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Summary: while 50 micron thinsats are probably manufacturable, they may be a bit too thin, perhaps by a factor of 2. Light-pressure-related eccentricity gets very large if the thinsats are too light. THIS NEEDS FURTHER STUDY. ------ |
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| For "heavy" thinsats 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 $ N_{PR} \approx -3 J_2 ( Y / P ) ( R_e / R_s ) ^ 2 ~ ~ $ where | For "heavy" thinsats 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. For small eccentricity and inclination, the precession, expressed as the number complete precessions per year, is $ N_{pr} \approx -3 J_2 ( Y / T ) ( R_e / A ) ^ 2 ~ ~ $ where |
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| ||$ N_{PR} $ ||precessions per year caused by oblateness || | ||$ N_{pr} $ ||precessions per year caused by oblateness || |
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| ||$ P $ ||orbit period in seconds || ||$ Y/P $ ||number of orbits per year || ||$ R_e $ ||earth equatorial radius = 6378k || ||$ R_s $ ||orbit equatorial radius || |
||$ T $ ||orbit period in seconds || ||$ Y/T $ ||number of orbits per year || ||$ R_e $ ||earth equatorial radius = 6378 km || ||$ A $ ||orbit semimajor axis (radius if circular) || |
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| || orbit || radius (RE) || orbits/year ||$N_{PR}$|| | Note: Pisacane's formula (5.47b) is for $ d \omega / dt $ in radians per second, and is in terms of $ n = 2 \pi / T $ . Dividing both sides by $ 2 \pi $ and multiplying by $ Y $ seconds per year gives the number of complete 360° (=2π radian) precessions per year. His more accurate formula uses $ a ( 1 - e^2 ) $ instead of $ A $ and has a correction for inclination, but both inclination and ellipsicity are small for these near-equatorial near-circular orbits, so we can use the above approximation to a few percent accuracy. || orbit || radius (RE) || orbits/year ||$N_{pr}$|| |
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To keep a thinsat orbit in the same orbit and orientation to the sun, the light pressure must retard the precession at m288 and m360, and advance it at m480 and m720, so that the total precessions per year is one. More complicated orbit evolution may allow different ratios, however, eccentricity should not accumulate over many years or the apogee and perigee will eventually intersect other orbital bands, causing collisions. |
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| $ \vec{ r } = \left( \matrix{ x \\ y \\ z } \right) = { { a( 1 - e^2 ) } \over { 1 + 3 cos~\nu } } \left( \matrix { cos~\Omega~cos( \omega+\nu ) - sin{ \Omega } sin( \omega+\nu ) cos~i \\ sin~\Omega~cos( \omega+\nu ) - cos~\Omega sin( \omega+\nu ) cos~i \\ sin( \omega+\nu ) sin~i } \right) $ . . . Soop pg 25 | $ \vec{ r } = \left( \matrix{ x \\ y \\ z } \right) = { { a( 1 - e^2 ) } \over { 1 + 3 cos~\nu } } \left( \matrix { cos~\Omega~cos( \omega+\nu ) - sin~\Omega~sin( \omega+\nu )~cos~i \\ sin~\Omega~cos( \omega+\nu ) - cos~\Omega~sin( \omega+\nu )~cos~i \\ sin( \omega+\nu )~sin~i } \right) $ . . . Soop pg 25 |
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| $ \vec{ e } = ( e~cos( \Omega + omega ), e~sin( \Omega + omega ) ) $ . . . Soop pg 27 | $ \vec{ e } = ( e~cos( \Omega + \omega ), e~sin( \Omega +\omega ) ) $ . . . Soop pg 27 |
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| $ \sigma = { light pressure } / { mass } $ . . . after Soop pg 93 | $ \lambda $ = 3.65e-5 s^-2^ is the light pressure acceleration, which is the solar pressure ( 4.56e-6 N/m^2^ ) times the area to mass ratio, 8 m^2^ = ( 0.024 m^2^ / 0.003 kg ) for a 50 μ m thinsat. . . similar to Soop pg 93 |
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| $ { { \partial \vec{ e } } \over { \partial t } } = { { P \sigma } \over { 2 \pi V } } \int_0^{2\pi} \left[ 2 \left( \matrix{ cos~s \\ sin~s } \right) sin( s-s_s ) - \left( \matrix{ sin~s \\ -cos~s } \right) cos(s-s_s) \right] ds $ . . . Soop pg 93 | $ { { \partial \vec{ e } } \over { \partial t } } = { \lambda \over { 2 \pi V } } \int_0^{2\pi} \left[ 2 \left( \matrix{ cos~s \\ sin~s } \right) sin( s-s_s ) - \left( \matrix{ sin~s \\ -cos~s } \right) cos(s-s_s) \right] ds $ . . . Soop pg 93 |
