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Server satellites spend their entire orbit near zero magnetic latitude. The flux is higher, but the server satellites present half the cross section to isotropic radiation flux - $ 2 \pi r^2 $ (front and back) for a disk rather than $ 4 \pi r^2 $ for a sphere. The flux is not isotropic, as the particles are gyrating around field lines. Very complex - but as a crude approximation, assume all these factors cancel out within a factor of 2 or so. Thinsats spend their entire orbit near zero magnetic latitude. The flux is higher, but the thinsats present half the cross section to isotropic radiation flux - $ 2 \pi r^2 $ (front and back) for a disk rather than $ 4 \pi r^2 $ for a sphere. The flux is not isotropic, as the particles are gyrating around field lines. Very complex - but as a crude approximation, assume all these factors cancel out within a factor of 2 or so.
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The big question is the amount of the flux of momentum is actually absorbed. Since server satellites are so thin, much of the flux will pass right through without much momentum change. Is the fraction close to 1.0? 0.1? This needs calculation. Imagine a sandwich of server satellites thick enough to absorb all the particles. That whole sandwich will have a fraction close to 1.0, but the amount absorbed by each server satellite will be that fraction divided by the number of server satellites. So the fraction for an isolated, individual server satellite is likely to be small. The big question is the amount of the flux of momentum is actually absorbed. Since thinsats are so thin, much of the flux will pass right through without much momentum change. Is the fraction close to 1.0? 0.1? This needs calculation. Imagine a sandwich of thinsats thick enough to absorb all the particles. That whole sandwich will have a fraction close to 1.0, but the amount absorbed by each thinsat will be that fraction divided by the number of thinsats. So the fraction for an isolated, individual thinsat is likely to be small.
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If we assume complete absorption, a 50 micron server satellite will have a much larger area to mass ratio, and will slow down much faster than LAGEOS. If we assume complete absorption, a 50 micron thinsat will have a much larger area to mass ratio, and will slow down much faster than LAGEOS.
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While this analysis is quite crude, it does show that radiation flux drag will eventually bring down server satellites, far more quickly than the neutral gas density might suggest. While this analysis is quite crude, it does show that radiation flux drag will eventually bring down thinsat, far more quickly than the neutral gas density might suggest.

The Lageos measurement satellites

Two passive satellites are used for laser geodesy. These are 60 centimeter, 400 kilogram solid metal spheres covered with retroreflectors, used for precisely measuring the geometry of the Earth's surface and the Earth's gravitational field.

m288 is 1200km above LAGEOS 1. Server sky orbits should stay above LAGEOS!

These satellites are interesting because their orbits are very precisely defined, and we can use them to verify both our orbital math and learn about drag at those extreme altitudes.

\large \mu = 3.98600448e14 {m^3}/{s^2} ~ \buildrel{?}\over{=} ~ \omega^2 a^3 = 3.988154435e14 = 1.000539376 \mu

The difference is probably due to J2 perturbation dragging forwards both perigee and mean motion. However, because LAGEOS 1 is in a highly inclined and slightly retrograde orbit, the equations I have (from Piscane, page 113) suggests it should get dragged backwards. Meanwhile, those same equations suggest that M288 must be about 3.4 kilometers higher to counteract the extra acceleration due to J_2

MORE LATER

Drag, and implications for Server Sky

LAGEOS 1 is slowed down by drag, mostly drag against the high flux of radiation particles. It is dense enough to absorb essentially all the van Allen belt particles that pass through, and change momentum as a result. The "encounter rate" is not limited by ram speed, as it is with gas molecules in the dense atmosphere, but by the speed at which the radiation particles themselves move. If \rho_p is the mass density of particles in the van Allen belt in kg/m3, and v_p is their average speed in m/s, then LAGEOS sheds power into them proportional to 0.5 \rho_p v_p v^2 A , where v^2 = \mu / r and A is the effective cross section. The power reduces the orbital radius r by 0.5 M a_g dr / dt where a_g is gravity and M % is the LAGEOS mass. Since dr \ dt $ = 1.3e-8 m/s, the mass flux per unit area is:

\large Flux = \rho v_p = { M \over { A ~ r } } { { d r } \over { d t } } = 5.1E-12 kg/m2 s

That is the radiation flux from all directions. Since LAGEOS is in a highly inclined orbit, the radiation flux changes character in a complex way as a function of magnetic latitude. At high latitudes, the flux is lower.

Thinsats spend their entire orbit near zero magnetic latitude. The flux is higher, but the thinsats present half the cross section to isotropic radiation flux - 2 \pi r^2 (front and back) for a disk rather than 4 \pi r^2 for a sphere. The flux is not isotropic, as the particles are gyrating around field lines. Very complex - but as a crude approximation, assume all these factors cancel out within a factor of 2 or so.

