At a given time, the geocentric equatorial position vector of the International Space Station is
$r =-2032.4 \hat{ I }+4591.2 \hat{ J }-4544.8 \hat{ K }( km )$
Determine the azimuth and elevation angle relative to a sea-level (H =0) observer whose latitude is φ=−40° and local sidereal time is θ =110°.

Question

At a given time, the geocentric equatorial position vector of the International Space Station is

[latex]r =-2032.4 \hat{ I }+4591.2 \hat{ J }-4544.8 \hat{ K }( km )[/latex]

Determine the azimuth and elevation angle relative to a sea-level (H =0) observer whose latitude is φ=−40° and local sidereal time is θ =110°.

Accepted Answer

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

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

[latex]R =-1673 \hat{ I }+4598 \hat{ J }-4078 \hat{ K }( km )[/latex]

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

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

or, in matrix notation,

[latex]\{\varrho\}_{X}=\left\{\begin{array}{l}-359.0 \-6.342 \-466.9\end{array} ight\}( km )[/latex]

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

[latex][ Q ]_{X x}=\left[\begin{array}{ccc}-\sin heta & \cos heta & 0 \-\sin \phi \cos heta & -\sin \phi \sin heta & \cos \phi \\cos \phi \cos heta & \cos \phi \sin heta & \sin \phi\end{array} ight][/latex]
[latex]=\left[\begin{array}{ccc}-\sin 110^{\circ} & \cos 110^{\circ} & 0 \-\sin \left(-40^{\circ} ight) \cos 110^{\circ} & -\sin \left(-40^{\circ} ight) \sin 110^{\circ} & \cos \left(-40^{\circ} ight) \\cos \left(-40^{\circ} ight) \cos 110^{\circ} & \cos \left(-40^{\circ} ight) \sin 110^{\circ} & \sin \left(-40^{\circ} ight)\end{array} ight][/latex]

Thus,

[latex]\{\varrho\}_{x}=[ Q ]_{X x}\{\varrho\}_{X}=\left[\begin{array}{ccc}-0.9397 & -0.3420 & 0 \-0.2198 & 0.6040 & 0.7660 \-0.2620 & 0.7198 & -0.6428\end{array} ight]\left\{\begin{array}{l}-359.0 \-6.342 \-466.9\end{array} ight\}[/latex]
[latex]=\left\{\begin{array}{c}339.5 \-282.6 \389.6\end{array} ight\}( km )[/latex]

or, reverting to vector notation,

[latex]\varrho=339.5 \hat{ i }-282.6 \hat{ j }+389.6 \hat{ k }( km )[/latex]

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

[latex]\hat{\varrho}=\frac{\varrho}{\varrho}=0.5765 \hat{ i }-0.4787 \hat{ j }+0.6615 \hat{ k }[/latex]

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

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

[latex]a=\sin ^{-1} 0.6615=\underline{41.41^{\circ}}[/latex]

Furthermore

[latex]\sin A=\frac{0.5765}{\cos a}=0.7687[/latex]
[latex]\cos A=\frac{-0.4787}{\cos a}=-0.6397[/latex]

It follows that

[latex]A=\cos ^{-1}(-0.6397)=129.8^{\circ}[/latex] (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°

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

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