# [amsat-bb] Conventions for apogee and perigee altitudes?

Alan wa4sca at gmail.com
Sun Oct 1 11:48:48 UTC 2017

```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: AMSAT-BB [mailto:amsat-bb-bounces at amsat.org] On Behalf Of Phil
<Karn
<Sent: Sunday, October 01, 2017 04:31 AM
<To: amsat-bb at amsat.org
<Subject: [amsat-bb] 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 * (1-e) - 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.
<
<
<
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```