The east longitude and latitude of an observer near San Francisco are Λ=238° and φ=38°, respectively. The local sidereal time, in degrees, is θ =215.1° (12hr 42min). At that time the planet Jupiter is observed by means of a telescope to be located at azimuth A=214.3° and angular elevation a=43°

Question

The east longitude and latitude of an observer near San Francisco are Λ=238° and φ=38°, respectively. The local sidereal time, in degrees, is θ =215.1° (12hr 42min). At that time the planet Jupiter is observed by means of a telescope to be located at azimuth A=214.3° and angular elevation a=43° .What are Jupiter’s right ascension and declination in the topocentric equatorial system?

Accepted Answer

The given information allows us to formulate the matrix of the transformation from topocentric horizon to topocentric equatorial using Equation 5.62b,

[latex][ Q ]_{x X}=\left[\begin{array}{ccc}-\sin heta & -\sin \phi \cos heta & \cos \phi \cos heta \\cos heta & -\sin \phi \sin heta & \cos \phi \sin heta \0 & \cos \phi & \sin \phi\end{array} ight][/latex] (5.62b)

[latex][ Q ]_{x X}=\left[\begin{array}{ccc}-\sin 215.1^{\circ} & -\sin 38^{\circ} \cos 215.1^{\circ} & \cos 38^{\circ} \cos 215.1^{\circ} \\cos 215.1^{\circ} & -\sin 38^{\circ} \sin 215.1^{\circ} & \cos 38^{\circ} \sin 215.1^{\circ} \0 & \cos 38^{\circ} & \sin 38^{\circ}\end{array} ight][/latex]
[latex]=\left[\begin{array}{ccc}0.5750 & 0.5037 & -0.6447 \-0.8182 & 0.3540 & -0.4531 \0 & 0.7880 & 0.6157\end{array} ight][/latex]

From Equation 5.58 we have

[latex]\hat{\varrho}=\cos a \sin A \hat{ i }+\cos a \cos A \hat{ j }+\sin a \hat{ k }[/latex]
[latex]=\cos 43^{\circ} \sin 214.3^{\circ} \hat{ i }+\cos 43^{\circ} \cos 214.3^{\circ} \hat{ j }+\sin 43^{\circ} \hat{ k }[/latex]
[latex]=-0.4121 \hat{ i }-0.6042 \hat{ j }+0.6820 \hat{ k }[/latex]

Therefore, in matrix notation the topocentric horizon components of [latex]\hat{\varrho}[/latex] are

[latex]\{\hat{\varrho}\}_{x}=\left\{\begin{array}{c}-0.4121 \-0.6042 \0.6820\end{array} ight\}[/latex]

We obtain the topocentric equatorial components [latex]\{\hat{ \varrho }\}_{X}[/latex] by the matrix operation

[latex]\{\hat{ \varrho }\}_{X}=[ Q ]_{x X}\{\hat{ \varrho }\}_{x}=\left[\begin{array}{ccc}0.5750 & 0.5037 & -0.6447 \-0.8182 & 0.3540 & -0.4531 \0 & 0.7880 & 0.6157\end{array} ight]\left\{\begin{array}{c}-0.4121 \-0.6042 \0.6820\end{array} ight\}[/latex]
[latex]=\left\{\begin{array}{c}-0.9810 \-0.1857 \-0.05621\end{array} ight\}[/latex]

so that

[latex]\hat{\varrho}=-0.9810 \hat{ I }-0.1857 \hat{ J }-0.05621 \hat{ K }[/latex]

Recall Equation 5.57,

[latex]\hat{\varrho}=\cos \delta \cos \alpha \hat{ I }+\cos \delta \sin \alpha \hat{ J }+\sin \delta \hat{ K }[/latex]

Comparing the Z components of these two expressions, we see that

sin δ = −0.0562

which means the topocentric equatorial declension is

[latex]\delta=\sin ^{-1}(-0.0562)=\underline{-3.222^{\circ}} [/latex]

Equating the X and Y components yields

[latex]\sin \alpha=\frac{-0.1857}{\cos \delta}=-0.1860[/latex]
[latex]\cos \alpha=\frac{-0.9810}{\cos \delta}=-0.9825[/latex]

Therefore,

[latex]\alpha=\cos ^{-1}(-0.9825)=169.3^{\circ}[/latex] (second quadrant) or 190.7° (fourth quadrant)

Since sinα is negative, α is in the fourth quadrant, which means the topocentric equatorial right ascension is
α = 190.7°
Jupiter is sufficiently far away that we can ignore the radius of the earth in Equation 5.53. That is, to our level of precision, there is no distinction between the topocentric equatorial and geocentric equatorial systems:

r = R + ϱ (5.53)

r ≈ ϱ
Therefore the topocentric right ascension and declination computed above are the same as the geocentric equatorial values.

Question 5.8: The east longitude and latitude of an observer near San Fran...

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