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∘, the geocentric equatorial position vector of the International Space Station is
r=−5368I^−1784J^+3691K^(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∘Substituting Re=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=[1−(2f−f2)sin2ϕRe+H]cosϕ(cosθI^+sinθJ^)+[1−(2f−f2)sin2ϕRe(1−f)2+H]sinϕK^ (5.56)
R=−5955I^−699.5J^+2168K^(km)Having found R, we obtain the position vector of the space station relative to the site from Equation 5.53:
ϱ=r–R
=(−5368I^−1784J^+3691K^)−(−5955I^−699.5J^+2168K^)
=586.8I^−1084J^+1523K^(km)
The magnitude of this vector is ϱ = 1960 km, so that
ϱ^=ϱϱ=0.2994I^−0.5533J^+0.7773K^Comparing this equation with Equation 5.57 we see that
ϱ^=cosδcosαI^+cosδsinαJ^+sinδK^ (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=sin−10.7773=51.01∘ (a)
as well as
sinα=cosδ−0.5533=−0.8795
cosα=cosδ0.2994=0.4759
Thus
α=cos−1(0.4759)=61.58∘ (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 and declination δ0, which were computed in Example 4.2,
Table 4.3 Oblateness and second zonal harmonics | ||
Planet | Oblateness | J2 |
Mercury | 0.000 | 60×10−6 |
Venus | 0.000 | 4.458×10−6 |
Earth | 0.003353 | 1.08263×10−3 |
Mars | 0.00648 | 1.96045×10−3 |
Jupiter | 0.06487 | 14.736×10−3 |
Saturn | 0.09796 | 16.298×10−3 |
Uranus | 0.02293 | 3.34343×10−3 |
Neptune | 0.01708 | 3.411×10−3 |
Pluto | 0.000 | – |
(Moon) | 0.0012 | 202.7×10−6 |