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| $ { { \partial \vec{ e } } \over { \partial t } } = { { 3 P \sigma } \over { 2 V } } \left( \matrix{ - sin~s_s\\cos~s_s } \right) $ | $ { { \partial \vec{ e_{lambda} } } \over { \partial t } } = { { 3 \lambda } \over { 2 V } } \left( \matrix{ - sin~s_s\\cos~s_s } \right) $ perpendicular to the Sun, perturbing the eccentricity vector eastward . . . Soop page 95 $ Y $ = one year . . . Soop page 95 $ \vec{ e_{\lambda} } ( t ) = { { 3 P \sigma Y } \over { 4 \pi V } } \left( \matrix{ cos~s_s\\sin~s_s } \right) $ . . . Soop page 95 Without the J2 modifications, a light-pressure-modified orbit(with $ \vec e $ and perigee pointing towards the sun ) for a 3 gram, 240cm^2^ thinsat ( \sigma = 8 m^-1^s^-2> would have a light pressure eccentricity of: $ e_{\lambda} = { { 3 \lambda Y } \over { 4 \pi V } } = { { 3\times3.65e-5\times3.156e7 } \over { 4\times\pi\times5582.74 } } $ = 0.049 The eccentricity of the light modified orbit is around 0.05 . If the thinsat is thinner, that gets larger. It also gets larger for more distant orbits ( M360, M480, etc.). Proportional to period T^1/3^ === Inclination to the Ecliptic === Because of the inclination of the orbit relative to the ecliptic, some of the light pressure is towards the south during summer and the north during winter. The eccentricity-modifying light pressure is a function of the sun's declination $ \delta_\odot $ : $ \lambda = \lambda_0 ~ cos ~ \delta_\odot $ where: $ \delta_\odot = \arcsin \left [ \sin \left ( -23.44^\circ \right ) \cdot \cos \left ({{360^{\circ}\times(day+10)}\over{365}} \right ) \right ] ~ ~ ~ $ where $ day $ is the day of the year starting with Jan 1 = 0 thus $ \lambda = \lambda_0 \sqrt{ 1 - \left [ \sin \left ( -23.44^\circ \right ) \cdot \cos \left ({{360^{\circ}\times(day+10)}\over{365}} \right ) \right ]^2 } $ $ \langle \lambda \rangle \approx 0.9592 ~ \lambda_0 $ so $ e_{\lambda} $ is reduced somewhat, to 0.047 . ---- === Modifications for earth shadow eclipse time === The eclipse time has a small effect, because we are only subtracting an acceleration approximately equal to sin^2^ of the +/- 30° angle behind the earth. Integrating from -150° to 150° subtracts approximately 1/24 of the effect, so $ e_{\lambda} $ is reduced to about 0.045 . ---- === Combining with the J2 Modification === The simplest case for M288 is an eccentricity vector diagram like so: {{ attachment:LightOrbit288.png | | width=640 }} For a bounded elliptical M288 orbit, $ e_{J2} = 1.771 \times e $ and $ e_{ year } = e $ , so with apogee towards the sun, $ e_{ \lambda } = 0.771 \times e $. That means that for a 50μm 8 m^2^/kg thinsat, $ e $ = 0.045 / 0.771 or 0.058, or perhaps a little larger because of the twice-annual variation in tangential light acceleration. Assume that the distance to orbits near M288 (for example, LAGEOS) should be at least 0.06*12789 km, so there should be no valuable equatorial crossers between 12000 and 13300 km . FIXME ... the LAGEOS satellites cross the equator at around 12230km - WE MAY NEED TO MAKE THE 50μm THINSAT HEAVIER!!! Or maybe our rising and descending nodes can be made to match LAGEOS in some manner, so we always stay many kilometers away. Needs more analysis!!! |
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| == Old Light Pressure Analysis == | M360 orbits have a J2 excess of 0.157 instead of 0.771, which means that the eccentricity to mass ratio will be almost 5 times larger than M288. M360 does not look like a practical orbit for low mass thinsats. Cheap launch and ballast will change this. |
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| {{{#!wiki caution '''THE FOLLOWING ANALYSIS NEEDS FIXING.''' THE ANALYSIS BELOW IS INCORRECT because there is no $ \ddot x = k x $ force. |
