Jan 31 2021.
The objective of this project is to assess orbit choices for an orbiting calibration source. There are two things we desire: 1) satellite is in the
Jan 31 2021.
The objective of this project is to assess orbit choices for an orbiting calibration source. There are two things we desire: 1) satellite is in the shadow of the Earth, so that we avoid reflected sunlight issues, and 2) angular rate not to exceed 30 arcsec/sec, i.e. twice the sidereal rate, so that large telescopes can track the source easily. Two limiting cases that don't work: i) geosynch orbits put source at fixed azimuth and elevation (angluar rate =0) but except for equinox these objects are in the sunlight, ii) low earth orbits, few 100 km above surface, are in shadow but have really high angular rate as seen from observatory.
Ideal would be 15.04 arcsec/sec of angular rate so that stars are fixed relative to this source. (this is sidereal rate, or in other words, the rate you would get for a 24 hour orbital period) - youd want this rate at the part where its overhead at night. If
1 radian / second = 206,264.806 arc seconds / second ; w = 2pi/T ; newtonian mechanics - How do I convert tangential speed to angular speed in an elliptic orbit? - Physics Stack Exchange ---- or conservation of energy & angular momentum in a comet orbit (duke.edu)
Steo 1: analysis of basic physics-
- for a circular orbit of radius R_orbit, what is apparent angular rate at zenith as seen from ground? Ans- it's v_orbit/(R_sat-R_Earth) in rad/sec. Orbit at 5000 km above-earth orbit has 5 km/s vel so 1 mrad/sec ~ 200 arcsec/sec so 10X sidereal. this means its going really fast, way faster than telescopes can track, cant rotate and resolve fast enough, too blurry. 13.3X sidereal?
- LEO Radius: 2000 km; period is about 90-120 min; so dividing 24 hrs/2 hrs = 12X sidereal rate, roughly; speed of satellites are 28000 km/h → 7 km/s
- geostationary orbit is 35786 km away (+7000 km if counting radius of earth); period is one sidereal day. speed is 3 km/s
- Polar orbits take the satellites over the Earth’s poles. The satellites travel very close to the Earth (as low as 200 km above sea level), so they must travel at very high speeds (nearly 8 km/s).
- R_ earth = 6,378 km at equator, 6,357 km at pole
- M earth = 6 × 10 24 kg
- For an elliptical orbit, conservation of energy implies
v2 = GM(2/r - 1/a) where r is distance to center of mass (i.e. center of Earth) and a is semi-major axis of orbit.
We need v of order 0.6 km/s when r=11,000 km = 1.1e7m to hit the desired angular rate. What's the implied value of a? More generally, what family of orbits does what we want? - this is a back of the envelope calc - this is a rough
- What is the sunlight intensity vs. distance from Earth (umbra vs. penumbra), in the shadow? What is angular size of shadow as seen from observatory? https://en.wikipedia.org/wiki/Umbra,_penumbra_and_antumbra
- Geometric Aspects | Solar Radiation (greenrhinoenergy.com) - this might help, but have to figure out how to do it for one day and distance from the shadow; however, it varies throughout the year as well
- it also matters where the observatory is placed?
- Energy (a-ghadimi.com) chp 7
- SkyMarvels.com Reference - flux measurements but on earth
- lecture.2.global.energy_cycle (uci.edu) - solar flux at earth
- Geometric Aspects | Solar Radiation (greenrhinoenergy.com) - this might help, but have to figure out how to do it for one day and distance from the shadow; however, it varies throughout the year as well
- Is there a sun-synch orbit that does the right thing? Sun-synch orbits have really high inclinations, like 80 degrees, and precess once per year so the lock plane of the orbit relative to the sun
- Is there a strobed-source approach that would work, if we could make a constant-integrated-photon-flux-per-pulse strobed source? What about scintillation?
Ionospheric scintillation is the rapid modification of radio waves caused by small scale structures in the ionosphere. Severe scintillation conditions can prevent a GPS receiver from locking on to the signal and can make it impossible to calculate a position. Less severe scintillation conditions can reduce the accuracy and the confidence of positioning results .Scintillation of radio waves impacts the power and phase of the radio signal. Scintillation is caused by small-scale (tens of meters to tens of km) structure in the ionospheric electron density along the signal path and is the result of interference of refracted and/or diffracted (scattered) waves. ??
- strobe - regular flashes of light; flicker
...
So technically, we want a dusk to dawn orbit! At some orbit that is Molnya like such that the angular rate issue is resolved. See new wiki page for details
Precession - In astronomy, precession refers to any of several gravity-induced, slow and continuous changes in an astronomical body's rotational axis or orbital path. Precession is a change in the orientation of the rotational axis of a rotating body. In an appropriate reference frame it can be defined as a change in the first Euler angle, whereas the third Euler angle defines the rotation itself. The sun synchronous orbit slowly precesses around Earth to always be in the dark at a given time over a part of the planet. Formula for precession is here: Sun-synchronous orbit - Wikipedia
Why does it do precession? - It's mostly physics magic! The Earth's rotation causes it to bulge slightly at the equator, which means the Earth is trying to twist the Orbit over on to its side. How this causes the orbit plane to rotate isn't very intuitive, but you can recreate the same effect by spinning up an old bike wheel and holding it on one side of the hub while it is upright. For the Earth's rotation to cause exactly the right rate of precession, we just have to periodically tweak the orbital inclination and altitude. We use around 20kg of fuel per year for this, and that's mainly to tweak the inclination. If we tried to make the orbit turn through the year without the help of the equatorial bulge, we'd burn through our 300kg fuel budget within a few hours!
...
Angular rate - rate at which the thing moves across the sky. Angular velocity is the rate of velocity at which an object or a particle is rotating around a center or a specific point in a given time period.
