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Some Reference Julian Dates Julian day 2451178.5 2451543.5 2451909.5 2452274.5 2452639.5 2453004.5 2453370.5 2453735.5 2454100.5 2454465.5 2454831.5 2455196.5
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January 0.0 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010
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For convenience in calculations the day number of the year is given in Table 2.3.
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Example 2.10 Find the Julian day for 13 h UT on 18 December 2000.
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The year 2000 is a leap year, and from Table 2.3, December 18 is day number 335 18 353. This is for midnight December 17/18. UT 13 h as a fraction of a day is 13/24 0.5416667. From Table 2.2, the Julian date for January 0.0, 2000 is 2451543.5, and therefore the required Julian date is 2451543.5 353 0.5416667 2451897.0417.
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In Sec. 2.9.7, certain calculations require a time interval measured in Julian centuries, where a Julian century consists of 36,525 mean solar days. The time interval is reckoned from a reference time of January 0.5, 1900, which corresponds to 2,415,020 Julian days.
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Day Number for the Last Day of the Month Day number for start of day (midnight) Numbers in parentheses are for leap years 31 59 (60) 90 (91) 120.5 (121.5) 151 (152) 181 (182) 212 (213) 243 (244) 273 (274) 304 (305) 334 (335) 365 (366)
Date January 31 February 28 (29) March 31 April 30 May 31 June 30 July 31 August 31 September 30 October 31 November 30 December 31
Orbits and Launching Methods
Denoting the reference time as JDref, the Julian century by JC, and the time in question by JD, then the interval in Julian centuries from the reference time to the time in question is given by T JD JDref (2.20)
JC This is illustrated in the following example.
Example 2.11 Find the time in Julian centuries from the reference time January
0.5, 1900 to 13 h UT on 18 December 2000.
Solution
JDref 2415020 days; JC 36525 days. From Example 2.10: JD 2451897.0417 days. Equation (2.20) gives T 2451897.0417 36525 1.00963838 2415020
Note that the time units are days and T is dimensionless.
2.9.4 Sidereal time
Sidereal time is time measured relative to the fixed stars (Fig. 2.7). It will be seen that one complete rotation of the earth relative to the fixed stars is not a complete rotation relative to the sun. This is because the earth moves in its orbit around the sun.
A sidereal day, or one rotation of the earth relative to fixed stars, is shorter than a solar day.
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The sidereal day is defined as one complete rotation of the earth relative to the fixed stars. One sidereal day has 24 sidereal hours, 1 sidereal hour has 60 sidereal minutes, and 1 sidereal minute has 60 sidereal seconds. Care must be taken to distinguish between sidereal times and mean solar times, which use the same basic subdivisions. The relationships between the two systems, given in Bate et al. (1971), are 1 mean solar day 1.0027379093 mean sidereal days 24 h 3 m 56.55536 s sidereal time 86,636.55536 mean sidereal seconds 1 mean sidereal day 0.9972695664 mean solar days 23 h 56 m 04.09054 s mean solar time (2.22) 86,164.09054 mean solar seconds Measurements of longitude on the earth s surface require the use of sidereal time (discussed further in Sec. 2.9.7). The use of 23 h, 56 min as an approximation for the mean sidereal day will be used later in determining the height of the geostationary orbit. (2.21)
2.9.5 The orbital plane
In the orbital plane, the position vector r and the velocity vector v specify the motion of the satellite, as shown in Fig. 2.8. For present purposes,
Perifocal coordinate system (PQW frame).
Orbits and Launching Methods
only the magnitude of the position vector is required. From the geometry of the ellipse (see App. B), this is found to be r a(1 1 e2) e cos (2.23)
The true anomaly is a function of time, and determining it is one of the more difficult steps in the calculations. The usual approach to determining proceeds in two stages. First, the mean anomaly M at time t is found. This is a simple calculation: M n(t Tp) (2.24)
Here, n is the mean motion, as previously defined in Eq. (2.8), and Tp is the time of perigee passage. The time of perigee passage Tp can be eliminated from Eq. (2.24) if one is working from the elements specified by NASA. For the NASA elements, M0 Therefore, Tp t0 M0 n (2.25) n(t0 Tp)
Substituting this in Eq. (2.24) gives M M0 n(t t0 ) (2.26)
Consistent units must be used throughout. For example, with n in degrees/day, time (t t0) must be in days and M0 in degrees, and M will then be in degrees.
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