{{ attachment:LightOrbit480.png | | width=640 }} |
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| Light pressure increases apogee in the eastward orbital direction, decreases perigee in the west direction. The eclipse and J_2 and sun rotation perturbations must be made to match the light pressure perturbations, perhaps with some additional help from sideways thrust from non-perpendicular orientation. }}} |
The M480 orbit ( 16756km ) is slower so $ e_{\lambda} is larger: |
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| I will show that for 3 gram 50 micron thick thinsats in m288 orbits, the orbital perturbations from light pressure are small, 180 meters front to back oscillations along the line of the orbit at the spring and fall equinoxes, and 165 meters front to back at the summer and winter solstices. 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. | $ e_{\lambda488} = { { 3 \lambda Y } \over { 4 \pi V } } = { { 3\times3.65e-5\times3.156e7 } \over { 4\times\pi\times 4877.51 } } $ = 0.056, reduced to perhaps 0.55 with eclipse effects. |
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| Previously I was not using the correct math for fictitious 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} $. Their rotating frame does not include gravity, so the centrifugal force described here is stronger in the radial direction. Assume a circular orbit - toroidal orbits do not deviate much from that, and we can ignore Euler forces. | For a bounded elliptical M480 orbit, $ e_{J2} = 0.688 \times e $ and $ e_{ year } = e $ , so with perigee towards the sun, $ e_{ \lambda } = 0.312 \times e $. That means that for a 50μm 8 m^2^/kg thinsat, $ e $ = 0.055 / 0.312 or 0.176 . Assume that the distance to orbits near M480 should be at least 0.18*16756 km, so there should be no valuable equatorial crossers between 13700 and 19800 km . This is a very wide band; we probably must make m480 thinsats heavier. |
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| The fictitious forces are related to the radial and tangential position from mean perturbation center, and to the tangential velocity relative to that center. | ---- |
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| An m288 orbit has the following parameters (subject to verification, please help me check them): | === Inclination - Flattened Toroidal Orbit === |
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| || 3.986004418e14 m^3^/s^2^ || $ \mu_{\oplus} $ || Earth gravitational parameter || || 23.439281° || $ \phi $ || Earth axial tilt || || 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 || || 3.648e-5 m/s^2^ || $ a_{\lambda} $ || total light pressure acceleration @ 3g, 50$\mu$m glass || || 191.4 m || $ \lambda $ || related light pressure displacement, see below || || 0.161 || || Equinox eclipse fraction || || 0.111 || || Solstice eclipse fraction || |
Because of the eccentricity, we will incline the orbit, so that the central orbit can be set aside for heavier thinsats. We should consider a flattened torus so that this orbit does not go too far south or north. If we make the inclination for the M288 orbit ( $ e $ = 0.058 ) 1°, then the orbit passes above and below the 12789 km central orbit by 223 km, compared to the 740 km sideways distance. The circular orbital speed at M288 is 5583 m/s. The relative velocity of the two orbits at crossing ( r = A, $ cos( \theta ) = -e $ = 0.058, or $ \theta $ = 93.3° ) is: |
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| In a circular orbit, the centrifugal acceleration balances the centripedal gravitational acceleration: | || tangential || $ e V cos( theta ) = e^2 V $ || 18.8 m/s || || radial || $ e V sin( theta ) $ || 323.3 m/s || || total || $ e V $ || 323.8 m/s || |
