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| For "heavy" server-sats relatively close to the earth, such as 7 gram satellites at m288, the dominant deviation from a perfect Kepler orbit is caused by the $ J_2 $ spherical harmonic of the gravity field, in turn caused by the oblateness of the spinning Earth. For small eccentricities, the eastward precession of the perigee of one elliptical equatorial orbit is proportional to the $ J_2 $ term of the WGS84 model ( -1.082626683E-03 , see Pisacane 2008 ) and the inverse of the orbit radius squared. The precession, expressed as the number complete precessions per year, is $ J \approx -3 J_2 ( Y / P ) ( R_e / R_s ) ^ 2 $ where | For "heavy" server-sats relatively close to the earth, such as 3 gram satellites at m288, the dominant deviation from a perfect Kepler orbit is caused by the $ J_2 $ spherical harmonic of the gravity field, in turn caused by the oblateness of the spinning Earth. For small eccentricities, the eastward precession of the perigee of one elliptical equatorial orbit is proportional to the $ J_2 $ term of the WGS84 model ( -1.082626683E-03 , see Pisacane 2008 ) and the inverse of the orbit radius squared. The precession, expressed as the number complete precessions per year, is $ J \approx -3 J_2 ( Y / P ) ( R_e / R_s ) ^ 2 $ where |
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| This page needs reworking - previously I was not using the correct math for fictious forces in a rotating frame. http://en.wikipedia.org/wiki/Rotating_frame has a good discussion, although they use $ \Omega $ where I use $ \omega_{m288} $. | This page is being reworked. I need to include a z component perpendicular to the orbital plane, because the sun is not in the equatorial plane! |
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| It turns out that for 3 gram 50 micron thick server satellites in m288 orbits, the orbital perturbations from light pressure are small, 360 meter elliptical oscillations along the line of the orbit. If the satellites get thinner, the perturbations increase proportional to the area to mass ratio. If the orbits move further out, the perturbations increase proportional to the cube of the orbit radius, because the orbit period grows, and the perturbations are proportional to the square of the orbit period. There is also a cumulative perturbation caused by the eclipse time, which is probably small enough to be corrected by optical maneuvering. | FIXME Previously I was not using the correct math for fictious forces in a rotating frame. http://en.wikipedia.org/wiki/Rotating_frame has a good discussion, although they use $ \Omega $ where I use $ \omega_{m288} $. I will show that for 3 gram 50 micron thick flat-sats in m288 orbits, the orbital perturbations from light pressure are small, 181 meter front to back oscillations along the line of the orbit. If the satellites get thinner, the perturbations increase proportional to the area to mass ratio. If the orbits move further out, the perturbations increase proportional to the cube of the orbit radius, because the orbit period grows, and the perturbations are proportional to the square of the orbit period. There is also a cumulative perturbation caused by the eclipse time, which is probably small enough to be corrected by optical maneuvering. |
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| || 3.986004418e14 m^3^/s^2^ || $ \mu_{\oplus} $ || Earth gravitational parameter || || 12,788,866 m || $ r_{m288} $ || m288 orbit radial distance || || 80,354,815 m || || orbit circumference || || 17,280 sec || || orbit period relative to Earth surface || || 14,400 sec || || orbit period relative to Sun || || 14,393.432 sec || $ T_{m288} $ || sidereal orbit period relative to stars || || 5,582.7418 m/s || $ v_{m288} $ || orbital velocity || || 2,290.7858 sec/rad || $ 1/\omega_{m288} $ || reciprocal of angular velocity || || 4.3653142e-4 rad/sec || $ \omega_{m288} $ || angular velocity || |
|| 3.986004418e14 m^3^/s^2^ || $ \mu_{\oplus} $ || Earth gravitational parameter || || 12,788,866 m || $ r_{m288} $ || m288 orbit radial distance || || 80,354,815 m || || orbit circumference || || 17,280 sec || || orbit period relative to Earth surface || || 14,400 sec || || orbit period relative to Sun || || 14,393.432 sec || $ T_{m288} $ || sidereal orbit period relative to stars || || 5,582.7418 m/s || $ v_{m288} $ || orbital velocity || || 2,290.7858 sec/rad || $ 1/\omega_{m288} $ || reciprocal of angular velocity || || 4.3653142e-4 rad/sec || $ \omega_{m288} $ || angular velocity || |
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| $ a_x = -( a / r ) x = \omega^2 x $ | $ a_x = -( a / r ) x = - \omega^2 x $ |
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| $ \partial a_r = \omega^2 + 2 \mu_{\oplus} / r^3 \partial r = 3 \omega^2 \partial r $ | $ \partial a_r = ( \omega^2 + 2 \mu_{\oplus} / r^3 ) \partial r = 3 \omega^2 \partial r $ |
