Position Transforms

Note that in the version of this paper published in Planetary&Space Science, 50, 217ff, the date taken for this example is erroneously given as Aug 28, 1996 16:46:00 UTC, not TT.

We assume that a spacecraft position is given in true geographic coordinates ($GEO_T$) on the date Aug 28, 1996 16:46:00 TT (JD 2450324.19861111). Numerical results are given in Tab.8 (We have chosen this date and position because software by M. Hapgood [personal communication] uses these values as a reference set.) The Julian century for this date is $T_0 = -0.0334237204350195$ (eqn.2). In the following we apply the formulas of section 3.3.3. To convert from $GEO_T$ to $GEI_T$ we calculate $\theta_{GMST} = 228^\circ.68095$ by eqn.20. To convert from the true equator of date to the mean equator of date we have to apply the nutation matrix (eqn.7) with $\epsilon_{0D} = 23^\circ.439726, \Delta\psi = 0^\circ.0011126098, \Delta{\epsilon} = -0^\circ.0024222837$. Then we apply $E(0,\epsilon_{0D},0)$ to transform to the mean ecliptic of date ($HAE_D$), the precession matrix (eqn.9) to transform to the ecliptic of J2000 ($HAE_{J2000}$) and $E(0,\epsilon_{0J2000},0)$ to transform to the equator of J2000 ($GEI_{J2000}$). (The vector is still geocentric since we did not apply a translation. So it might better be called $GAE_D$ etc. but we stick with the $H$ to avoid more acronyms.)

We use $T(GEI_{J2000},HGC_{J2000})$ and $T(HAE_D,HCD)$ of section 3.2.2 to transform to the heliographic systems. To transform to geocentric Earth Ecliptic ($HEE_D$) coordinates we use $T(HAE_{D},HEE_{D})$ from section 3.2.2, for $HEEQ_D$ we use $\theta_\odot = 259^\circ.89919$ (eqn.17). To transform to $GSE_D$ with low precision we use the ecliptic longitude of the Earth $\lambda_{geo} = -24^\circ.302838$ (eqn.36). To transform to $GSM_D$ we use the Earth dipole position $\lambda_D = 288^\circ.58158, \Phi_D = 79^\circ.411145$ (eqn.22) and angles $\psi_D = -21^\circ.604166, \mu_D = 20^\circ.010247$.

To proceed to position dependent systems we now determine the Earth position to a higher precision using the orbital elements of the EMB from Tab.4 corrected by Tab.6 of Simon et al. (1994) (values available on our website):

$$a = 1.0000025, \lambda = -22^\circ.769425, e = 0.016710039, \... ...irc.92657, i = -0^\circ.00043635047,\Omega = 174^\circ.88123.$$ (45)

Using eqns.26 and 30 with $\mu_E = 1/332946$(Tab.4), we get the EMB position in $HAE_{J2000}$:

$$\lambda_{EMB}= -24^\circ.305587, \beta_{EMB} = -0^\circ.00014340633, r_{EMB} = 1.0099340\mbox{[AU]}.$$ (46)

Using the Delauney argument $D = -184^\circ.63320$ we get the Earth position in $HAE_{J2000}$ (eqn.32):

$$\lambda_{E}= -24^\circ.305442, r_{E} = 1.0099033\mbox{[AU]}.$$ (47)

We apply the precession matrix (eqn.9) to get the Earth position vector in $HAE_D$ (1AU = 149 597 870km, 1$R_E$ =6378.14km):

$$X_E = 21579.585 [R_E], Y_E = -9767.205[R_E], Z_E = 0.000016[R_E]$$ (48)

Adding this vector to the geocentric position ($GAE_D$) and transforming to $HCD$ we get the $HCD$ longitude and latitude of the spacecraft:

$$\Phi_{S/C} = -100^\circ.11050, \Theta_{S/C} = 7^\circ.1466473,$$ (49)

from which we calculate the S/C-centered position vector of the Earth $HGRTN_E$.