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| $ 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. This can be divided into two components, the planar light pressure parallel to the plane and the light pressure perpendicular to the plane. These are a function of the time of year $ \beta $, and the Earth's axial tilt of $ \phi = $ 23.439281° . We can compute the components from the cross product of the sunwards light pressure vector $ a_{\lambda} ~ \hat j $ and the normal vector of the equatorial plane given by $ \hat i = \sin( \phi ) \cos( \beta ) ~ $, $ ~ ~ \hat j = \sin( \phi ) \sin( \beta ) $, and $ \hat k = \cos ( \phi ) $. The perpendicular pressure is $ a_{\lambda ~ z} = a_{\lambda} \sin( \phi ) \sin( \beta ) $, and the planar pressure is $ a_{\lambda ~ p} = a_{\lambda} sqrt{ 1 - ( \sin( \phi ) \sin( \beta ) )^2 } $. At the equinoxes, $ a_{\lambda ~ z} = 0 $ and $ a_{\lambda ~ p} = a_{\lambda} $. At the solstices, $ a_{\lambda ~ z} = \pm a_{\lambda} \sin( \phi ) $ and $ a_{\lambda ~ p} = a_{\lambda} \cos( \phi ) $. The perpendicular component does not change as the object orbits. The object is displaced above or below the equatorial plane by $ z ~ = ~ a_{\lambda ~ z} / \omega^2 ~ = ~ ( a_{\lambda} / { \omega^2 } ) \sin ( \phi ) \sin( \beta ) $. Define the parameter $ \lambda ~ \equiv ~ a_{\lambda} / { \omega^2 } = $ 191.4 meters for a 3 gram, 50 $\mu$m thick glass flat sat at m288 . The z displacement varies sinusoidally between $\pm$ 76.1 meters over the course of a year, far smaller than the cross section of the toroidal orbit. $ \omega^2 = \mu_{\oplus} / r^3 $, so $ \lambda = a_{\lambda} r^3 / \mu_{\oplus} $. A thinner thinsat has higher $ a_{\lambda} $ and higher $ \lambda $. A more distant orbit also has higher $ \lambda $. For sufficiently thin satellites or large distances, the approximations above break down, and light pressure will push the satellites out of orbit, perhaps to escape velocity. The planar light pressure 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 ~ p} \sin( \omega t ) $ $ a_{{\lambda} ~ x } = - a_{\lambda ~ p } \cos( \omega t ) $ Assume that $ x = k \lambda \cos( \omega t ) ~ ~ $ Then $ ~ ~ ~ v_x = \dot { x } = - k \lambda \omega \sin( \omega t ) $ and $ ~ ~ ~ \ddot { x } = - k \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 ~ p } \cos( \omega t ) = \ddot { x } + a_x = - \omega^2 x + - \omega^2 x = -2 \omega^2 k \lambda \cos( \omega t ) $ Dividing both sides by $ \omega^2 \cos( \omega t ) $: $ - a_{\lambda ~ p } / \omega^2 = \lambda = -2 k \lambda $. Thus, $ k = -0.5 $. Therefore $ x = - 0.5 \lambda \cos( \omega t ) $ and $ v_x = - 0.5 a_{\lambda} / \omega \sin( \omega t ) $ The acceleration of y is the $ a_y $ fictitious force plus the light pressure: $ \ddot{ y } = a_y + a_{{\lambda ~ p } ~ y } = 3 \omega^2 y + 2 \omega v_x + a_{\lambda ~ p} \sin( \omega t ) = 3 \omega^2 y + 2 \omega ( - a_{\lambda ~ p} / ( 2 \omega ) ) \sin( \omega t ) + a_{\lambda ~ p} \sin( \omega t ) = 3 \omega^2 y - a_{\lambda ~ p} \sin( \omega t ) + a_{\lambda ~ p} \sin( \omega t ) = 3 \omega^2 y $ If $ \ddot{ y } = 3 \omega^2 y $ , the only stable solution is $ y = 0 $ !!! The satellite oscillates back and forth along the path of the orbit, but is not displaced radially by light pressure. The centrifugal acceleration of the tangential velocity exactly balances the light pressure. == Perturbations because of eclipse == The above would be accurate if the earth was transparent. However, the satellite passes behind the earth for as much as 16.1% of its orbit during the equinoxes, and 11.1% at the solstices. |
The "velocity shear" across the orbit is 1.45 m/s per km or 1.45e-3 / second. If we imagine a series of increasing-weight thinsats around the central orbit, lighter at the edges and denser near the central orbit, then M288 orbits that differ in velocity by 1.45 m/s will always more than 1 km apart. |
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--------------- Old (incorrect) analysis removed. |
Light Pressure Modified Orbits
Summary: while 50 micron thinsats are probably manufacturable, they may be a bit too thin, perhaps by a factor of 2. Light-pressure-related eccentricity gets very large if the thinsats are too light. THIS NEEDS FURTHER STUDY.
