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.
...
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 |
| 3 days | 87,705 | 161,410 | 7000 | 0.0682 | 0.2945 | ? | 2.2E-6 (really?)
|
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 i.e. an elliptical orbit with period of more than 4 hours.
v/v_geo is