Question 5.9: At a given time, the geocentric equatorial position vector o...

Using Equation 5.56 we find the position vector of the observer to be

R =\left[\frac{R_{e}}{\sqrt{1-\left(2 f-f^{2}\right) \sin ^{2} \phi}}+H\right] \cos \phi(\cos \theta \hat{ I }+\sin \theta \hat{ J })+\left[\frac{R_{e}(1-f)^{2}}{\sqrt{1-\left(2 f-f^{2}\right) \sin ^{2} \phi}}+H\right] \sin \phi \hat{ K }              (5.56)

For the position vector of the space station relative to the observer we have (Equation 5.53)

\varrho= r – R
=(-2032 \hat{ I }+4591 \hat{ J }-4545 \hat{ K })-(-1673 \hat{ I }+4598 \hat{ J }-4078 \hat{ K })
=-359.0 \hat{ I }-6.342 \hat{ J }-466.9 \hat{ K }( km )

or, in matrix notation,

To transform these geocentric equatorial components into the topocentric horizon system we need the transformation matrix [ Q ]_{X x}, which is given by Equation 5.62a,

Thus,

or, reverting to vector notation,

The magnitude of this vector is ϱ=589.0km. Hence, the unit vector in the direction of ϱ is

Comparing this with Equation 5.58 we see that sin a = 0.6615, so that the angular elevation is

\hat{ \varrho }=\cos a \sin A \hat{ i }+\cos a \cos A \hat{ j }+\sin a \hat{ k }            (5.58)

Furthermore

\sin A=\frac{0.5765}{\cos a}=0.7687
\cos A=\frac{-0.4787}{\cos a}=-0.6397

It follows that

A=\cos ^{-1}(-0.6397)=129.8^{\circ} (second quadrant) or 230.2° (third quadrant)

A must lie in the second quadrant because sin A > 0. Thus the azimuth is
A = 129.8°