Light pressure effects modify thinsat array orbits. In the nominal orbit, thinsats 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 thinsat 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" thinsats 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. For small eccentricity and inclination, the precession, expressed as the number complete precessions per year, is N_{pr} \approx -3 J_2 ( Y / T ) ( R_e / A ) ^ 2 ~ ~ where
N_{pr} |
precessions per year caused by oblateness |
J_2 |
spherical harmonic of gravity causing oblateness |
Y |
year period in seconds, |
T |
orbit period in seconds |
Y/T |
number of orbits per year |
R_e |
earth equatorial radius = 6378 km |
A |
orbit semimajor axis (radius if circular) |
Note: Pisacane's formula (5.47b) is for d \omega / dt in radians per second, and is in terms of n = 2 \pi / T . Dividing both sides by 2 \pi and multiplying by Y seconds per year gives the number of complete 360° (=2π radian) precessions per year. His more accurate formula uses a ( 1 - e^2 ) instead of A and has a correction for inclination, but both inclination and ellipsicity are small for these near-equatorial near-circular orbits, so we can use the above approximation to a few percent accuracy.
orbit |
radius (RE) |
orbits/year |
N_{pr} |
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 |
To keep a thinsat orbit in the same orbit and orientation to the sun, the light pressure must retard the precession at m288 and m360, and advance it at m480 and m720, so that the total precessions per year is one. More complicated orbit evolution may allow different ratios, however, eccentricity should not accumulate over many years or the apogee and perigee will eventually intersect other orbital bands, causing collisions.
Thinsat characteristics
Light pressure parameters |
|||
Light Power |
1367 |
W/m2 |
|
Speed of Light |
2.998e+8 |
m/s |
|
Light pressure |
4.56e-6 |
kg/m-s2 |
|
Thinsat parameters |
|||
mass |
3e-3 |
kg |
|
thickness |
5e-5 |
m |
|
density |
2.5e+3 |
kg/m3 |
|
volume |
1.2e-6 |
m3 |
|
area |
2.4e-2 |
m2 |
|
length |
18.5e-2 |
m |
rounded thruster top |
force |
1.0944e-7 |
kg-m/s2 |
|
acceleration |
3.648e-5 |
m/s2 |
|
Light Pressure
A useful starting analysis is in E. M. Soop, "Handbook of Geostationary Orbits". Soop's analysis is for geostationary orbits, which are rarely in eclipse and much less subject to J2. perturbations. However, Soop's analysis is a good starting point. The book is practical, focused on satellite operation, the math is moderate, and the references are rather skimpy.
Soop analyzes the geostationary orbit in the cartesian MEGSD (Mean Equatorial Geocentric System of Date) coordinate system (pg. 15). This system is approximate and quasi-inertial; very accurate analyses will require full numerical solutions. The X-Y plane of MEGSD is the Earth's equatorial plane (which slowly precesses 0.014° per year), with the x direction oriented sidereally, towards the Vernal Equinox or First Point of Aries, where the equatorial plane and the ecliptic plane intersect. The Z direction is north.
Soop describes the various orbital parameters as both scalars (pg 21) and vectors. Unlike Soop, I will use \vec{ x } instead of \overline{ x } .