Sidereal day just refers to when the earth completes one revolution relative to itself in a vacuum. a solar day it when the same spot on earth faces the sun again. the sidereal rate is the rate at which the earth spins on its axis, and consequently the rate at which the 'fixed' stars appear to move in the sky.A sidereal day – 23 hours 56 minutes and 4.1 seconds – is the amount of time needed to complete one rotation, so there are 4 minutes left that carry over. sidereal time at any given place and time will gain about four minutes against local civil time, every 24 hours, until, after a year has passed, one additional sidereal "day" has elapsed compared to the number of solar days that have gone by.
Using sidereal time, it is possible to easily point a telescope to the proper coordinates in the night sky. Briefly, sidereal time is a "time scale that is based on Earth's rate of rotation measured relative to the fixed stars. This is because you are measuring and going by the absolute rotation of the Earth so the stars must be in the same place every day, not shifted around ever 4 minutes.
In the picture of the sun and distant star above: Sidereal time vs solar time. Above left: a distant star (the small orange star) and the Sun are at culmination, on the local meridian m. Centre: only the distant star is at culmination (a mean sidereal day). Right: a few minutes later the Sun is on the local meridian again. A solar day is complete.
the sidereal rate is the rate at which the earth spins on its axis, and consequently the rate at which the 'fixed' stars appear to move in the sky.
equinox - the time or date (twice each year) at which the sun crosses the celestial equator, when day and night are of equal length (about September 22 and March 20).
The umbra (Latin for "shadow") is the innermost and darkest part of a shadow, where the light source is completely blocked by the occluding body. An observer within the umbra experiences a total eclipse. The penumbra (from the Latin paene "almost, nearly") is the region in which only a portion of the light source is obscured by the occluding body. An observer in the penumbra experiences a partial eclipse. https://mysite.du.edu/~jcalvert/astro/shadows.htm
However, the umbra and penumbra of course aren't just triangles, they're cones; and as such, treat them. Question is : are earth and the sun in the same plane?
- What is the sunlight intensity vs. distance from Earth (umbra vs. penumbra), in the shadow? What is angular size of shadow as seen from observatory? https://en.wikipedia.org/wiki/Umbra,_penumbra_and_antumbra
- this matters because if we are in the penumbra, some sunlight will come through, defeat the purpose of being a calibration source; or we'd have to subtract that intensity, technically, to get a good reading.
- Is there a strobed-source approach that would work, if we could make a constant-integrated-photon-flux-per-pulse strobed source? What about scintillation?
- wdym?
apparent parallax - an apparent change in the position of an object resulting from a change in position of the observer. astronomy the angle subtended at a celestial body, esp a star, by the radius of the earth's orbit
Parallax makes it seem to move faster than background stars, in opposite direction.
Another way to see how this effect works is to hold your hand out in front of you and look at it with your left eye closed, then your right eye closed. Your hand will appear to move against the background.
...
This simple relationship is why many astronomers prefer to measure distances in parsecs.
ellipticity vs eccentricity
The orbital eccentricity of an astronomical object is a dimensionless parameter that determines the amount by which its orbit around another body deviates from a perfect circle. A value of 0 is a circular orbit, values between 0 and 1 form an elliptic orbit, 1 is a parabolic escape orbit, and greater than 1 is a hyperbola. Orbital eccentricity - Wikipedia
For elliptical orbits eccentricity e can also be calculated from the periapsis and apoapsis since rp = a(1 − e) and ra = a(1 + e), where a is the semimajor axis. look at wikipedia
Linear speed is the measure of the concrete distance travelled by a moving object. it would be the tangent of the circle in this case.
- we'd be at the focus of the ellipse, and the ellipse is the orbit of the satellite; the fourth picture is the side view of what would be going on, with earth being M. The angular rate at apoapsis needs to be about 15 arcsec/sec, and the period in total needs to be about 24 hrs.
Feb 20, 2021. Stubbs
I thought about this some more. To make an "astro-stationary" orbit, the line of sight between the surface of the Earth and the satellite has to not rotate in inertial space. The only way for that to happen (for simplicity assume satellite overhead at midnight, at its apogee) is to have the velocity components perpendicular to the line of sight at the two ends be equal in both magnitude and direction. That means the satellite tangential velocity has to match that of the Earth's surface, which is a meagre 0.46 km/s. An orbit that does that has a semi-major axis of 2 million km, which makes things awkward in terms of orbital period.
...
A quick photon flux estimate: Imagine we had a 1 Watt optical source, broadcasting into 1/4 of 4pi. If it's at a distance of 1000 km then a 1 meter radius collector (2m dia telescope) subtends a surface area fraction of that segment of the sphere that is f= pi*1m^2/(pi*1E6^2) =1E-12 of the illuminated surface area at that distance. That means we'd intercept 1 pW of optical power, which generates (rough numbers) 1 pA of photocurrent. That's a photoelectron rate of 1E-12/1.6E-19 ~ 6 million photons per second.
How bright, in astronomical magnitudes, is this thing? Use sun as a benchmark:
Sun emits around 3.8E26 Watts, at a distance of 150E9 meters. Sun has an apparent magnitude of -26.7. If we made it 1W it becomes 2.5 log (3.8e26) mags fainter, or 66.4 mag fainter, making it mag 39.7. Now move it closer by a factor of 150E9/1e6 = 150E3. Magnitude change is 5*log(D1/D2) which makes our 1 Watt source at a distance of 1000 km be comparable to a star at around 13th apparent magnitude.
That is roughly consistent with this exposure time calculator, for 4 meter DECam system, downloaded from http://www.ctio.noao.edu/noao/content/Exposure-Time-Calculator-ETC
refs
Adaptive optics guide star and calibration satellite, ORCAS:
...