the 3.5e16 Pa at the core of our Sun, though much smaller than the 1.6e35 Pa pressure at the core of a neutron star or even the 2e29 Pa pressure at the core of a white dwarf star.

Angular Momentum

But the end will not turn that fast, and the pressure will not be that high. The earth has a limited amount of angular momentum, and the mass at the end of the tall silo would absorb a heck of a lot of it. The angular inertia of the earth is about 8e37 kg-m². The angular momentum of the mass at the end of our too-tall grain elevator is the radius (in meters) squared times the mass, or 1.27e42 kg-m², 16,000 times that of the earth alone. Angular momentum is conserved, and almost all of it would transfer to the mass of the grain; the sidereal day would become 16,000 times longer, and the pressure at the end becomes a mere 8.7e13 Pa, 87 trillion Pascals, 87 TPa.

Even at 87 TPa, we can expect the grain to be very compressed - "incompressable" water compresses to a density of 2.4 at 1 TPa. So if our grain was 10 times denser, we could make our space grain elevator 10 times shorter, which makes the day only 1600 times longer - and the pressure at the end 10 times higher yet, increasing the density still more, making it shorter and faster, etc.

(see http://www1.lsbu.ac.uk/water/water_phase_diagram.html)

We will iterate towards a consistent rotation rate, length, and compression.

Degeneracy Pressure

How much compression? Thanks to our rigid unobtanium container, quite a bit. The grain will be compressed well beyond any pressure we can create in a lab; strength of material considerations no longer apply, this will be a melange of nuclei and electrons squeezed well beyond solid/liquid/gas, a highly conductive "degenerate" material, because we are squeezing the electrons into higher energy states than found in ordinary materials. Since our container is narrow, the materials may heat to hyperincandescence as they are squeezed, but they will likely reach black body equilibrium (250K?) with the sun and deep space fairly quickly. No idea what the heat capacity would be, but we can presume every electron in the material can store {3 \over 2} kT of energy, and thermal conductivity is higher than a metal.

Typical dry plant composition from [http://www.nzic.org.nz/ChemProcesses/soils/2A.pdf] Table 1 on page 4

So, a gram of wheat contains about 0.52 moles of electrons. The electron degeneracy pressure is

P = 35.23 h^2 ( 0.52 * Avagadro * \rho )^{5/3} / m_e

P = 2.45e9 * {\rho}^{5/3}

\rho = 37 kg/m^3 ( P / TPa )^{0.6}

Lets add this "degeneracy density" to the density of ice at 1 TPa, 2400 kg/m³

Assume density starts at 700 kg/m³ at zero pressure, increases linearly to 2400 kg/m³ at 1 TPa, then follows the table above. For a crude approximation,

Lets compute the mass by numerically integrating mass and pressure outwards from GEO, ignoring Earth's gravity. Our equation is simpler:

dP ~ = ~ \rho(P) ~ {\omega}^2 ~ r ~ dr

The accumulating mass is:

dM ~ = ~ \rho(P) ~ Area ~ dr

or

dM / Area ~ = ~ \rho(P) ~ dr

Where Area = 0.004 m². When M = 4.3e14 kg, or M/Area = 1.08e17 kg/m², stop integrating, we've reached the end of this integration pass. However, we are not done calculating. The accumulated moment of inertia I of our column is:

dI ~ = ~ r^2 ~ dM

dI / Area ~ = ~ \rho(P) r^2 ~ dr

And this will be significant. This is added to the Earth's angular momentum I_e = 8.034e37 kg m², and \omega can be computed from the original \omega_e = 7.271e-5 rad/s as:

\omega ~ = ~ \omega_e \left( I_e \over { I_e + I } \right)

That will change the pressure and density and angular momentum calculation. Rinse and repeat until \omega converges, possibly using stepwise averaging or Newton's method to avoid divergences. I just did it manually. Here's the first pass at a program sge1.c

Here's the results:

     0.953639070  Reduced angular velocity factor
     0.048614755  Elevator fraction of earth moment of inertia
    1.669384e+11  meters radius of elevator
        1.115914    AU radius of elevator
    1.228919e+06  kg/m3 density of grain at end
    3.420918e+19  Pa pressure on grain at end
    1.160897e+07  m/s velocity of end
        0.038723    fraction of speed of light
    2.152120e+14  kg mass of elevator
    2.152120e+14    kg target mass of elevator
          200.00    m integration step

The end will emit a LOT of heat - it is radiating away 10% of the earth's rotational energy, and probably some nuclear fusion energy, too. The end mass will also be increased by a relativistic factor of about 0.07%. I expect this will generate gravity waves detectable within the solar system. I expect the elevator slicing through the Moon, Sun, Mars, Venus, and Mercury will be more noticable.

