practice problem 1

Geosynchronous Satellite
There is a special class of satellites that orbit the Earth with a period of one day.
  1. Determine the orbital radius at which the period of a satellite's orbit will equal one day. State your answer as…
    1. an approximate fraction of the moon's orbital radius
    2. an approximate multiple of the Earth's radius
    3. an "exact" number of kilometers
  2. How will the satellite's motion appear when viewed from the surface of the Earth?
  3. What type of satellites use this orbit and why is it important for them to be located in this orbit? (Keep in mind that this is a relatively high orbit. Satellites not occupying this band are normally kept in much lower orbits.)


  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  =  r3moon
      T2satellite T2moon
      r3satellite  =  r3moon
      (1 day)2 (27 days)2
      rsatellite ≈  1  earth-moon distance
    2. The earth-moon distance is about sixty times the radius of the Earth.
      rsatellite ≈  60  ≈ 7 earth radii
    3. Repeat the first part of this problem using more "exact" values.
      r3satellite  =  r3moon  
      T2satellite T2moon  
      r3satellite  =  (3.944 × 108 m)3  
      (0.9973 day)2 (27.32 day)2  
      rsatellite  =  4.230 × 107 m  
      rsatellite  ≈  42,000 km  
    You might have noticed some odd values in this solution.
    • The period of the Earth's rotation is approximately equal to the mean solar day (1 day = 24 x 3,600 = 86,400 s), but for best results the sidereal day (86,164 s or 0.9973 days) should be used. A mean 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.
    • 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 29.5 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. 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.

practice problem 2

Kirkwood Gaps
The asteroids are a group of small rocky bodies orbiting the sun in relatively circular orbits. (In comparison, comets are small icy bodies orbiting the sun in highly elliptical orbits.) The number of asteroids currently identified is something on the order of 200,000 but only about half of these are in orbits that are known with enough certainty to receive an official catalog entry in the Minor Planet Center Orbit Database (MPCORB). The vast majority of asteroids lie in the region between the orbits of mars and jupiter known as the main asteroid belt. The graph below shows the distribution of asteroids in the densest part of this region. The values highlighted in red show orbital radii for which there are few or even no corresponding asteroids. The values highlighted in blue show the effective edges of this part of the main belt. These features were discovered by the American astronomer Daniel Kirkwood (1814–1895) in the Nineteenth Century and are now known as Kirkwood gaps.

These orbits are empty because they share a simple harmonic relationship with the orbit of jupiter; that is, the ratio of the period of an unoccupied or under-occupied orbit in a Kirkwood gap forms a simple whole number ratio with the period of an orbit of jupiter (something like 2:1 or 3:2 or 5:3). Because of this synchrony the point of closest approach between the two bodies -- the moment when their mutual gravitational attraction is the greatest -- will always take place at the same phase in the asteroid's orbit. Small perturbations applied at just the right moment over and over again reinforce one another until eventually the asteroid enters a new orbit. Repeat this procedure for many simple harmonic ratios and a series of gaps will open up in an asteroid belt that would otherwise be randomly populated.

Using a statistical or spreadsheet application determine…

  1. the radii of all possible resonant orbits that can be generated using the numbers 1 through 9
  2. the resonance ratios responsible for each of the seven Kirkwood gaps identified above


  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  = 
    5.2 AU   y  
    rresonant orbit = 5.2 AU

    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.

    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
    Resonant orbits with Jupiter (AU)
  2. Filter out the junk and make a new table.
    orbital radius
    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
    Kirkwood gaps and their ratios

    and a new graph…

Some additional comments:

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

orbital radius (AU) feature harmonic ratio
~1.91 hungaria group 9:2
2.06 inner edge of main belt 4:1
2.06~2.50 main belt i  
2.50 gap 3:1
2.50~2.70 main belt iia  
2.70 gap 8:3
2.70~2.82 main belt iib  
2.82 gap 5:2
2.82~3.03 main belt iiia  
3.03 gap 9:4
3.03~3.28 main belt iiib  
3.28 outer edge of main belt 2:1
~3.58 cybele group 4:7
~3.96 hilda group 3:2
4.29 thule group 4:3
4.2~5.0 big empty region  
~5.20 trojan group 1:1
Features of the asteroid belt shaped by orbital resonance with Jupiter

practice problem 3

Write something different.


practice problem 4

Write something different.