Deviation of semi-major axis

Question

If I calculate the semi-major axis of Molniya-1T with $a = \sqrt[3]{\dfrac{GM}{n^2}}$ with $n=3.18683728\text{ }d^{-1}$, I get another apoapsis ($a=19505.7$ km) as noted at Heavens Above:

apogee height: 25659 km

perigee height: 595 km

which is $a=\frac{25659+595}{2}=13127$ km.

This is my approach in Octave/MATLAB:

# Computes the semi major axis a from n with constant GM
# @params:
#   GM constant (cubic km per square second)
#   n: mean motion  (revs/day)
# @return:
#   a: semi major axis (km)
function a = getSemiMajorAxis(GM,n)
  n = n2pi/(246060); #conversion (revs/day) --> (rad/s)
  a = (GM/(power(n,2))).^(1./3);
endfunction
getSemiMajorAxis(398600.44,3.18683728) # Test with Molniya-1T

So, what is the origin of the deviation? Isn't apogee the same as apoapsis basically?

I saw your profile; here are some geoinformatics questions you might find interesting: Shape and dimensions of the Moon's reference surface for selenographic latitude/longitude? and How do we define geographical coordinates on non-spheroid celestial bodies? — uhoh, Dec 28 '18 at 02:22

score 10 · Answer 1 · edited Dec 28 '18 at 00:50

10

Ok, that's embarrassing:

You just have to add earth's radius (traditionally the equatorial radius) of about 6378 or 6378.137 km to apogee or perigee heights to get distances to the center.

edited Dec 28 '18 at 00:50

uhoh

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answered Dec 27 '18 at 17:48

sequoia

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An endless source of confusion. Get used to it :) – SF. Dec 27 '18 at 21:17
4

It's good that you wrote an answer, others may well have the same issue! – Organic Marble Dec 27 '18 at 21:50
The clock in San Dimas, is always running... you have to dial one number higher. – Mazura Dec 28 '18 at 07:23

Deviation of semi-major axis

1 Answers1