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 = n*2*pi/(24*60*60); #conversion (revs/day) --> (rad/s)
  a = (GM/(power(n,2))).^(1./3);

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?


1 Answer 1


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.

  • 3
    $\begingroup$ An endless source of confusion. Get used to it :) $\endgroup$
    – SF.
    Dec 27, 2018 at 21:17
  • 4
    $\begingroup$ It's good that you wrote an answer, others may well have the same issue! $\endgroup$ Dec 27, 2018 at 21:50
  • $\begingroup$ The clock in San Dimas, is always running... you have to dial one number higher. $\endgroup$
    – Mazura
    Dec 28, 2018 at 7:23

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