Question 4.9: A satellite is to be launched into a sun-synchronous circula......

We find the altitude z from the period relation for a circular orbit, Eqn (2.64):

T_{\mathrm{circular}}={\frac{2\pi}{\sqrt{\mu}}}r^{\frac{3}{2}}              (2.64)

T={\frac{2\pi}{\sqrt{\mu}}}(R_{E}+z)^{\frac{3}{2}}\Rightarrow100\times60={\frac{2\pi}{\sqrt{398,600}}}(6378+z)^{\frac{3}{2}}\Rightarrow z = 758.63 km

For a sun-synchronous orbit, the ascending node must advance at the rate

{\dot{\Omega}}={\frac{2\pi\,\operatorname{rad}}{365.26\times24\times3600\,{\mathrm{s}}}}=1.991\times10^{-7}\,\mathrm{rad/s}

Substituting this and the altitude into Eqn (4.47), we obtain,

[\mathbf{Q}]_{X\bar{x}}=[\mathbf{R}_{3}(\omega)][\mathbf{R}_{1}(i)][\mathbf{R}_{3}(\Omega)]          (4.47)

1.991\times10^{-7}=-\left[{\frac{3}{2}}{\frac{\sqrt{398.600}\times0.00108263\times6378^{2}}{(1-0^{2})^{2}(6378+758.63)^\frac{7}{2}}}\right]\cos i\Rightarrow \cos i = – 0.14658

Thus, the inclination of the orbit is

i = \cos^{-1} ( – 0.14658) = 98.43°

This illustrates the fact that sun-synchronous orbits are very nearly polar orbits (i = 90°).