Conventions for apogee and perigee altitudes?
I've been returning to satellite tracking software after a while (I wrote some early AMSAT tools in the early 1980s) and am wondering if there has ever been a resolution of the exact definitions of "apogee height" and "perigee height".
The simple geometric definitions of "perigee" and "apogee" are the points where the spacecraft is the closest to or farthest from the center of the earth. This is easy if you assume the earth is a perfect sphere; all you need is the semimajor axis a and the eccentricity e:
apogee = a * (1+e)  earth_radius perigee = a * (1e)  earth_radius
But reality is more complicated than that. For a nonequatorial orbit the apogee and perigee usually occur over some point off the equator where the earth's radius is smaller than at the equator. You can correct for this given the inclination and the argument of perigee, which together tell you the latitude at which apogee and perigee occur; one will occur in the northern hemisphere and the other will occur in the southern hemisphere at the opposite latitude.
There's a complication here in that this is geocentric latitude, while we more often use geodetic latitude on a daily basis. Converting geodetic latitude to geocentric is fairly easy, but converting in the other direction is like Kepler's equation: apparently there's no closed form solution so you have to iterate.
But this is a relatively minor detail. The real problem comes when you have a satellite with a relatively high inclination and an argument of perigee close to 0 or 180 degrees; in this situation the satellite can easily be farther from the earth's surface than *either* the calculated apogee or perigee!
The ISS is a case in point at the moment. Using element set 906, which has an argument of perigee of about 329 degrees I calculate an apogee of 408 km and a perigee of 402.4 km assuming an oblate earth and ignoring the distinction between geodetic and geocentric latitude (which is relatively small for this argument of perigee). But near the southernmost point of its orbit, I calculate an altitude of about 421 km, well above both the perigee and apogee heights because the earth's surface through the poles is more elliptical than the ISS's orbit.
So what conventions do people use? How meaningful do people expect these figures to be? For a low altitude orbit like the ISS, the difference in drag between 402 and 421 km is actually quite significant. At the very least it would be nice if everybody used the same convention.
Phil,
One of my hobbies is to periodically calculate the apogee and perigee of a few satellites to monitor their decay rates. (The total energy of the orbit can be parameterized in terms of the sum of the apogee and perigee, which is ultimately what you want for this purpose.) When I started this, I surveyed the results of tracking programs, and found the same chaos you report. At least some use an average, others a mean radius. This is certainly _part_ of the explanation for hearing satellites when they are reported to be below the horizon, diffraction and refraction being other factors.
I decided that for my purposes, trends, it didn't matter so long as I stuck with one. I finally chose the NASA Debris Assessment Software program which is the official tool for calculating lifetimes, etc. Unfortunately the program is a black box, or at least not documented in the program notes, but there are hints that is uses a more sophisticated model than a simple sphere.
I expect for amateur purposes, a simple sphere is fine.
73s,
Alan WA4SCA
<Original Message <From: AMSATBB [mailto:[email protected]] On Behalf Of Phil <Karn <Sent: Sunday, October 01, 2017 04:31 AM <To: [email protected] <Subject: [amsatbb] Conventions for apogee and perigee altitudes? < <I've been returning to satellite tracking software after a while (I <wrote some early AMSAT tools in the early 1980s) and am wondering if <there has ever been a resolution of the exact definitions of "apogee <height" and "perigee height". < <The simple geometric definitions of "perigee" and "apogee" are the <points where the spacecraft is the closest to or farthest from the <center of the earth. This is easy if you assume the earth is a perfect <sphere; all you need is the semimajor axis a and the eccentricity e: < <apogee = a * (1+e)  earth_radius <perigee = a * (1e)  earth_radius < <But reality is more complicated than that. For a nonequatorial orbit the <apogee and perigee usually occur over some point off the equator where <the earth's radius is smaller than at the equator. You can correct for <this given the inclination and the argument of perigee, which together <tell you the latitude at which apogee and perigee occur; one will occur <in the northern hemisphere and the other will occur in the southern <hemisphere at the opposite latitude. < <There's a complication here in that this is geocentric latitude, while <we more often use geodetic latitude on a daily basis. Converting <geodetic latitude to geocentric is fairly easy, but