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| $ a_{{\lambda} ~y } = a_{\lambda} \sin( \omega t ) $ | $ a_{{\lambda} ~ y } = a_{\lambda} \sin( \omega t ) $ |
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| $ a_{{\lambda} ~ x } = a_{\lambda} \cos( \omega t ) $ | $ a_{{\lambda} ~ x } = - a_{\lambda} \cos( \omega t ) $ |
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| Assume that $ a_{{\lambda} ~ x } = a_x $, so that: | Assume that $ x = \lambda \cos( \omega t ) ~ ~ $ where $ \lambda \equiv $ the light pressure distance. |
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| $ a_x = \ddot x = a_{\lambda} \cos( \omega t ) $ | Then $ ~ ~ ~ v_x = \dot { x } = - \lambda \omega \sin( \omega t ) $ |
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| Integrating for x: | and $ ~ ~ ~ \ddot { x } = - \lambda \omega^2 \cos( \omega t ) = - \omega^2 x $ |
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| $ v_x = \dot x = ( { a_{\lambda} / \omega } ) \sin( \omega ~ t ) $ | The sum of the inertial force and the displacement force is equal to the light pressure: |
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| Integrate again for x: | $ - a_{\lambda} \cos( \omega t ) = \ddot { x } + a_x = - \omega^2 x + - \omega^2 x = -2 \omega^2 \lambda \cos( \omega t ) $ |
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| $ x = - ( a_{\lambda} / \omega^2 ) \cos ( \omega ~ t ) $ | Therefore $ ~ ~ ~ \lambda = a_{\lambda} / ( 2 \omega^2 ) ~ ~ ~ = $ 90.6 meters and $ v_x = - a_{\lambda} / ( 2 \omega ) \sin( \omega t ) $ |
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| That is the x position relative to orbit center. | |
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| Assume that $ a_{{\lambda} ~ y } = a_y $ so that | |
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| $ 3 \omega^2 y + 2 \omega v_x = a_{\lambda} \sin( \omega t ) $ | |
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| $ 3 \omega^2 y + 2 a_{\lambda} \sin( \omega ~ t ) = a_{\lambda} \sin( \omega t ) $ | The acceleration of y is the $ a_y $ fictitious force plus the light pressure" |
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| $ 3 \omega^2 y = - a_{\lambda} \sin( \omega ~ t ) $ | $ \ddot{ y } = a_y + a_{{\lambda} ~ y } = 3 \omega^2 y + 2 \omega v_x + a_{\lambda} \sin( \omega t ) $ |
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| $ y = -( { a_{\lambda} } / 3 \omega^2 ) \sin( { \omega ~ t } ) $ | $ \ddot{ y } = 3 \omega^2 y + 2 \omega ( - a_{\lambda} / ( 2 \omega ) ) \sin( \omega t ) + a_{\lambda} \sin( \omega t ) $ |
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| Define $ \lambda $ as the light pressure distance: | $ \ddot{ y } = 3 \omega^2 y - a_{\lambda} \sin( \omega t ) + a_{\lambda} \sin( \omega t ) $ |
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| $ \lambda \equiv a_{\lambda} / \omega^2 = 181.2 meters $ | $ \ddot{ y } = 3 \omega^2 y $ The only solution for this equation is $ y = 0 $ !!! The satellite oscillates back and forth along the path of the orbit. Since $ r^3 = \mu_{\oplus}/ \omega^2 $, $ \lambda = a_{\lambda} r^3 / \mu_{\oplus} $The light pressure distance increases as the cube of the orbital radius. Very thin server satellites at very high distances will get perturbed right out of orbit, the approximations break down. In fact, if these satellites are tilted slightly off axis from the sun, they can hover near the Earth-Sun line, as suggested by Roger Angel ( NEED REFERENCE HERE ). However, if control is lost, they can also get perturbed into the low night sky and cause biologically damaging light pollution, so such ultra-light flat-sats are probably not a very good idea near the Earth. |
Light Pressure Modified Orbits
Light pressure effects modify server-sat array orbits. In the nominal orbit, server-sats are at "half thrust", with each thruster half mirror and half transparent. Variations to full or zero reflectivity, and full or zero thrust, allow each server-sat to maneuver in relation to the array, or for the array as a whole to maneuver around its assigned centerpoint. The following is an analysis of two effects, earth oblateness and nominal half-thrust light pressure, on the orbit. We will assume that the arrays maintain a constant, slightly elliptical orbit that precesses once per year in the equatorial plane. We will assume continuous illumination tangential to the equatorial plane, and zero light pressure effects from Earth albedo or black-body radiation, and no solar or lunar tides. These assumptions are somewhat crude approximations to get us into the ballpark of a solution. Precise solutions will probably demand accurate numerical solutions simulating many years of orbital evolution.
