# Q on satellite orbits + orbital drift

1. Jul 12, 2008

### JamesGoh

Im aware that due to the oblate nature of the Earth, the satellite's line of apside and line of nodes rotate according to the following formulas (where i = angle of inclination of orbit to Earth and K is mean motion per day)

Variance in right ascension of ascending node ($$\Omega$$) due to rotation of line of nodes

d$$\Omega$$/dt = -Kcos(i)

Variance in argument of perigree ( $$\omega$$ )due to rotation of line of apsides

d$$\omega$$/dt = K(2 - 2.5sin(i)*sin(i) )

New argument of perigee (taking into consideration rotation of line of apsides)

$$\omega$$ = $$\omega$$o + d$$\omega$$/dt(t-to)

where $$\omega$$o is argument of perigee at epoch, to is time at epoch

In a textbook example I am reading, the author is trying to visualise the drift of the orbits as a result of just the rotation of the line of apsides (in other words i=90 degrees and the orbit is polar).

He starts of by assuming the situation where the perigee is exactly over the ascending node (in other words $$\omega$$ = 0 degrees) and d$$\omega$$/dt = - K/2. One orbital period (Pa) later, he states that the perigee would appear south of the equator (in other words $$\omega$$ = -KPa/2.

Mathematically, he is correct, but I cannot seem to visualise how the perigee would appear south of the equator. Would anyone be able to help me out here ?

If I am not clear, could anyone let me know ?