converting in the <other direction is like Kepler's equation: apparently there's no closed <form solution so you have to iterate. < <But this is a relatively minor detail. The real problem comes when you <have a satellite with a relatively high inclination and an argument of <perigee close to 0 or 180 degrees; in this situation the satellite can <easily be farther from the earth's surface than *either* the calculated <apogee or perigee! < <The ISS is a case in point at the moment. Using element set 906, which <has an argument of perigee of about 329 degrees I calculate an apogee of <408 km and a perigee of 402.4 km assuming an oblate earth and ignoring <the distinction between geodetic and geocentric latitude (which is <relatively small for this argument of perigee). But near the <southernmost point of its orbit, I calculate an altitude of about 421 <km, well above both the perigee and apogee heights because the earth's <surface through the poles is more elliptical than the ISS's orbit. < <So what conventions do people use? How meaningful do people expect these <figures to be? For a low altitude orbit like the ISS, the difference in <drag between 402 and 421 km is actually quite significant. At the very <least it would be nice if everybody used the same convention. < < < <_______________________________________________ <Sent via [email protected]. AMSATNA makes this open forum available <to all interested persons worldwide without requiring membership. Opinions <expressed <are solely those of the author, and do not reflect the official views of AMSAT <NA. <Not an AMSATNA member? Join now to support the amateur satellite <program! <Subscription settings: http://www.amsat.org/mailman/listinfo/amsatbb
On 10/1/17 04:48, Alan wrote:
I expect for amateur purposes, a simple sphere is fine.
I'm interested in what others are doing so I can be at least somewhat consistent with their results.
My current method for computing apogee and perigee seems to give results that match many of the online tracking programs. That is, find the geometric apogee and perigee and subtract the radius of the earth under those points. I can't easily tell if they make the correction for geodetic latitude because the effect is small and the results are usually not given with a lot of precision. But you do end up with weird oddities like seeing the ISS at an altitude that exceeds calculated apogee. As long as everybody is happy with that little anomaly, I'm happy too.
If you want to compute the energy in a (decaying) orbit, there's a simple and straightforward formula:
specific orbital energy = GM/(2*a)
where 'GM' (also known as mu) is the earth's gravitational parameter = 3.986004418e14 m^3/s^2 and 'a' is the orbit's semimajor axis in meters.
The specific energy is the sum in joules of the satellite's kinetic and potential energy per kilogram of its mass. Because of energy conservation, it will remain constant for any satellite in any trajectory unless it is gaining or losing energy from thrust, drag, or gravitational exchanges with a third body. It is always negative for a closed orbit because potential energy is zero at infinity and increasingly negative as you get closer. A positive specific energy means a hyperbolic (escape) trajectory.
For the ISS I calculate a specific energy of about 29.38 MJ/kg.
Problem is, you can't recover the semimajor axis from the apogee and perigee unless you know how they're computed. And if they're done the way I described, you also need the latitude of apogee and perigee so you can add the right earth radius, and that in turn usually requires knowing the inclination and argument of perigee, the latter changing steadily with time.
But if you have the mean motion or period you can compute the semi major axis from it using Kepler's third law:
a = (GM*P^2)^1/3
where the period P is expressed in seconds/radian = seconds/revolution / (2*pi), or
a = (GM/N^2)^1/3
where the mean motion N is expressed in radians/sec = rev/day * 2 * pi/86400.
So all you really need to compute specific energy and track decay is the satellite's period. That's it! Forget apogee and perigee...
You can also track the specific angular momentum of the satellite, which must also remain constant absent external forces besides simple 2body gravity:
h = sqrt(GM * a(1e^2))
The units of specific angular momentum are m^2/s, or area per time. This is where Kepler's second law comes from: the satellite sweeps out equal areas in equal time because that represents the satellite's (conserved) specific angular momentum.
In fact, we could describe an orbit with the angular momentum vector, which points normal to the orbit plane and defines RAAN and inclination, and the eccentricity vector, which points at perigee and defines the argument of perigee and the eccentricity. The two together define the mean motion, and if the epoch corresponds to a specific point on the orbit that sets the mean anomaly. That's all 6 Keplerian elements.
Phil
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Alan

Phil Karn