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{ Practical astronomy | Astronomy | The Moon | Physical ephemeris }


Physical ephemeris

Subsections:

By ephemeris is primarily meant the position of a celestial body at a given time. Physical ephemeris would describe not where but how the body appears to us. For the planets and Moon these parameters are

The apparent radius is calculated from the equatorial radius (USNO et al. 1998a, p.E88) and the topocentric distance:

ρ = asin(0.001738 Gm/r)

The elongation and phase angle are calculated from the triangle formed by the topocentre, the topocentric position of the Sun and the topocentric position of the Moon. The elongation is therefore correct, but the phase angle is not properly corrected for light time effects. Both are calculated as the spherical distance angle, but then their sign is adjusted according to the three-vector product of the vectors to Sun, Moon and North Celestial Pole:

sgn(E) = sgn((rSun × rMoon) · k) = sgn(ySun xMoon − xSun yMoon)

sgn(φ) = sgn(E)

Hence a western elongation is negative and the phase angle is then also negative.

The illuminated fraction results from the phase angle as

L = (1 + cos φ) / 2

The V magnitude is calculated from the V(0,1) values listed in USNO et al. (1998a, p.E88) These are first corrected for heliocentric and topocentric distance, rS and rE, then for the phase angle φ according to Harris (1961a, p.272ff):

A = |φ|/100°
V = 0.21 + 5 lg(rS/au) + 5 lg(rE/au) + 3.05 A − 1.02 A2 + 1.05 A3

To give an indication of the orientation of the terminator, its position angle is calculated. By this we mean the following. The terminator is approximately an elliptic arc on one side of the major axis of the ellipse. By position angle of the terminator we mean the PA of the major axis of this ellipse. It is perpendicular to the position angle, on our celestial sphere, of the Sun as measured at the Moon.

To calculate the our aspect of the Moon we use the expressions of Davies et al. (1996a) for the celestial coordinates of the lunar pole of rotation and the prime meridian W:

T0 = JD 2451545.0 d

τ = (t − T0) / 36525 d

E1 = 125.045° − 0.0529921°/d (t−T0)
E2 = 250.089° − 0.1059842°/d (t−T0)
E3 = 260.008° − 13.0120009°/d (t−T0)
E4 = 176.625° + 13.3407154°/d (t−T0)
E5 = 357.529° + 0.9856003°/d (t−T0)
E6 = 311.589° + 26.4057084°/d (t−T0)
E7 = 134.963° + 13.0649930°/d (t−T0)
E8 = 276.617° + 0.3287146°/d (t−T0)
E9 = 34.226° + 1.7484877°/d (t−T0)
E10 = 15.134° − 0.1589763°/d (t−T0)
E11 = 119.743° + 0.0036096°/d (t−T0)
E12 = 239.961° + 0.1643573°/d (t−T0)
E13 = 25.053° + 12.9590088°/d (t−T0)

α1 = 269.9949° + 0.0031° τ − 3.8787° sin E1 − 0.1204° sin E2
     + 0.0700° sin E3 − 0.0172° sin E4 + 0.0072° sin E6
     − 0.0052° sin E10 + 0.0043° sin E13

δ1 = 66.5392° + 0.0130° τ + 1.5419° cos E1 + 0.0239° cos E2
     − 0.0278° cos E3 + 0.0068° cos E4 − 0.0029° cos E6
     + 0.0009° cos E7 + 0.0008° cos E10 − 0.0009° cos E13

W = 38.3213° + 13.17635815°/d (t − T0 − r/c) − 1.4 10-12 °/d2 (t − T0 − r/c)2
     + 3.5610° sin E1 + 0.1208° sin E2 − 0.0642° sin E3
     + 0.0158° sin E4 + 0.0252° sin E5 − 0.0066° sin E6
     − 0.0047° sin E7 − 0.0046° sin E8 + 0.0028° sin E9
     + 0.0052° sin E10 + 0.0040° sin E11 + 0.0019° sin E12
     − 0.0044° sin E13

The inclination i of the rotation axis toward the observer, the position angle PA of the rotation axis on the observer's sky, and the central meridian CM (selenographic longitude of the centre of the apparent disc of the Moon) can then be calculated (USNO et al. 1998a, p.E87):

sin(i) = −sin(δ1) sin(δ) − cos(δ1) cos(δ) cos(α1−α)

sin(PA) = sin(α1−α) cos(δ1) / cos(i)
cos(PA) = [sin(δ1) cos(δ) − cos(δ1) sin(δ) cos(α1−α)] / cos(i)

sin(K) = [−cos(δ1) sin(δ) + sin(δ1) cos(δ) cos(α1−α)] / cos(i)
cos(K) = sin(α1−α) cos(δ) / cos(i)
CM = K − W