Lifting the grain to the silo fill port above GEO will require a lot of energy, about 50 MJ per kilogram; interestingly, also the pressure of grain near the earth for a more slowly rotating grain elevator. About 1e22 Joules total, 16 years of 20 TW global energy production, or about 200 days of 1000W/m² solar energy (10% efficiency) on the entire earth. Blocking sunlight to the entire planet would interfere with wheat production.

The force holding the elevator down is about 1.4e18 Newtons, the end pressure times the space grain elevator area. The mass of the earth is 6e24 kg, and the whole planet will be displaced in the opposite direction from the elevator by a distance of F / M_e \omega^2 or about 40 meters. I'd guess that about 100 meters of ocean will slosh away from the space grain elevator side of the planet to the opposite side, but I haven't done the calculations.

Note - I have not computed the additional force and volume of the silos over the nearby chess squares with 1/2, 1/4, 1/8, etc. of the mass. These will have additional effects, but together much less than the heaviest square, since the effects are approximately proportional to mass per square cubed.

Plan B - A Pyramidal Frustum

If instead of a 40 cm² square shaft all the way up, let's project the chessboard square as a tall thin pyramid, with the apex at the center of the earth. A pyramid truncated towards the apex (in this case, by a plane at GEO altitude) is called a "frustum". The sides of this frustum are directly above the edges of the square, but the frustum gets much wider at higher radius and is much roomier than the narrow square shaft. A frustrum, looking like a pyramidal arrowhead thrust into the earth, will hold far more grain in a shorter length. Which means this option is much more boring than the straight shaft grain elevator, but in engineering, boring is good.

How far into space does this frustum extend? The math is somewhat more complicated, because the force of the mass pushing from closer to the earth spreads out over an increasing area. The easiest way to handle this is to integrate the force over the distance, then divide by the (steadily increasing) area to arrive at the pressure, then the density (which will be much smaller at the much closer end).

The area goes up as the square of the radius. If the area of the chessboard square is A_e = 40 cm² at radius r_e = 6.378e6 m, then the area A(r) as a function of radius r $ is:

A(r) ~ = ~ A_e ~ ( r / R_e )^2

At GEO, the cross sectional area is 0.175 m², 43.7 times larger than the earth surface chessboard square, and at the radius of the moon, the area is 14.5 m², 3630 times the area of the surface square. This space grain elevator will be much shorter.

For the frustum mass per radius is:

dM(r) ~ = ~ \rho(P) ~ A(r) ~ dr ~ = ~ \rho(P) ~ A_e ( r / R_e )^2 ~ dr

The additional force is:

dF(r) ~ = ~ dM(r) ~ {\omega}^2 ~ r

dF(r) ~ = ~ \rho(P)~ {\omega}^2 ~ A_e ~ r^3 / {R_e}^2 ~ dr

And the pressure is:

P(r) ~ = ~ F(r) / A(r)

Lower Pyramidal Space Elevator

Unlike the square elevator, we can store a significant amount of grain below GEO, and it will provide downwards anchoring forces. Not too much, or the gravitational weight of the space grain elevator will not be supported against gravity. Since we are operating closer to the earth, the gravity AND radial acceleration matter, with gravity dominating below GEO. This will complicate our program, and we must calculate once downwards to find out how much grain is stored below, pressure increasing as we approach the ground.

MoreLater

Here's the first pass at a frustrum program sgef1.c (to be modified for gravity )

Here's the results for one tall square:

     0.999996673  Reduced angular velocity factor
     0.000003327  Elevator fraction of earth moment of inertia
    1.410584e+09  meters radius of elevator
        3.669572    relative to moon
    2.475272e+03  kg/m3 density of grain at end
    6.359932e+12  Pa pressure on grain at end
    1.028611e+05  m/s velocity of end
    7.500722e+00  m/s2 acceleration at end
      195.653959  m2 area at end
    1.244346e+15  N force at end
    2.152120e+14  kg mass of elevator
    2.152120e+14    kg target mass of elevator
            5.00    m integration step

This calculation is more sensitive to neigboring squares - the 61st square space grain elevator is about half as tall. We can also store mass in the bottom of all of the squares, supporting it with centrifugal force from the taller elevators.

Here is the mass of the uppermost squares:

MoreLater

Conclusion

I conclude that this would not be the best use for expensive unobtainium. I concur with the King's beheading of the mathematician.