Versions Compared

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.

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. 

...

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

Precession rate is

omega_p=(-3/2) R_e^2 (a(1-e^2))^-2 J_2 Omega cos (inclination)

 

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

orbitaapogeeperigeeflux/geo=(42164/apogee)2velliptcal/vcircular at apogeev/vgeoangular rate/geo rate = (geo/apo)*v/vgeo
geosynch42,16442,16442,1641111
double-tundra, 48 h66,931126,86070000.1150.23490.18650.0620, i.e. 0.4 arcsec/sec at dec= - 64
tundra, 24 hr42,16477,6586,6700.29480.29310.29310.1591
molnya, 12 hour26,56245,7307,3780.85010.40210.50660.4671
8 hour20,27033,8706,6701.54970.44380.64000.7967
6 hour16,73326,7966,670 2.47600.49890.79201.1046
4 hour12,76918,8686,670 4.9938 0.59461.08042.4144
3 days87,705161,41070000.06820.2945?

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/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