Question 4.8: A spacecraft is in a 280 km by 400 km orbit with an inclinat......

The perigee and apogee radii are

$r_{p}=6378+280=6658\,{\mathrm{km}}\quad r_{a}=6378+400=6778\,{\mathrm{km}}$

Therefore, the eccentricity and semimajor axis are

$e={\frac{r_{a}-r_{\mathrm{p}}}{r_{a}+r_{\mathrm{p}}}}=0.008931$

$a={\frac{1}{2}}\left(r_{a}+r_{\mathrm{p}}\right)=67\,18\,\mathrm{km}$

From Eqn (4.52), we obtain the rate of node line regression.

$\dot{Ω}=-\left[{\frac{3}{2}}{\frac{\sqrt{\mu}J_{2}R^{2}}{{(1-e^{2})}^{2}a^\frac{7}{2}}}\right]\cos i$        ( 4.52)

${\dot{\Omega}}=-\left[{\frac{3}{2}}{\frac{\sqrt{398.600}\times 0.0010826 \times6378^{2}}{\left(1-0.0089312^{2}\right)^{2}\times6718^{7/2}}}\right]\cos51.43^{\circ}=-1.0465\times10^{-6}\,\,\mathrm{rad/s}$

or

${\dot{\Omega}}= 5.181°$ per day to the west

From Eqn (4.54),

$\dot{\omega}=\dot{\Omega}\frac{(5/2)\sin^{2}i-2}{\cos i}$              (4.54)

$\dot{\omega}=-1.0465\times10^{-6} . \left(\frac{5}{2}\mathrm{sin}^{2}51.43^{\circ}-2\right)=+7.9193\times10^{-7}\mathrm{~rad/s}$

or

$\dot{\omega}=  3.920°$  per day in the flight direction