Thinsat Charging and Array Spreading Forces

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 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 800V, the charge is Q_1 = C_1 V = 5223 picoColomb per thinsat

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?

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