The big question is the amount of the flux of momentum is actually absorbed. Since thinsats are so thin, much of the flux will pass right through without much momentum change. Is the fraction close to 1.0? 0.1? This needs calculation. Imagine a sandwich of thinsats thick enough to absorb all the particles. That whole sandwich will have a fraction close to 1.0, but the amount absorbed by each thinsat will be that fraction divided by the number of thinsats. So the fraction for an isolated, individual thinsat is likely to be small.

This analysis also assumes that the absorbed particles have zero average momentum. They protons actually drift towards the west, with a velocity of 700km/sec at 10MeV, 7 km/sec at 100keV, with the bulk of the particle flux at the lower energies. If most particle energies are below 10keV, there will be little effect from this.

If we assume complete absorption, a 50 micron thinsat will have a much larger area to mass ratio, and will slow down much faster than LAGEOS.

\large { { d r } \over { d t } } = { { A ~ r } \over M } Flux ~ = ~ { { 0.024 * 12789000 * 5.1E-12 } \over 0.003 } m/s ~ ~ = 0.5 mm/sec = 1.6 km/year

While this analysis is quite crude, it does show that radiation flux drag will eventually bring down thinsat, far more quickly than the neutral gas density might suggest.

MORE LATER

Orbital elements from various sources

LAGEOS 1

LAGEOS 2

- nyo

NORAD ID

8820

22195

Int'l Code

1976-039A

1992-070B

Perigee

5,845.9 km

5,622.3 km

Apogee

5,954.4 km

5,959.4 km

Inclination

109.8°

52.6°

Period

225.5 min

222.5 min

Semi major axis

12,271.2 km

12,161.9 km

Launch date

May 4, 1976

October 22, 1992

Source

United States

Italy

- JPL

mass

411kg

405kg

diameter

0.6m

0.6m

eccentricity

0.0045

- NSSDC

periapsis

5837.0 km

5900.0 km

apoapsis

5946.0 km

5900.0 km

period

225.41000366210938 min

225.0 min

Inclination

109.80000305175781°

54.0°

eccentricity

0.00443999981507659

0.009999999776482582

- springerlink

decay

1.1mm/day 1.5V

- Innovateus

mass

406.965

405.38

spin

0.61s

0.906s

period( sec )

13246.002197265628

\large\omega

4.743457847e-4

Two Line Elements

LAGEOS1 TLE:

  • 1 08820U 76039A 11094.82079485 +.00000016 +00000-0 +10000-3 0 01141
  • 2 08820 109.8462 077.3556 0044210 050.5232 309.9115 06.38664784558975

LAGEOS2 TLE:

  • 1 22195U 92070B 11095.53472161 -.00000009 00000-0 10000-3 0 3192
  • 2 22195 052.6462 175.4139 0138581 279.8295 078.6642 06.47294193436229

#

chars

description

LAGEOS 1

LAGEOS 2

1

01-01

Line number

1

1

2

03-07

Satellite number

08820

22195

3

08-08

Classification (U=Unclassified)

U

U

4

10-11

International Designator (Last two digits of launch year)

76

92

5

12-14

International Designator (Launch number of the year)

039

070

6

15-17

International Designator (Piece of the launch)

A

B

7

19-20

Epoch Year (Last two digits of year)

11

11

8

21-32

Epoch (Day of the year and fractional portion of the day)

094.82079485

095.53472161

9

34-43

First Time Derivative of the Mean Motion divided by two

+.00000016

-.00000009

10

45-52

Second Time Derivative of Mean Motion
divided by six (decimal point assumed)

+00000-0

00000-0

11

54-61

BSTAR drag term (decimal point assumed)

+10000-3

10000-3

12

63-63

The number 0 (formerly "Ephemeris type")

0

0

13

65-68

Element number

0114

319

14

69-69

Checksum (Modulo 10)

1

2

1

01-01

Line number

2

2

2

03-07

Satellite number

08820

22195

3

09-16

Inclination [Degrees]

109.8462

052.6462

4

18-25

Right Ascension of the Ascending Node [Degrees]

077.3556

175.4139

5

27-33

Eccentricity (decimal point assumed)

0044210

0138581

6

35-42

Argument of Perigee [Degrees]

050.5232

279.8295

7

44-51

Mean Anomaly [Degrees]

309.9115

078.6642

8

53-63

Mean Motion [Revs per day]

06.386647845

06.472941934

9

64-68

Revolution number at epoch [Revs]

58975

36229

10

69-69

Checksum (Modulo 10)

5

9

. . . 11094 = April 4, 2011 . . . 11095 = April 5, 2011

References:

Lageos (last edited 2013-08-29 04:24:13 by KeithLofstrom)