Question 4.10: Determine the perigee and apogee for an earth satellite whos......

The period determines the semimajor axis,

T={\frac{2\pi}{\sqrt{\mu}}}a^{\frac{3}{^{2}}}\Rightarrow3.3600={\frac{2\pi}{\sqrt{398_{.}600}}}\,a^{\frac{3}{{2}}}\Rightarrow a=10.560{\mathrm{~km}}

For the apse line to be stationary, we know from Eqn (4.53) that i = 64.435° or i = 116.57°. However, an inclination of less than 90° causes a westward regression of the node, whereas a sun-synchronous orbit requires an eastward advance, which i = 116.57° provides. Substituting this, the semimajor axis, and the \dot{\Omega}   in radians per second for a sun-synchronous orbit (cf. Example 4.9) into Eqn (4.52), we get

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

         (4.53)

1.991\times10^{-7}=-{\frac{3}{2}}{\frac{\sqrt{398}\,600\times0.0010826\times6378^{2}}{(1-e^{2})^{2}\times10.560^{7/2}}}\cos116.57^{\circ}\Rightarrow e=0.3466

Now we can find the angular momentum from the period expression (Eqn (2.82))

{\frac{v^{2}}{2}}-{\frac{\mu}{r}}=-{\frac{\mu}{2a}}(2.82)

T={\frac{2\pi}{\mu^{2}}}{\bigg(}{\frac{h}{\sqrt{1-e^{2}}}}{\bigg)}^{3}\Rightarrow3.3600={\frac{2\pi}{398.600^{2}}}{\bigg(}{\frac{h}{\sqrt{1-0.34655^{2}}}}{\bigg)}^{3}\Rightarrow h=60.850\,\mathrm{km}^{2}/s

Finally, to obtain the perigee and apogee radii, we use the orbit formula:

z_{p}+6378=\frac{h^{2}}{\mu}\frac{1}{1+e}=\frac{60.860^{2}}{398600}\frac{1}{1+0.34655}⇒ z_{p} = 522.6 km

z_{a}+6378={\frac{h^{2}}{\mu}}{\frac{1}{1-e}}\,\Rightarrow z_{a}= 7842 km