# Thinsat Charging and Array Spreading Forces

Thinsats intercept electrons and protons in the van Allen belt; they can be discharged by Fowler-Nordheim tunnelling through carefully aligned arrays of carbon nanotubes at the corners, spaced far enough apart on those corners to maximize the electric field. The charging rate is about 180 nA/m² for protons and 3400 nA/m² for electrons. Thinsats stop low energy protons more effectively, but the low mass electrons move faster and consequently the flux is higher. The thinsat will charge negatively.

Mass-sorted, calibrated CNT in solution will be used, these are commercially available and used for CNT-enabled silicon substrate integrated circuits. Presumably some kind of integrated circuit tool-chip will be used to pluck the CNT macromolecules out of solution, align them, and prepare them to be bonded to a carrier chip. Unlike an electronic circuit, the yield of this bonding process does not need to be perfect or even high; what is most important is that most of the CNT are bonded radially for charge emission into the concentrated field near the corners of the thinsat. Per-CNT currents can be fairly high, and is crudely exponential with voltage; what is most important for reducing voltage is maximizing the electric field gradient and lowering the work function of the CNT tips.

Hydrogen (proton) "contamination" of the CNT tips lowers their work function and lowers thinsat voltage further.

What is the field gradient at the tip of a CNT on the corner of a thinsat? Watcherone (2008) suggests a field gradient enhancement β over the "free space" field at the rounded corner of a graphene square peaks at around 400. The graphene sheet modelled in that paper is 1 nm thick and 1 μm square; This is less sharp than a 70 μm thick and 160 mm square thinsat, so we can presume that the corner β is better for the thinsat.

The CNTs off the corner provide additional field concentration factor; Figure 2 of Eletskii (2010) suggests a β factor of 700 for a CNT with a height to radius factor of 700. If the CNT has a diameter of 2 nm, the height would be about 1.4 μm. If the CNT on each corner are spaced 5 μm apart, that might reduce the β factor to 500. The multiplied β factor might be 2e5. Assuming 50% fill and an array 70 by 70 μm, that is 100 CNT per corner or 400 CNT per thinsat.

A 250 cm² thinsat will collect a net charge of about 80 nA, or about 200 pA per CNT.

Lets assume the free space field is proportional to the thinsat voltage divided by thinsat edge length. If the voltage is 100 V, then the field would be around 500 volts per meter. Times the multiplied beta factor, that is 1e7 volts per meter.

So, what's the voltage?

The capacitance of a square (pointy corner!) thinsat with side S is C = 0.367 × 4πε₀ × S; for a 16 cm thinsat, C₁ = 0.367 × 111.265 pF/m × 0.16 m = 6.530 pF. If the voltage on the thinsat is 200V, the charge is Q_1 = C_1 V = 5223 picoColomb per thinsat.

Assume N thinsats are charged to voltage V_th and are spaced L units apart in an approximately triangular geodesic grid mapped on a sphere. The area of the unit cell is √3/2 L², the area of the sphere is √3/2 L² N = 4 π R², so

R ~=~ \sqrt{ { \Large { \sqrt{ 3 } \over { 8 \pi } } } N } ~ L ~\approx~ 0.26252 \sqrt{ N } ~ L

If the thinsat sphere is 7842 thinsats (icosahedral geodesic sphere with tiling ratio V = 28) and L = 5 meters, then R = 116 meters .

The charge on the whole sphere of thinsats creates a Colomb force on a single thinsat. This resembles the math that shows that the gravitational force at (or above) a spherical shell is like the force from a concentrated mass at the center; we can move the charge to the center, and the effect on a charge at the edge will be the same. This simplifies the math; the force on a thinsat is approximately the simple Colomb force between the charge of N thinsats in the center and the charge of one thinsat distance R away:

F ~=~ Q_1 Q_N / 4 \pi \epsilon_0 R^2 ~=~ Q_1^2 N / 4 \pi \epsilon_0 R^2 ~=~ { \Large {2 \over { \sqrt{3} \epsilon_0}} \left( Q_1 \over L \right)^2 } ~ \approx ~ 24.56 \epsilon_0 { \Large \left( { S V } \over L \right )^2 } ~ \approx ~ (0.2175 ~ nanoNewtons/Volt^2) { \Large \left( { S V } \over L \right )^2 }

If V=800 volts, S = 0.16 meters, and L = 5 meters, then F = 143 nanoNewton. The light pressure on a thinsat (ignoring albedo and thermal re-radiation) is P/c = 1368W / 2.998e8 m/s = 4.56 nN/m² times the S^2 area, or 0.117 nanoNewton, much smaller than the electrostatic pressure. This doesn't look good ...

Suppose we make S 4 times larger, increasing thinsat area by a factor of 16, reducing the tiling ratio V from 28 to 7 and N from 7842 to 492. Thinsat area increases to 0.41 m² and light pressure to 1.87 nanoNewton. That will increase turn and stop time by the square root of the thinsat length, from 15 minutes to about 30 minutes for a 45 degree turn. Further, let's increase the array radius by a factor of 4 (increasing cross-array maneuvering time by 2) and spacing L by a factor of 16. The electrostatic force is reduced by a factor of ( 4/16 )² to 8.9 nanoNewtons. That is still almost 5 times larger than the light pressure ...

MoreLater - how does this actually affect array rotation? what are the tidal forces via the Clohessy-Hill equation, and how does this force affect the array rotation rate?

MoreLater - where does the 800V come from? Can we lower the voltage? *this may actually be differential voltage, which we can prevent by design.*

MoreLater - a positive voltage on a thinsat implies it is either missing electrons or has stopped some extra protons from the radiation belt that it hasn't boiled off. How many protons is that?

#### Capacitance of a Square

- Improved Extrapolation Technique in the Boundary Element Method to Find the Capacitances of the Unit Square and Cube
- F. H. Read
- Schuster Laboratory, University of Manchester, Manchester M13 9PL, United Kingdom
E-mail: !Frank.Read@man.ac.uk

- JOURNAL OF COMPUTATIONAL PHYSICS ARTICLE NO .133, 1–5 (1997). doi.org/10.1006/jcph.1996.5519
- square = 0.366787 × 4πε₀ × S ... cube = 0.8806785 × 4πε₀ × S ... 4πε₀ = 111.265006 pF/m