Heliocentrism

1. Use Kepler's third law of planetary motion: the square of the period of a satellite in a circular orbit is proportional to the cube of its radius.

   r3satellite	=	r3moon
   T2satellite		T2moon

   This problem can also be solved using Newton's law of universal gravitation with the centripetal force formula. That solution is presented another section of this book.

   1. Compare the radius and period of the geosynchronous satellite's orbit to the radius and period of the Moon's orbit. Use convenient "approximate" values.

      r3satellite T2satellite	=	r3moon T2moon
      r3satellite	=	r3moon
      (1 day)2		(27 days)2

      r_satellite = ¹₉ Earth-Moon distance

   2. The Earth-Moon distance is about sixty times the radius of the Earth.

      rsatellite = 19(60 Earth radii)

      rsatellite = 7 Earth radii

   3. Repeat the first part of this problem using more precise values.

      r3satellite T2satellite	=	r3moon T2moon
      r3satellite	=	(3.8400 × 108 m)3
      (0.99727 day)2		(27.32 day)2
      rsatellite	=	4.226 × 107 m
      rsatellite	=	42,260 km

   You might have noticed some odd values in this last solution.

   • The period of the Earth's rotation is approximately equal to the mean solar day (24 × 60 × 60 = 86,400 s), but for best results use the sidereal day (23h 56m 04s or 86,164 s or 0.99727 day). A solar day is the time between local noon one day and local noon the next day. (Local noon is the time when the Sun is at its highest position in the sky when viewed from a particular location on the Earth.) The sidereal day is the time it takes for the Earth to return to its original orientation with respect to the distant stars. The mean solar day is a bit longer since the Earth is moving around the Sun. After the Earth has completed one rotation relative to the stars, it has to spin for an extra four minutes to line up with the Sun again (on average).
   • The period of the Moon's orbit used above was not the 30 days found in a typical calendar month, nor was it the 28 day lunar cycle familiar to 51% of the Earth's population, nor was it the more precise 29½ day period between full moons. These numbers are all related to the synodic period of the Moon — the time it takes for the Sun, Earth, and Moon to line up in the same relative positions. The number we used was the sidereal period — the time it takes for the Moon to return to its same orientation with respect to the distant stars (27d 7h 43m 12s or 2,360,592 s or 27.32166 days). The Moon has to travel an additional 2⅙ days after it has already completed one orbit in order to catch up with the moving Earth.

2. The answer appears elsewhere in this book.

3. The answer appears elsewhere in this book.

1. Start with Kepler's harmonic law. Use the orbital radius of Jupiter (5.2 au) and let x and y be the orbital periods of the asteroid and Jupiter respectively.

   ⎛ ⎜ ⎝	rresonant orbit	⎞3 ⎟ ⎠	=	⎛ ⎜ ⎝	x	⎞2 ⎟ ⎠
   	5.2 au				y

   rresonant orbit = 5.2 au	⎛ ⎜ ⎝	x	⎞⅔ ⎟ ⎠
   		y

   Cycle both x and y through all possible combinations of the whole numbers from 1 to 9. The results are compiled in the table below. The values nearest to the observed gap radii are highlighted in red and blue. Repeated ratios like 2:2 (which is a repeat of 1:1) or 3:9 (which is a repeat of 1:3) have been grayed out.

   Resonant orbits with Jupiter (au)

   x	1:x	2:x	3:x	4:x	5:x	6:x	7:x	8:x	9:x
   1	5.200	3.276	2.500	2.064	1.778	1.575	1.421	1.300	1.202
   2	8.254	5.200	3.968	3.276	2.823	2.500	2.256	2.064	1.908
   3	10.816	6.814	5.200	4.293	3.699	3.276	2.956	2.704	2.500
   4	13.10	8.254	6.299	5.200	4.481	3.968	3.581	3.276	3.028
   5	15.20	9.578	7.310	6.034	5.200	4.605	4.155	3.801	3.514
   6	17.17	10.82	8.254	6.814	5.872	5.200	4.692	4.293	3.968
   7	19.03	11.99	9.148	7.551	6.508	5.763	5.200	4.757	4.398
   8	20.80	13.10	10.00	8.254	7.113	6.299	5.684	5.200	4.807
   9	22.50	14.17	10.82	8.929	7.695	6.814	6.148	5.625	5.200

2. Filter out the junk and make a new table.

   Kirkwood gaps and their ratios

   orbital radius (au)		number of orbital periods
   observed	theoretical	asteroid	:Jupiter
   2.060	2.064	4	:1
   2.500	2.500	3	:1
   2.710	2.704	8	:3
   2.822	2.823	5	:2
   2.956	2.956	7	:3
   3.030	3.028	9	:4
   3.280	3.276	2	:1

   and a new graph…

   Magnify

Some additional comments:

• Not all resonant orbits produce gaps and sometimes the exact opposite happens. Certain harmonic ratios seem to attract asteroids resulting in the formation of groups at some resonant orbits. The most famous of these are the Trojan asteroids, which are locked in a 1:1 resonance with Jupiter and are clumped 60° on either side of it at the Lagrange points. (Lagrange orbits will be discussed in more detail in the two sections on orbital mechanics in this book.) The second most important example of this effect is the Hilda group, which occupies the 3:2 resonant orbit at 3.58 au. Why some resonances produce groups while others produce gaps is a subject that still hasn't been fully resolved.
• Some resonant orbits appear to have only one occupant. The asteroid 279 Thule is the lone member of the Thule group in 4:3 resonance with Jupiter It is also one of only a half dozen or so objects which lie in an otherwise big empty region from 4.2 to 5.0 au. If Thule is lonely today it can only get lonelier tomorrow. The other residents of this forbidden zone are all in orbits that appear to be unstable. Jupiter's gravity is gradually sweeping this region clean.
• Sometimes the connection made between a particular group or gap and an orbital resonance is loose.
  • The Cybele group is reported to be in a 7:4 resonance with Jupiter The core of this group orbits 3.38 au from the Sun, but the resonance it's assigned to lies at 3.58 au — a 6% difference.
  • Many reliable sources show a prominent gap at the 7:2 resonance (2.256 au), but in my own analysis of the MPCORB data there is barely a dip in the population at this radius.

To finish this problem off, here's a table identifying the key orbital resonance features of the asteroid belt.