Earth Oblateness
For "heavy" server-sats relatively close to the earth, such as 3 gram satellites at m288, the dominant deviation from a perfect Kepler orbit is caused by the J_2 spherical harmonic of the gravity field, in turn caused by the oblateness of the spinning Earth. For small eccentricities, the eastward precession of the perigee of one elliptical equatorial orbit is proportional to the J_2 term of the WGS84 model ( -1.082626683E-03 , see Pisacane 2008 ) and the inverse of the orbit radius squared. The precession, expressed as the number complete precessions per year, is J \approx -3 J_2 ( Y / P ) ( R_e / R_s ) ^ 2 where J \equiv precessions per year caused by oblateness, J_2 \equiv spherical harmonic of gravity causing oblateness, Y \equiv year period in seconds, P \equiv orbit period in seconds, Y/P \equiv number of orbits per year, R_e \equiv earth equatorial radius = 6378km, R_s \equiv orbit equatorial radius.
orbit |
radius (RE) |
orbits/year |
J |
LEO |
1.047 |
5758.5 |
17.061 |
m288 |
2.005 |
2191.5 |
1.771 |
m360 |
2.264 |
1826.2 |
1.157 |
m480 |
2.627 |
1461.0 |
0.688 |
m720 |
3.182 |
1095.7 |
0.351 |
m1440 |
4.168 |
730.5 |
0.137 |
GEO |
6.611 |
365.2 |
0.027 |
Light Pressure
This page is being reworked. I need to include a z component perpendicular to the orbital plane, because the sun is not in the equatorial plane!
FIXME
Previously I was not using the correct math for fictious forces in a rotating frame. http://en.wikipedia.org/wiki/Rotating_frame has a good discussion, although they use \Omega where I use \omega_{m288} .
I will show that for 3 gram 50 micron thick flat-sats in m288 orbits, the orbital perturbations from light pressure are small, 181 meter front to back oscillations along the line of the orbit. If the satellites get thinner, the perturbations increase proportional to the area to mass ratio. If the orbits move further out, the perturbations increase proportional to the cube of the orbit radius, because the orbit period grows, and the perturbations are proportional to the square of the orbit period. There is also a cumulative perturbation caused by the eclipse time, which is probably small enough to be corrected by optical maneuvering.
The fictitous forces are proportional to the radial and tangential position from mean perturbation center, and the tangential velocity relative to that center. Assume a circular orbit - toroidal orbits do not deviate much from that.
