# Heliocentrism

## Practice

### practice problem 1

There is a special class of satellites that orbit the Earth with a period of one day.

- Determine the orbital radius at which the period of a satellite's orbit will equal one day. State your answer as…
- an approximate fraction of the moon's orbital radius
- an approximate multiple of the Earth's radius
- an "exact" number of kilometers

- How will the satellite's motion appear when viewed from the surface of the Earth?
- 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.)

#### solution

- 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.
*r*^{3}_{satellite}= *r*^{3}_{moon}*T*^{2}_{satellite}*T*^{2}_{moon}- 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.
*r*^{3}_{satellite}= *r*^{3}_{moon}*T*^{2}_{satellite}*T*^{2}_{moon}*r*^{3}_{satellite}= *r*^{3}_{moon}(1 day) ^{2}(27 days) ^{2}*r*≈_{satellite}1 earth-moon distance 9 - The earth-moon distance is about sixty times the radius of the Earth.
*r*≈_{satellite}60 ≈ 7 earth radii 9 - Repeat the first part of this problem using more "exact" values.
*r*^{3}_{satellite}= *r*^{3}_{moon}*T*^{2}_{satellite}*T*^{2}_{moon}*r*^{3}_{satellite}= (3.944 × 10 ^{8}m)^{3}(0.9973 day) ^{2}(27.32 day) ^{2}*r*_{satellite}= 4.230 × 10 ^{7}m*r*_{satellite}≈ 42,000 km

- 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.

- 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.
- The answer appears elsewhere in this book.
- The answer appears elsewhere in this book.

### practice problem 2

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…

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

#### solution

- 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.⎛

⎝*r*_{resonant orbit}⎞

⎠^{3}= ⎛

⎝*x*⎞

⎠^{2}5.2 AU *y**r*= 5.2 AU_{resonant orbit}⎛

⎝*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 - Filter out the junk and make a new table.
Kirkwood gaps and their ratios orbital radius

(AU)number of

orbital periodsobserved 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…

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 lonlier 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 very 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 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 |

### practice problem 3

#### solution

- Answer it.

### practice problem 4

#### solution

- Answer it.