Orbits for calibrators.
mu=GM=398600.4418 km3/s2
Found some MATLAB code on the web that does crude orbit determination. Revised it to also compute zenith angle to satellite from Pachon.
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Period depends only on semi major axis, T=1pi sqrt(a^3/GM).
geosynch has radius of 42,164 km
so T= one day *(a/42,164)3/2
a=42,164 * (T)2/3
semimajor axis is the average of apogee and perigee
| orbit | a | apogee | perigee | flux/geo=(42164/apogee)2 | velliptcal/vcircular at apogee | v/vgeo | angular rate/geo rate = (geo/apo)*v/vgeo |
|---|---|---|---|---|---|---|---|
| geosynch | 42,164 | 42,164 | 42,164 | 1 | 1 | 1 | 1 |
| double-tundra, 48 h | 66,931 | 126,860 | 7000 | 0.115 | 0.2349 | 0.1865 | 0.0620, i.e. 0.4 arcsec/sec at dec= - 64 |
| tundra, 24 hr | 42,164 | 77,658 | 6,670 | 0.2948 | 0.2931 | 0.2931 | 0.1591 |
| molnya, 12 hour | 26,562 | 45,730 | 7,378 | 0.8501 | 0.4021 | 0.5066 | 0.4671 |
| 8 hour | 20,270 | 33,870 | 6,670 | 1.5497 | 0.4438 | 0.6400 | 0.7967 |
| 6 hour | 16,733 | 26,796 | 6,670 | 2.4760 | 0.4989 | 0.7920 | 1.1046 |
| 4 hour | 12,769 | 18,868 | 6,670 | 4.9938 | 0.5946 | 1.0804 | 2.4144 |
Conservation of angular momentum implies that L=Iomega=MR2omega=constant=MRv=constant so v=vperigee(R/Rp).
IF we require perigee be an altitude of over 300 km, then perigee has to be larger than 6370+300=6670. Then apogee must satisfy apogee=2a-perigee=2a-6670
velocity as a function of r,a is v=sqrt(GM(2/r-1/a)).
What's ratio of apparent angular rate for a given period, for elliptical vs. circular orbit? Do this as seen from center of the Earth, for now.
ve/vc=sqrt((2a/raapo-1)).
What about angular size? For a 10m telescope, theta=10/apo(in meters). For this to be under 0.1 arc sec, we require 4.86E-7=10/apo so apo(m) >= 10/4.86E-7 = 20,000 km. Wow.
That requires a circular orbit with a period of greater than 8 hours, or else an elliptical orbit with period of more than 4 hours.