Orbits

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.

Options: Circular or elliptical versions

Special inclinations:

63.4 degrees has zero precession, as seen from Earth, apogee stays locked over single place on the Earth. So precession rate matches 24 hours.

zero degrees should also have zero precession, esp if circular orbit.

sun-synch: nearly polar for LEO/HEO altitudes.

-30: goes overhead, but only for limited amount of night time due to precession

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	v_elliptcal/v_circularat apogee	v/v_geo	angular rate/geo rate = (geo/apo)*v/v_geo
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=v_perigee(R/R_p).

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/apo-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, i.e. an elliptical orbit with period of more than 4 hours.

v/v_geo is