A satellite is to be launched into a sun-synchronous circular orbit with period of 100 minutes. Determine the required altitude and inclination of its orbit.

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Accepted Answer

We find the altitude z from the period relation for a circular orbit, Equation 2.54:

[latex]T_{ ext {circular }}=\frac{2 \pi}{\sqrt{\mu}} r^{\frac{3}{2}}[/latex] (2.54)

[latex]T=\frac{2 \pi}{\sqrt{\mu}}\left(R_{E}+z ight)^{\frac{3}{2}} \Rightarrow 100 \cdot 60=\frac{2 \pi}{\sqrt{398 600}}(6378+z)^{\frac{3}{2}} \Rightarrow \underline{z=758.63 km }[/latex]

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

[latex]\dot{\Omega}=\frac{2 \pi rad }{365.26 \cdot 24 \cdot 3600 s }=1.991 imes 10^{-7} rad / s[/latex]

Substituting this and the altitude into Equation 4.47, we obtain,

[latex]\dot{\Omega}=-\left[\frac{3}{2} \frac{\sqrt{\mu} J_{2} R^{2}}{\left(1-e^{2} ight)^{2} a^{\frac{7}{2}}} ight] \cos i[/latex] (4.47)

[latex]1.991 imes 10^{-7}=-\left[\frac{3}{2} \frac{\sqrt{398 600} \cdot 0.00108263 \cdot 6378^{2}}{\left(1-0^{2} ight)^{2}(6378+758.63)^{\frac{7}{2}}} ight] \cos i \Rightarrow \cos i=-0.14658[/latex]

Thus, the inclination of the orbit is

[latex]\underline{i=\cos ^{-1}(-0.14658)=98.43^{\circ}} [/latex]

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

Question 4.7: A satellite is to be launched into a sun-synchronous circula...

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