Astronomy 11 - Fall 1998 (S.T. Myers)

1. The crater Daedalus is on the far side of the moon at lunar longitude 179.4 E and latitude 5.9 S (this has no impact on the problem). If the Sun is overhead at Daedalus, and at an angle of 31.5 degrees from the zenith (as given by the shadow) at Dante, the arc from Daedalus to Dante subtends the fraction of the circumference 31.5/360, and thus

   A = ( 31.5 / 360 ) x 2 R_moon = 956 km

   for R_moon = 1738 km (Appendix 3). It does not matter that Dante is due North, since this gives the arc length along the great circle between Daedalus and Dante no matter the orientation, since the Sun is directly overhead at Daedalus (circular symmetry about the Sun-Daedalus axis).

2. You should first draw a picture of the orbits of Mars and Jupiter from above the ecliptic. Remember to make the line of greatest elongation tangent to the circular orbit of Mars from Jupiter, which makes a right angle with the radius from the Sun to Mars (1.52 AU) which is the opposite side of the right triangle from the elongation angle. The Jupiter-Sun line is the hypoteneuse (5.20 AU) and thus

   sin = 1.52 / 5.20 = 0.292
   => = 17°

   is the Greatest Elongation of Mars as seen from Jupiter.

3. The time interval between any consecutive similar configurations between two planets and the Sun is the synodic period S, where

   1 / S = 1 / P_in - 1 / P_out

   for the sidereal periods P of the inner and outer planets of the pair. For Jupiter and Saturn

   P_in = 11.86 yrs (Jupiter), P_out = 29.46 yrs (Saturn) => S = 19.85 yrs.

   Since S is the synodic period, it is the time between two oppositions of Saturn from Jupiter, or between inferior conjunctions of Jupiter from Saturn (or any other similar configurations). Furthermore, Saturn at opposition and Jupiter at inferior conjunction are the same from the respective viewpoints of Jupiter and Saturn!

4. The major axis is given by the sum of r_p and r_a

   r_p + r_a = 80 AU + 100 AU = 180 AU = 2 a

   so the semimajor axis is a = 90 AU. The eccentricity can be calculated using the aphelion for example,

   r_a = a ( 1 + e ) => e = r_a/a - 1 = 100/90 - 1 = 1/9

   and thus the eccentricity of the orbit is e=0.11.

5. From Kepler's Third Law, normalized to the Earth's orbit

   ( P / 1 yr )² = ( a / 1 AU )³

   which of course only holds for orbits around the Sun. Thus,

   P = ( a / 1 AU )^3/2 yrs

   which for our Planet X

   P = ( 90 )^3/2 yrs = 854 yrs.

   (How hard would this be to detect? It would move a full 360° around its orbit in 854 years, and thus has an angular velocity of 25' per year, which is half the full moon diameter. This would be fairly easy to detect photographically. In a later problem set, we will compute its brightness.)

smyers@nrao.edu   Steven T. Myers