t S 18. Determining Planetary Co-ordinates

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1 8. Determining Planetary Co-ordinates θ θ 0 ω R In the heliocentric reference frame which rotates with the Earth s orbital motion, suppose that initially a planet s unplanet line makes an angle θ 0 with the stationary un-earth line. After a time interval t, the planet s relative orbital motion will cause it to have rotated around the un by an additional angle θ. The planet s relative angular speed ω R ω P - ω E can be determined from its synodic period using equations (7.3) and (7.4). Therefore the additional angle will be given by π t o θ ωr t t 360 [8.] If we know the starting angle θ 0 from observations of the initial planetary configuration, we can then determine the full angle θ in the final un-earth-planet triangle EP, i.e. θ θ 0 + θ ince we know sides and their P internal angle (E AU; P a, the planet s orbital d radius in AU; angle PE θ ), a we can use the cosine formula θ η for plane triangles, equation E (6.6) to calculate the current straight-line distance d between the planet and the Earth. Thereafter, we can use the sine formula for plane triangles, equation (6.5), to calculate the planet s current elongation η. If desired, we can also calculate the planet s phase angle. A Positional Astronomy Page 55 Lecture 0

2 In our simplifying assumptions (see ection 7), we asserted that all planets have inclination i 0, i.e. they lie exactly on the ecliptic. Therefore, for any planet the ecliptic latitude β 0. The elongation η of a planet is the angular separation between the planet and the un, as viewed from Earth. Therefore, η gives the angular distance along the ecliptic between the planet and the un for a terrestrial observer. Eastern elongation is usually considered to be positive, and western elongation to be negative. The ecliptic longitude of the un λ can be determined from the current date using the system of simplified solar motion discussed in ection 5, i.e. λ 0 on the vernal equinox, λ 90 on the summer solstice, etc. The ecliptic longitude λ of the planet will then be the ecliptic longitude of the un plus the planet s elongation: λ λ + η [8.] With the planet s ecliptic latitude β and ecliptic longitude λ thus determined, they can be converted to right ascension α and declination δ (and then if required altitude a and azimuth A) in the usual way by employing appropriate spherical triangles on the celestial sphere. 9. Phases and Illumination While the elongation of an inferior planet is limited to between 0 and a maximum value η max, the phase angle can take any value from 0 at superior conjunction to 80 at inferior conjunction, i.e. all phase angles are possible. η A Positional Astronomy Page 56 Lecture 0

3 η Conversely, for a superior planet the elongation can take any value from 0 at conjunction to 80 at opposition, but the phase angle is restricted to lie between 0 and some maximum phase angle max. From equation (7.), for any planet we have sinη asin For superior planets, η can range freely between 0 and 80, so the left hand side of this equation will reach its maximum value at η 90 (i.e. at quadrature) where sin η. Thus a sin max sin max a arcsin max a ( ) [9.] Therefore, superior planets have their maximum phase angle at quadrature, and this maximum angle depends only on their orbital radius a. For Mars a.54 AU, so max 4.0 For Jupiter a 5.04 AU, so max. For aturn a 9.58 AU, so max 6.0 For Uranus a AU, so max.9 For Neptune a AU, so max.9 A Positional Astronomy Page 57 Lecture 0

4 D night hemisphere A C F B A E B r E to un to Earth day hemisphere viewed from above Consider a planet with a non-zero phase angle. The hemisphere of this planet which faces towards the un is fully illuminated (the day hemisphere DAE), and the hemisphere which faces away from the un is completely dark (the night hemisphere DBE). When we view this planet from Earth, we see the hemisphere which faces towards us, AEB. This is illuminated along the arc AE, and dark along the remaining arc EB. The width of the full planetary disc as seen from Earth is the diameter AB. The width of the illuminated portion of the disc visible from Earth, projected onto this diameter, is the line AF. The phase q of the planet is the ratio of these lengths, i.e. q AF AB uppose the radius of the planet is r. Then the diameter AB r The length CF is r cos, so AF AC + CF r + r cos Then r + r cos r viewed face-on from Earth ( + cos) q [9.] A Positional Astronomy Page 58 Lecture 0

5 ome notable phases are 0, q ½ ( + cos 0 ) ½ ( + ), full moon 90, q ½ ( + cos 90 ) ½ ( + 0) 0.5, half moon 80, q ½ ( + cos 80 ) ½ ( + -) 0, new moon As it orbits the Earth, our moon shows the full range of phases: Image credit: University of California, Lick Observatory imilarly, although the relative configuration of object, observer and illumination is different, inferior planets such as Venus also show the full range of phases: Image credit: University of Arizona However, all the superior planets have a maximum phase angle of less than 45. Therefore, they never show a phase of less than q ½ ( + cos 45 ) 0.85 i.e. superior planets are never seen at less than gibbous phase. A Positional Astronomy Page 59 Lecture 0

6 Example: Ceres is a dwarf planet whose orbit lies between Mars and Jupiter at an orbital radius of a.766 AU. How many days elapse between successive instances of maximum phase for Ceres? Maximum phase is q, and occurs where (from equation 9.) q + cos cos o 0 ( + cos ) Ceres is a superior planet, so can have any value of elongation. A phase angle of 0 will occur twice per observed revolution around the un, once at opposition and once at conjunction. Therefore, the time between successive maximum phases will be half the time it takes for Ceres to appear to travel once around the un as seen from Earth, i.e. half of Ceres synodic period. By Kepler s 3 rd Law (see Dynamical Astronomy), a 3 T o the sidereal period of Ceres is T P a years Now, using equation (7.4) for the synodic period, T P T E years 33 days Answer: The time elapsed between successive maximum phases of Ceres is half of its synodic period, or 33 days. A Positional Astronomy Page 60 Lecture 0