CalSat Chris Meeting 4/3 Takeaways

Scratch Notes below

1. The minimum angular rate will always be slowest when farthest away from the orbit, given angular momentum is conserved, it angular rate goes with 1/r^2, as velocity goes with 1/r and v/r = w = 1/r^2. 
2. There is not simple function for the mean anomaly that describes the xy position as a function of time - the ones that do exist are very complicated and nasty. 
3. As we've seen, the angular rate is as we've calculated in the code. This is known as the apparent angular rate, where we assume we are at the foci of the ellipse in question, allowing us to assume this 1/r^2 relation. 
   1. However, we aren't at the center of the foci - we are 6000 km away from it at the surface of the earth, and because of this, we perceive the angular rate differently. This angular rate is known as the relative angular rate, and it's different from the apparent for multiple reasons:
      1. We are not at the foci of the ellipse. 1/r^2 relationship only true at foci. 
      2. We are not stationary. The Earth is rotating along with the satellite. If you are at the foci, this wouldn't matter since a point rotating doesn't introduce a difference. But rotating where we are introduces a difference in what we perceive.
      3. The satellite is orbiting, too. 
      4. Aberration may shift the angular rate as well. Essentially, Starlight is a function of declination and ascension, and so stars higher above will rotate less. However, since most of the orbits we are looking at are going at speeds much faster than this rotation, we will see a minimal effect. 
   2. There is even a special case at the apogee, where we see the satellite straight ahead. The angular rate at this point increase by r/a, so it's easier to calculate from this position. 
4.  An inclination of 0 was brought up because it was thought that if not near the equatorial plane, the orbit would not be in the sun's shadow at some times of year. However, it was realized that you can orient the inclination in whatever way you desire (because J2 is symmetric across all of earth) so this wouldn't be a problem to worry about. Any inclination will do.
5. Next, we explained some things in the table:
   1. Isotropic Source Range Dilution
      1. Assumption: In a circular orbit 8000 km from Earth (+r_earth), this gives you one unit of flux (check if 8000 or 14000). Use this as a benchmark. 
      2. Isotropic means light going out in all directions. Eventually, we'll shoot a direct beam of light, but that'll be for later.
      3. Due to the inverse square law, flux decreases with 1/r^2, where r is the distance from the center of the earth to the satellite. 
      4. Calculate dilution with 8000 or 14000/d^2 for flux. 
   2. Time above 20 degrees
      1. Elevation angle of 20 degrees off the ground 
      2. How much time is it visible?
      3. Get the time over the zenith (maximal time) - STK can do this as well. 
   3. Airmass
      1. 1/ cos(a) 
      2. Rate of change of the airmass over the zenith angle  - to calculate, use formula above and divide by time over 20 deg. 
      3. This would be a useful quantity to seek out. If this variation is large enough, we can use CalSat not only to calibrate as a star source, but also use it to measure atmospheric variation!
   4. Relative angular rate
      1. Subtract all the sources of error above described
      2. Plane geometry
6. We use chemical propellant to get to an orbit, and ion thrusters to raise our orbit. 
   1. Getting to orbits and the right ones is a function of time and mass
7. New parameter: Time spent in Earth's shadow:
   1. The time spent in earth's shadow is a function of how much time the satellite will be in the dark. For orbits not on the axial plane (ecliptic), not seasonal variation, but it's a function of where it is on the plane of the ecliptic , the time you would be able to observe it throughout the night would change as it revolves around the sun
   2. If you set the inclination to the ecliptic (23.5 degrees), or the earth orbital plane, then this time would always be constant. 
      1. aka time is independent of season
   3. Be careful about definition of inclination (does 23.5 = 180-23.5?) yes for now, it's just they go in opposite directions. 
8. FOV Imager, 10 arcmin
   1. LSST FOV is about 10 minutes
   2. If I pointed at it and stayed stationary on my telescope, how much time will it take to go from one side of my imager to the other (entering and exiting a stationary FOV)
   3. Good = about 5 minutes to collect the data
9. Sun synchronous
   1. NOT 10 30 everywhere
   2. The orbit is synchronous, not the satellite. The satellite can be in any place at any time; the orbit itself is orienting relative to the sun. 
   3. This is always true: The orientation of the orbit (semi-major axis) is farthest away from the Sun at midnight.
10. To Do:
    1. Fix inclination to 22.5 deg (earth ecliptic) - what combination of a,e and other parameters are given for a sun sync orbit, optimizing for ang rate and speed. (and relative!)
    2. Verify the calculations that we now know how to do + add the time spent in earth shadow as well. 
    3. Think about the parallax correction a little bit more
    4. Can we see this from Antarctica? Employ the triangular approach Chris suggested. 
    5. L2 orbit calcs!