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

At the instant when the Greenwich sidereal time is θG=126.7\theta_{G}=126.7^{\circ}, the geocentric equatorial position vector of the International Space Station is

r=5368I^1784J^+3691K^(km)r =-5368 \hat{ I }-1784 \hat{ J }+3691 \hat{ K }( km )

Find the topocentric right ascension and declination at sea level (H =0), latitude φ=20° and east longitude Λ=60°.

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

θ=θG+Λ=126.7+60=186.7\theta=\theta_{G}+\Lambda=126.7+60=186.7^{\circ}

Substituting ReR_{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=[Re1(2ff2)sin2ϕ+H]cosϕ(cosθI^+sinθJ^)+[Re(1f)21(2ff2)sin2ϕ+H]sinϕK^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)

R=5955I^699.5J^+2168K^(km)R =-5955 \hat{ I }-699.5 \hat{ J }+2168 \hat{ K }( km )

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

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

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

ϱ^=ϱϱ=0.2994I^0.5533J^+0.7773K^\hat{\varrho}=\frac{\varrho}{\varrho}=0.2994 \hat{ I }-0.5533 \hat{ J }+0.7773 \hat{ K }

Comparing this equation with Equation 5.57 we see that

ϱ^=cosδcosαI^+cosδsinαJ^+sinδK^\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,

δ=sin10.7773=51.01\delta=\sin ^{-1} 0.7773=\underline{51.01^{\circ}}                 (a)

as well as

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

Thus

α=cos1(0.4759)=61.58\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 α0\alpha_{0} and declination δ0\delta_{0}, which were computed in Example 4.2,

α0=198.4δ0=33.12\alpha_{0}=198.4^{\circ} \quad \delta_{0}=33.12^{\circ}

 

Table 4.3 Oblateness and second zonal harmonics
Planet Oblateness J2J _{2}
Mercury 0.000 60×10660 \times 10^{-6}
Venus 0.000 4.458×1064.458 \times 10^{-6}
Earth 0.003353 1.08263×1031.08263 \times 10^{-3}
Mars 0.00648 1.96045×1031.96045 \times 10^{-3}
Jupiter 0.06487 14.736×10314.736 \times 10^{-3}
Saturn 0.09796 16.298×10316.298 \times 10^{-3}
Uranus 0.02293 3.34343×1033.34343 \times 10^{-3}
Neptune 0.01708 3.411×1033.411 \times 10^{-3}
Pluto 0.000
(Moon) 0.0012 202.7×106202.7 \times 10^{-6}

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