Fundamentals of astrodynamics and applications 4th Edition (2013)

230 COORDINATE AND TIME SYSTEMS 3.7
An example is useful to tie all the concepts together.
Example 3-15. Performing a IAU-76/FK5 Reduction.
GIVEN: r ITRF
= −1033.479 383 0 Î + 7901.295 275 4 Ĵ + 6380.356 595 8 Kˆ km
v = −3.225 636 520 Î −2.872 451 450 Ĵ + 5.531 924 446 Kˆ km/s
ITRF
FIND: r , v on April 6, 2004, 07:51:28.386 009UTC
GCRF GCRF
Calculate the time quantities before starting the problem. Get ΔAT (32.0 s ) directly from the Astronomical
Almanac (2004:K9). From the IERS’s EOPCO4 final data, find ΔUT1 = −0.439 961 9 s , x p =
–0.140 682'' and y p = 0.333 309'', δΔW 1980 = −0.052 195'' and δΔe 1980 = = −0.003 875'', and LOD =
0.001 556 3s. Although general practice interpolates these values with a spline, don’t interpolate
these values for this example.
UT1 = UTC + ΔUT1 = 07:51:28.386 009 − 0.439 961 9 = 07:51:27.946 047
TAI = UTC + ΔAT = 07:51:28.386 009 + 32 s = 07:52: 0.386 009
TT = TAI + 32.184 s = 07:52: 0.386 009 + 32.184 s = 07:52:32.570 009
DMY HMS JD TT = 2,453,101.828 154 745 (Be sure to keep all digits for accuracy.)
JD TT – 2,451,545.0 2,453,101.828 154 745 – 2,451,545.0
T TT = ----------------------------------------------- = ----------------------------------------------------------------------------------------- = 0.042 623 631 9
36,525
36,525
First, determine the effects of polar motion and you’ll get these initial PEF vectors:
r PEF = −1033.475 031 3 Î + 7901.305 585 6 Ĵ + 6380.344 532 8 Kˆ km
v PEF = −3.225 632 747 Î −2.872 442 511 Ĵ + 5.531 931 288 Kˆ km/s
Next, determine the v GMST1982 using the JD UT1 , Eq. (3-45), and Eq. (3-42):
v GMST1982 = 312.809 894 3°
Find the AST using Eq. (3-79) with the last two terms because the example occurs in 2004
v GAST1982 = v GMST1982 + ΔW 1980 COS( e 1980 ) = 312.806 765 4°
You’ll need the velocity correction from
q K × r PEF = −0.576 172 29 Î −0.075 362 19 Ĵ + 0.000 Kˆ km/s
After doing the rotation for sidereal time, you’ll find the vectors for true equinox and true equator of
date:
r tod
= 5094.514 780 4 Î + 6127.366 461 2 Ĵ + 6380.344 532 8 Kˆ km
v = −4.746 088 567 Î + 0.786 077 222 Ĵ + 5.531 931 288 Kˆ km/s
tod
Determine the angles for the Earth’s nutation (r = 360°). Notice these are similar to those in Example
3-14.
M z = 314.911 859 0°, M L = 91.937 993 1°, u Mz = 169.096 827 2°,
D L = 196.751 811 6°, Q z = 42.604 614 0°
Now find the obliquity of the ecliptic [Eq. (3-68), e 1980
= 23.438 736 8°] and the nutation corrections
for the true of date system, using the 106 coefficients from the IAU 1980 Theory of Nutation
[Eq. (3-83)].
ΔW 1980 = −0.003 410 8°, Δe 1980 = 0.002 031 6°