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Telling Time by the Stars - Sidereal Time
360o/day + 360o/(365.2422 days)
= 360o/day + 0.9856o/day
Thus, relative to the earth, the celestial sphere completes one rotation (sidereal day) in something less than a solar day. In fact, the sidereal day is just
(360/360.9856) x 24 hours = 23.9345 hours
= 23 hr 56 min 4.2 sec
i.e., approximately 3 min 56 sec shorter than the solar day. Another way to show the same thing is to form the ratio of the length of the sidereal day to that of the solar day:
(24 hr/solar day)/(23.9345 hr/sidereal day) = 1.0027 sidereal day/solar day
The tropical year is then
(365.2422 solar days) x 1.0027 = 366.2284 sidereal days
Now we may solve the problem of estimating the sidereal time on 3:00 pm, November 29. From noon on March 21 to noon on November 29 is 253 solar days. From noon to 3:00 pm on November 29 is an additional 3/24 = 0.125 solar days. Hence, the total elapsed time from the vernal equinox to 3:00 pm on november 29 is
253.125 solar days
Now 253 solar days = 253 x 1.0027 = 253.683 sidereal days. Thus, solar noon on November 29 is
(0.683 sidereal days) x (23.9345 hr/sidereal day) = 16.347 hr
= 16 hr 21 min.
(i.e., a sidereal clock at noon solar time on November 29 reads 4 hr 21 min ahead of a solar clock).
Also, 0.125 solar days = 0.125 x 1.0027 = 0.125 sidereal days (to within the accuracy of the calculation).
(0.125 sidereal days) x (23.9345 hr/sidereal day) = 2.992 hr
= 3 hr 00 min (to within rounding accuracy)
The sidereal time at 3:00 pm, November 29 is, therefore
(16 hr 21 min) + (3 hr 00 min) = 19 hr 21 min
An alternative approach would be to convert 253.125 solar days into sidereal days in a single step:
(253.125 solar days) x 1.0027 = 253.808 sidereal days
If all clocks start at noon on the vernal equinox, then the sidereal clock, at 3:00 pm solar, November 29, is reading 0.81 day past sidereal noon or
(0.808 sidereal day) x (23.9345 hr/sidereal day) = 19.339 hr
= 19 hr 20 min
Accuracy in the last digit is due to rounding error.
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