Apsidal Precession


The Earth's equatorial bulge causes orbits (perigee and apogee) to drift westward. From AE Roy Orbital Motion 1978:

\Omega

longitude of the ascending node

a

semimajor axis

e

eccentricity

i

inclination

\mu

standard gravitational parameter, 398600.4418 km³/s² for Earth

J_2

zonal ablateness factor, 1.08262668e-3 for Earth

p

p ~=~ a ( 1 - e^2 ) = r_p r_a / a

n_0

unperturbed mean motion

\bar{n}

perturbed mean motion

R

Earth Equatorial Radius = 6378.137 km

T

unperturbed orbital period

{n_0}^2 ~=~ \mu / {a_0}^3 . . . unperturbed mean motion

n_0 ~ \approx ~ { \large { { 2 \pi } \over T } }

\bar{n} ~=~ n_0 \left[ 1 + { \large { { 3 ~ J_2 R^2 } \over { 2 ~~~ p^2 ~ } } } \left( 1 - {\large { 3 \over 2 } } \sin( i )^2 \right) (1-e^2)^{1/2} \right] ~\approx~ n_0

{ \Large { { \partial \Omega } \over { \partial t } } } ~=~ -{ \large { { 3 ~ J_2 R^2 } \over { 2 ~~~ p^2 ~ } } } ~ \bar{n} ~ \cos( i ) ~\approx ~ -{ \large { { 3 \over 2 } ~ { { J_2 R^2 } \over { a ( 1 - e^2 ) } } ~ { { 2 \pi } \over T } } } ~ \cos( i )

Roy writes about \omega as if it is the angle from the orbiting body perpendicular the equatorial plane ... or something. Confusing.


An example

An 8 degree inclined orbit, r_p = 8378 km, r_a = 76000 km, period 86164.099 seconds

So, a = 42164 km, e = 0.80189, p = 15051.209 km,

{ \large { { 3 ~ J_2 R^2 } \over { 2 ~~~ p^2 ~ } } } ~= 2.92784e-4

\bar{n} ~= n_0 \times 1.000171899 which suggests the orbital period is reduced by 0.000171869 × 86164.099 or 14.809 seconds

The apsides will precess by about the same amount per orbit.


Wikipedia differs?

A different result from Wikipedia's Nodal precession page.

\omega_p ~=~ - { \Large { 3 \over 2 } ~ { { R^2 } \over { a (1-e^2) )^2 } } } ~ J_2 ~ \omega ~ \cos i