Use Algorithm 5.3 to find the local sidereal time (in degrees) of Tokyo, Japan, on 3 March 2004 at 4:30:00 UT. The east longitude of Tokyo is 139.80°. (This places Tokyo nine time zones ahead of Greenwich, so the local time is 1:30 in the afternoon.)

Question

Accepted Answer

Step 1:

[latex]J_{0}=367 \cdot 2004- INT \left\{\frac{7\left[2004+ INT \left(\frac{3+9}{12} ight) ight]}{4} ight\}+ INT \left(\frac{275 \cdot 3}{9} ight)[/latex] + 3 + 1721013.5
= 2453067.5 days

Recall that the .5 means that we are half way into the Julian day, which began at noon UT of the previous day.
Step 2:

[latex]T_{0}=\frac{2453067.5-2451545}{36525}=0.041683778[/latex]

Step 3:

[latex] heta_{G_{0}}=100.4606184+36000.77004(0.041683778)+0.000387933(0.041683778)^{2}-2.583\left(10^{-8} ight)(0.041683778)^{3}[/latex]
= 1601.1087°

The right-hand side is too large. We must reduce [latex] heta_{G_{0}}[/latex] to an angle which does not exceed 360°. To that end observe that

INT(1601.1087/360) = 4

Hence,

[latex] heta_{G_{0}}=1601.1087-4 \cdot 360=161.10873^{\circ}[/latex] (a)
Step 4:
The universal time of interest in this problem is

[latex]U T=4+\frac{30}{60}+\frac{0}{3600}=4.5 hr[/latex]

Substitute this and (a) into Equation 5.51 to get the Greenwich sidereal time:

[latex] heta_{G}= heta_{G_{0}}+360.98564724 \frac{U T}{24}[/latex] (5.51)

[latex] heta_{G}=161.10873+360.98564724 \frac{4.5}{24}=228.79354^{\circ}[/latex]

Step 5:
Add the east longitude of Tokyo to this value to obtain the local sidereal time,

θ = 228.79354 + 139.80 = 368.59°

To reduce this result into the range 0 ≤ θ ≤ 360° we must subtract 360° to get

θ = 368.59 − 360 = 8.59° (0.573hr)

Observe that the right ascension of a celestial body lying on Tokyo’s meridian is 8.59°.

Question 5.6: Use Algorithm 5.3 to find the local sidereal time (in degree...

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