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.

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 seconds. The angular velocity ω is 2 π/86164 or 7.29e-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,

\rho = 700 + 1700 ( P / Tpa ) . . . . . if P/TPa < 1 \rho = 2400 + 37 ( P / TPa )^0.6 . . . if P/TPa > 1

Lets compute the mass by integrating mass and pressure outwards from the center of the earth, ignoring Earth's gravity. In reality, gravitational acceleration