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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 "J2" spherical harmonic of the gravity field, in turn caused by the oblateness of the spinning Earth. For small eccentricities, the J2 term of the WGS84 model ( -1.082626683E-03 , see Pisacane 2008 ), times the inverse of the orbit radius squared, scales the eastward precession of the perigee of an elliptical equatorial orbit. This precession can be expressed as the number complete precessions per year as $ N \approx -3 J_2 ( Y / P ) ( R_e / R_s ) ^ 2 $ where $ Y $ = year period in seconds, $ P $ = orbit period in seconds, $ Y/P $ = number of orbits per year, $ R_e $ = earth equatorial radius = 6378km, $ R_s $ = orbit equatorial radius. |
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. |
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| || orbit || radius (RE) || orbits/year || precessions/year || || 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 || |
|| 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 || |
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| It is convenient to design for an orbit that always appears in the same spot in the sky, every day. That would be an orbit that precesses once per year. As we will see, light pressure causes an eccentric orbit with apogee towards the sun to precess retrograde ( negative ) per year, while an orbit with perigee towards the sun precesses prograde (positive). We can use light pressure to make up the difference, plus or minus, if it is small. For orbits like m360, the light pressure effects must be small to make up the small difference, which will be difficult for low-mass server-sats with small eccentricity orbits. Perhaps this orbit should be reserved for high mass satellites like O3B. |
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| ||{{attachment:NavigationV01/light-shift1.png|orbit from the north|width=300px}}|| Solar light pressure is useful for maneuvering, but it distorts orbits. Light pressure slows objects orbiting towards the sun, and speeds up objects orbiting away from it. This raises and lowers portions of the orbit. A slightly elliptical orbit with perigee towards the sun will precess prograde, while an orbit with apogee towards the sun will precess retrograde. With proper matching of eccentricity, and added to the J2 precession the apogee and perigee will complete one precession per year. || ||{{attachment:orbitcircle.png|orbit circle|width=300px}}|| An orbit can be viewed in a rotating frame of reference centered on a point in circular orbit. For small eccentricities, an elliptical orbit traces a circle in this frame, rotating in the opposite direction from the orbit itself. That is, if the orbit is eastwards, rotating counterclockwise when viewed from the north, the orbit rotates clockwise in the rotating frame. If the frame is positioned with the earth below, the top of the circle is the apogee and the bottom of the circle is the perigee, so the orbit moves backwards (to the right) compared to the "circular center" at apogee (more slowly), and forwards (to the left) at perigee (more rapidly).<<BR>><<BR>>An elliptical orbit is distorted from a circular one by the distance between the focii of the ellipse. The distance from the center to each focus, or the eccentricity times the semimajor axis, is called the ''linear eccentricity'' of the ellipse, or $ \epsilon r_0 $. The linear eccentricity is the radius of the orbit circle in the rotating frame. || |
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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| If the orbit precesses once per year, then the vector of the linear eccentricity makes one complete rotation per year. Each orbit adds a little bit of rotation to that vector. |
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} $. |
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| Viewed closely, the displacement caused by light pressure looks like a cycloid ruffle on the orbit circle. In the rotating frame of reference, the sun appears to rotate around the orbit circle once per orbit period, with the light pressure acceleration pointing away from the sun. The sun "disappears" during the eclipsed part of the orbit, and this changes the math somewhat, but the major effects are caused when the object is moving towards or away from the sun, on the sides of the ellipse. This rough estimate ignores eclipses. | I will show that for 3 gram 50 micron thick server satellites in m288 orbits, the orbital perturbations from light pressure are small, 360 meter diameter circular oscillations crossing 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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| Referenced to a zero angle at apogee, the light pressure acceleration in the x direction is $ a_x = a_L sin( \omega t ) $ and $ a_y = a_L cos( \omega t ) $. $ a_L $ is approximately 17 $ \mu m / s ^ 2 $ for a 100 micron thick server-sat. Assuming we start at zero position and velocity, then doubly integrating each of these equations results in $ x = ( a_L / \omega^2 ) ( \omega t - sin ( \omega t ) ) $ and $ y = ( a_L / \omega^2 ) ( 1 - cos ( \omega t ) ) $. This is the formula for a cycloid. After one orbit period $P$ ( $\omega = 2 \pi / P $ ) x has shifted by $ a_0 P^2 / 2 \pi $. The sum of these increments add up to the orbit circle over one year $Y$, so $ x Y / P = 2 \pi \epsilon r_0 $. Solving for the linear eccentricity, $ \epsilon r_0 = ( a_L P Y )/( (2 \pi)^2 $. For $ a_L = 17 \mu m / s ^ 2 $ and the m288 orbit ( P = 4*3600 = 14400 seconds ) and a year of 31556926 seconds, this is around '''200 km'''. |
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. |
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| Lighter server-sats will have higher light pressure acceleration, and orbit with larger linear eccentricities. This will result in larger relative displacements. Regions of m288 with higher and lower server-sat area-to-mass ratios should be separated by hundreds or thousands of kilometers. | |
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| Perhaps the m360 and m480 orbits should be reserved for lighter and lighter server-sats, assuming that the trend over time will be towards lighter server-sats, better boosters, and more sophisticated latency management. Solar power arrays at lunar distances will be very light, but will have a lot more room for linear eccentricity displacements. | An m288 orbit has the following parameters (subject to verification, please help me check them): |
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| == Orbital Decay or Expansion == | || 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 || || 2.4371020 m/s^2^ || $ a_{m288} $ || gravitational force || || 3.436e-5 m/s^2^ || $ a_{\lambda} $ || light pressure acceleration @ 3g, 50$\mu$m glass || |
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| The above discussion assumes that the server-sat is under control, and face-forwards towards the sun for maximum power. This is best for computation productivity. However, it is also possible to raise or lower the orbit somewhat by tilting the server-sat at various parts of the orbit. For example, if the server-sat is tilted 60 degrees away from the sun while orbiting towards it (cutting light power and light pressure in half) and face-on towards the sun while orbiting away from it, it will slowly gain orbital speed, about 2cm/sec per orbit, or 42 meters per second per year, raising the orbit. After a few decades, it might be close enough to the moon to either be slingshotted into solar orbit, or sent plunging back to reenter. | 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 $ a_{{\lambda} ~ x } = a_x $, so that: $ a_x = \ddot x = - a_{\lambda} \cos( \omega t ) $ Integrating for x: $ v_x = \dot x = - ( { a_{\lambda} / \omega } ) \sin( \omega t ) $ Integrate again for x: $ x = ( a_{\lambda} / \omega^2 ) \cos ( \omega t ) $ That is the x position relative to orbit center. Assume that $ a_{{\lambda} ~ y } = a_y $ so that $ 3 \omega^2 y + 2 \omega v_x = a_{\lambda} \sin( \omega t ) $ $ 3 \omega^2 y - 2 a_{\lambda} \sin( \omega t ) = a_{\lambda} \sin( \omega t ) $ $ 3 \omega^2 y = 3 a_{\lambda} \sin( \omega t ) $ $ y = -( { a_{\lambda} } / \omega^2 ) \sin( { \omega t } ) $ Define $ \lambda $ as the light pressure distance: $ \lambda \equiv a_{\lambda} / \omega^2 = $ 181.2 meters Since $ r^3 = \mu_{\oplus}/ \omega^2 $, this is equivalent to $ \lambda = a_{\lambda} r^3 / \mu_{\oplus} $ So 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 |
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!
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 server satellites in m288 orbits, the orbital perturbations from light pressure are small, 360 meter diameter circular oscillations crossing 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 a_{{\lambda} ~ x } = a_x , so that:
a_x = \ddot x = - a_{\lambda} \cos( \omega t )
Integrating for x:
v_x = \dot x = - ( { a_{\lambda} / \omega } ) \sin( \omega t )
Integrate again for x:
x = ( a_{\lambda} / \omega^2 ) \cos ( \omega t )
That is the x position relative to orbit center.
Assume that a_{{\lambda} ~ y } = a_y so that
3 \omega^2 y + 2 \omega v_x = a_{\lambda} \sin( \omega t )
3 \omega^2 y - 2 a_{\lambda} \sin( \omega t ) = a_{\lambda} \sin( \omega t )
3 \omega^2 y = 3 a_{\lambda} \sin( \omega t )
y = -( { a_{\lambda} } / \omega^2 ) \sin( { \omega t } )
Define \lambda as the light pressure distance:
\lambda \equiv a_{\lambda} / \omega^2 = 181.2 meters
Since r^3 = \mu_{\oplus}/ \omega^2 , this is equivalent to
\lambda = a_{\lambda} r^3 / \mu_{\oplus}
So 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