|
scalar |
vector |
|
semimajor axis |
a |
||
eccentricity |
e |
\vec{ e } |
|
inclination |
i |
\vec{ I } |
|
right ascension of the ascending node |
\Omega |
||
argument of perigee |
\omega |
||
true anomaly |
\nu |
||
unperturbed orbit angle |
s |
||
radius |
r |
\vec{ r } |
|
velocity |
dr / dt |
d\vec{ r }/dt |
|
unperturbed orbit velocity |
V = \sqrt{ \mu / A } |
|
similar to Soop page 39 |
apogee |
r_a |
||
perigee |
r_p |
||
period |
T = 2 \pi \sqrt{ a^3/\mu } |
||
radial velocity |
V_r = V (e_x~sin~s~-~e_y~cos~s) |
|
similar to Soop page 39 |
tangential velocity |
V_t = 2 V (e_x~cos~s~+~e_y~sin~s) |
|
similar to Soop page 39 |
orthogonal (North) velocity |
V_o = V (i_x~sin~s~-~i_y~cos~s) |
|
similar to Soop page 39 |
\vec{ I } = \left( \matrix{ sin~i~sin~\Omega \\ -sin~i~cos~\Omega \\ cos~i } \right)
\vec{ r } = \left( \matrix{ x \\ y \\ z } \right) = { { a( 1 - e^2 ) } \over { 1 + 3 cos~\nu } } \left( \matrix { cos~\Omega~cos( \omega+\nu ) - sin~\Omega~sin( \omega+\nu )~cos~i \\ sin~\Omega~cos( \omega+\nu ) - cos~\Omega~sin( \omega+\nu )~cos~i \\ sin( \omega+\nu )~sin~i } \right) . . . Soop pg 25
A \equiv MXXX unperturbed orbit radius . . . similar to Soop page 26
\vec{ i } = ( i~sin( \Omega ), i~cos( \Omega ) ) . . . Soop pg 27
\vec{ e } = ( e~cos( \Omega + \omega ), e~sin( \Omega +\omega ) ) . . . Soop pg 27
r \approx A + \delta a~-~A e~cos~\nu . . . Soop pg 29
s_b is angle at thrust . . . Soop pg 53
\Delta \vec{ e } = { { 2 \Delta V } \over V } \left( \matrix{ cos~s_b \\ sin~s_b } \right) . . . Soup pg 54
\lambda = 3.65e-5 s-2 is the light pressure acceleration, which is the solar pressure ( 4.56e-6 N/m2 ) times the area to mass ratio, 8 m2 = ( 0.024 m2 / 0.003 kg ) for a 50 μ m thinsat. . . similar to Soop pg 93
s is sidereal angle of orbit . . . Soop pg 93 text
s_s is sidereal angle of Sun . . . Soop pg 93 text
{ { d\vec{ e } } \over dt } = { 2 \over V } \left( \matrix{ cos~s \\ sin~s } \right) { { d V_t } \over dt } + { 1 \over V } \left( \matrix{ sin~s \\ -cos~s } \right) { { d V_r } \over dt } . . . Soop pg 93
{ { \partial \vec{ e } } \over { \partial t } } = { \lambda \over { 2 \pi V } } \int_0^{2\pi} \left[ 2 \left( \matrix{ cos~s \\ sin~s } \right) sin( s-s_s ) - \left( \matrix{ sin~s \\ -cos~s } \right) cos(s-s_s) \right] ds . . . Soop pg 93
{ { \partial \vec{ e_{lambda} } } \over { \partial t } } = { { 3 \lambda } \over { 2 V } } \left( \matrix{ - sin~s_s\\cos~s_s } \right) perpendicular to the Sun, perturbing the eccentricity vector eastward . . . Soop page 95
Y = one year . . . Soop page 95
\vec{ e_{\lambda} } ( t ) = { { 3 P \sigma Y } \over { 4 \pi V } } \left( \matrix{ cos~s_s\\sin~s_s } \right) . . . Soop page 95
Without the J2 modifications, a light-pressure-modified orbit(with \vec e and perigee pointing towards the sun ) for a 3 gram, 240cm2 thinsat ( \sigma = 8 m-1s^-2> would have a light pressure eccentricity of:
e_{\lambda} = { { 3 \lambda Y } \over { 4 \pi V } } = { { 3\times3.65e-5\times3.156e7 } \over { 4\times\pi\times5582.74 } } = 0.049
The eccentricity of the light modified orbit is around 0.05 . If the thinsat is thinner, that gets larger. It also gets larger for more distant orbits ( M360, M480, etc.). Proportional to period T1/3
Inclination to the Ecliptic
Because of the inclination of the orbit relative to the ecliptic, some of the light pressure is towards the south during summer and the north during winter. The eccentricity-modifying light pressure is a function of the sun's declination \delta_\odot :
\lambda = \lambda_0 ~ cos ~ \delta_\odot
where:
\delta_\odot = \arcsin \left [ \sin \left ( -23.44^\circ \right ) \cdot \cos \left ({{360^{\circ}\times(day+10)}\over{365}} \right ) \right ] ~ ~ ~ where day is the day of the year starting with Jan 1 = 0
thus
\lambda = \lambda_0 \sqrt{ 1 - \left [ \sin \left ( -23.44^\circ \right ) \cdot \cos \left ({{360^{\circ}\times(day+10)}\over{365}} \right ) \right ]^2 }
\langle \lambda \rangle \approx 0.9592 ~ \lambda_0 so e_{\lambda} is reduced somewhat, to 0.047 .
