Question 5.7: At the instant when the Greenwich sidereal time is θG =126.7...

According to Equation 5.52, the local sidereal time at the observation site is

Substituting R_{e}=6378km, f =0.003353 (Table 4.3), θ =189.7° and φ=20° into Equation 5.56 yields the geocentric position vector of the site:

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)

Having found R, we obtain the position vector of the space station relative to the site from Equation 5.53:

\varrho= r – R
=(-5368 \hat{ I }-1784 \hat{ J }+3691 \hat{ K })-(-5955 \hat{ I }-699.5 \hat{ J }+2168 \hat{ K })
=586.8 \hat{ I }-1084 \hat{ J }+1523 \hat{ K }( km )

The magnitude of this vector is ϱ = 1960 km, so that

Comparing this equation with Equation 5.57 we see that

\hat{\varrho}=\cos \delta \cos \alpha \hat{ I }+\cos \delta \sin \alpha \hat{ J }+\sin \delta \hat{ K }                  (5.57)
cos δ cos α = 0.2997
cos δ sin α = −0.5524
sin δ = 0.7778
From these we obtain the topocentric declension,

\delta=\sin ^{-1} 0.7773=\underline{51.01^{\circ}}                 (a)

as well as

\sin \alpha=\frac{-0.5533}{\cos \delta}=-0.8795
\cos \alpha=\frac{0.2994}{\cos \delta}=0.4759

Thus

\alpha=\cos ^{-1}(0.4759)=61.58^{\circ} (first quadrant) or 298.4° (fourth quadrant)

Since sin α is negative, α must lie in the fourth quadrant, so that the right ascension is
α = 298.4°                 (b)
Compare (a) and (b) with the geocentric right ascension \alpha_{0} and declination \delta_{0}, which were computed in Example 4.2,