Question 5.6: Use Algorithm 5.3 to find the local sidereal time (in degree...
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.)
Step 1:
J_{0}=367 \cdot 2004- INT \left\{\frac{7\left[2004+ INT \left(\frac{3+9}{12}\right)\right]}{4}\right\}+ INT \left(\frac{275 \cdot 3}{9}\right) + 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:
Step 3:
\theta_{G_{0}}=100.4606184+36000.77004(0.041683778)+0.000387933(0.041683778)^{2}-2.583\left(10^{-8}\right)(0.041683778)^{3}
= 1601.1087°
The right-hand side is too large. We must reduce \theta_{G_{0}} to an angle which does not exceed 360°. To that end observe that
INT(1601.1087/360) = 4
Hence,
\theta_{G_{0}}=1601.1087-4 \cdot 360=161.10873^{\circ} (a)
Step 4:
The universal time of interest in this problem is
Substitute this and (a) into Equation 5.51 to get the Greenwich sidereal time:
\theta_{G}=\theta_{G_{0}}+360.98564724 \frac{U T}{24} (5.51)
\theta_{G}=161.10873+360.98564724 \frac{4.5}{24}=228.79354^{\circ}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°.