Modifications for earth shadow eclipse time
The eclipse time has a small effect, because we are only subtracting an acceleration approximately equal to sin2 of the +/- 30° angle behind the earth. Integrating from -150° to 150° subtracts approximately 1/24 of the effect, so e_{\lambda} is reduced to about 0.045 .
Combining with the J2 Modification
The simplest case for M288 is an eccentricity vector diagram like so:
For a bounded elliptical M288 orbit, e_{J2} = 1.771 \times e and e_{ year } = e , so with apogee towards the sun, e_{ \lambda } = 0.771 \times e . That means that for a 50μm 8 m2/kg thinsat, e = 0.045 / 0.771 or 0.058, or perhaps a little larger because of the twice-annual variation in tangential light acceleration. Assume that the distance to orbits near M288 (for example, LAGEOS) should be at least 0.06*12789 km, so there should be no valuable equatorial crossers between 12000 and 13300 km .
FIXME ... the LAGEOS satellites cross the equator at around 12230km - WE MAY NEED TO MAKE THE 50μm THINSAT HEAVIER!!! Or maybe our rising and descending nodes can be made to match LAGEOS in some manner, so we always stay many kilometers away. Needs more analysis!!!
MORE LATER
M360 orbits have a J2 excess of 0.157 instead of 0.771, which means that the eccentricity to mass ratio will be almost 5 times larger than M288. M360 does not look like a practical orbit for low mass thinsats. Cheap launch and ballast will change this.
The M480 orbit ( 16756km ) is slower so $ e_{\lambda} is larger:
e_{\lambda488} = { { 3 \lambda Y } \over { 4 \pi V } } = { { 3\times3.65e-5\times3.156e7 } \over { 4\times\pi\times 4877.51 } } = 0.056, reduced to perhaps 0.55 with eclipse effects.
For a bounded elliptical M480 orbit, e_{J2} = 0.688 \times e and e_{ year } = e , so with perigee towards the sun, e_{ \lambda } = 0.312 \times e . That means that for a 50μm 8 m2/kg thinsat, e = 0.055 / 0.312 or 0.176 . Assume that the distance to orbits near M480 should be at least 0.18*16756 km, so there should be no valuable equatorial crossers between 13700 and 19800 km . This is a very wide band; we probably must make m480 thinsats heavier.
Inclination - Flattened Toroidal Orbit
Because of the eccentricity, we will incline the orbit, so that the central orbit can be set aside for heavier thinsats. We should consider a flattened torus so that this orbit does not go too far south or north. If we make the inclination for the M288 orbit ( e = 0.058 ) 1°, then the orbit passes above and below the 12789 km central orbit by 223 km, compared to the 740 km sideways distance. The circular orbital speed at M288 is 5583 m/s. The relative velocity of the two orbits at crossing ( r = A, cos( \theta ) = -e = 0.058, or \theta = 93.3° ) is:
tangential |
e V cos( theta ) = e^2 V |
18.8 m/s |
radial |
e V sin( theta ) |
323.3 m/s |
total |
e V |
323.8 m/s |
The "velocity shear" across the orbit is 1.45 m/s per km or 1.45e-3 / second. If we imagine a series of increasing-weight thinsats around the central orbit, lighter at the edges and denser near the central orbit, then M288 orbits that differ in velocity by 1.45 m/s will always more than 1 km apart.
MORE LATER
Old (incorrect) analysis removed.