An m288 orbit has the following parameters (subject to verification, please help me check them):
3.986004418e14 m3/s2 |
\mu_{\oplus} |
Earth gravitational parameter |
12,788,866 m |
r_{m288} |
m288 orbit radial distance |
80,354,815 m |
|
orbit circumference |
17,280 sec |
|
orbit period relative to Earth surface |
14,400 sec |
|
orbit period relative to Sun |
14,393.432 sec |
T_{m288} |
sidereal orbit period relative to stars |
5,582.7418 m/s |
v_{m288} |
orbital velocity |
2,290.7858 sec/rad |
1/\omega_{m288} |
reciprocal of angular velocity |
4.3653142e-4 rad/sec |
\omega_{m288} |
angular velocity |
2.4371020 m/s2 |
a_{m288} |
gravitational force |
3.436e-5 m/s2 |
a_{\lambda} |
light pressure acceleration @ 3g, 50\mum glass |
In a circular orbit, the centrifugal acceleration balances the centripedal gravitational acceleration:
v^2 / r = \omega^2 r = a = \mu_{\oplus} / r^2 ~ ~ ~ ~ = 2.4371020 m/s^2 at m288
\omega^2 = \mu_{\oplus} / r^3
If x \equiv the tangential distance forward of orbit center, then for small x the triangle of tangential and radial accelerations is proportional to the tangential and radial distances, from congruent triangles:
\partial a_x / a = - \partial x / r
a_x = -( a / r ) x = - \omega^2 x
With no perturbations, the vertical acceleration is:
a_r = \omega^2 ~ r - \mu_{\oplus} / r^2 = v^2 / r - \mu_{\oplus} / r^2 = 0
If the tangential velocity is perturbed, the radial acceleration is perturbed:
\partial a_r = 2 v / r \partial v_x = 2 \omega ~ \partial v_x
If the radial distance is perturbed, the radial acceleration is also perturbed:
\partial a_r = ( \omega^2 + 2 \mu_{\oplus} / r^3 ) \partial r = 3 \omega^2 \partial r
So the total radial acceleration, for small perturbations of y \equiv \Delta r and x is:
a_y = 3 \omega^2 ~ y + 2 \omega ~ v_x
The radial and tangential accelerations are caused by the light pressure from the Sun. That makes one rotation per 14400 seconds around the guiding center, and is interrupted when the satellite is eclipsed by the Earth. We will compute the effect of this later. If we approximate the rotation as the sidereal period instead, and assume we can make up for eclipse and rotation changes by maneuvering (dangerous assumption), then we can approximate the light pressure components as:
a_{{\lambda} ~ y } = a_{\lambda} \sin( \omega t )
a_{{\lambda} ~ x } = - a_{\lambda} \cos( \omega t )
Assume that x = \lambda \cos( \omega t ) ~ ~ where \lambda \equiv the light pressure distance.
Then ~ ~ ~ v_x = \dot { x } = - \lambda \omega \sin( \omega t )
and ~ ~ ~ \ddot { x } = - \lambda \omega^2 \cos( \omega t ) = - \omega^2 x
The sum of the inertial force and the displacement force is equal to the light pressure:
- a_{\lambda} \cos( \omega t ) = \ddot { x } + a_x = - \omega^2 x + - \omega^2 x = -2 \omega^2 \lambda \cos( \omega t )
Therefore ~ ~ ~ \lambda = a_{\lambda} / ( 2 \omega^2 ) ~ ~ ~ = 90.6 meters and v_x = - a_{\lambda} / ( 2 \omega ) \sin( \omega t )
The acceleration of y is the a_y fictitious force plus the light pressure"
\ddot{ y } = a_y + a_{{\lambda} ~ y } = 3 \omega^2 y + 2 \omega v_x + a_{\lambda} \sin( \omega t )
\ddot{ y } = 3 \omega^2 y + 2 \omega ( - a_{\lambda} / ( 2 \omega ) ) \sin( \omega t ) + a_{\lambda} \sin( \omega t )
\ddot{ y } = 3 \omega^2 y - a_{\lambda} \sin( \omega t ) + a_{\lambda} \sin( \omega t )
\ddot{ y } = 3 \omega^2 y
The only solution for this equation is y = 0 !!! The satellite oscillates back and forth along the path of the orbit.
Since r^3 = \mu_{\oplus}/ \omega^2 , \lambda = a_{\lambda} r^3 / \mu_{\oplus} The light pressure distance increases as the cube of the orbital radius.
Very thin server satellites at very high distances will get perturbed right out of orbit, the approximations break down. In fact, if these satellites are tilted slightly off axis from the sun, they can hover near the Earth-Sun line, as suggested by Roger Angel ( NEED REFERENCE HERE ). However, if control is lost, they can also get perturbed into the low night sky and cause biologically damaging light pollution, so such ultra-light flat-sats are probably not a very good idea near the Earth.
MORE LATER, needs checking!
MORE LATER
Over a very long time, tumbling and completely out of control, random variations in up and down orbit acceleration would eventually drive it up and down, probably in something resembling a random walk. This process is likely to be very slow.
MORE LATER