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An 8 degree inclined orbit, $ r_p $ = 8378 km, $ r_a $ = 76000 km. An 8 degree inclined orbit, $ r_p $ = 8378 km, $ r_a $ = 76000 km, period 86164.099 seconds
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So, $ a $ = 42164 km, e = 0.80189 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.

Apsidal Precession


The Earth's equatorial bulge causes orbits to drift, with westward orbit perigee and apogee drifting 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

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

\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]

{ \Large { { \partial \Omega } \over { \partial t } } } ~=~ -{ \large { { 3 ~ J_2 R^2 } \over { 2 ~~~ p^2 ~ } } } ~ \bar{n} ~ \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.

ApsidalPrecession (last edited 2018-04-07 22:01:05 by KeithLofstrom)