Heliocentrism

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© 1998-2008 by Glenn Elert -- A Work in Progress
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Discussion

early

Aristarchus a.k.a. Αρίσταρχος (310-230 BCE) Greece (Samos)

However, Aristarchus' only remaining work on the topic, On the Sizes and Distances of the Sun and Moon, is geocentric!

Pythagoras a.k.a. Πυθαγόρας (582 BC-496 BCE) Greece (Ionia)

copernicus

Nicolaus Copernicus (1473-1543) Poland (latinized version of Mikołaj Kopernik). De Revolutionibus Orbium Cœlestium (On the Revolutions of Heavenly Spheres) 1543.

In medio uero omnium residet Sol.   In the center of all rests the sun.

Quotes

Notes

The Heliocentric System of Copernicus
Adapted from the
Latin first edition of 1543.		 Adapted from the
1939 English translation by Wallis.
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How heliocentrism deals with retrograde motion.


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bruno

Giordano Bruno a.k.a. Filippo Bruno a.k.a. Bruno Nolano (1548-1600) Italy (Naples)

galileo

Galileo Galilei (1564-1642) Italy

Notes and Quotes

Just Quotes

tycho

Tycho Brahe (1546-1601) Denmark

Notes

Quotes

kepler

Johannes Kepler (1571-1630) Holy Roman Empire (now Austria)

Kepler was a terrible high school teacher.

Kepler's first attempt fails

Mysterium Cosmographicum (Cosmic Mystery) 1596.

Before the universe was created, there were no numbers except the Trinity, which is God himself… For, the line and the plane imply no numbers: here infinitude itself reigns. Let us consider, therefore, the solids. We must first eliminate the irregular solids, because we are only concerned with orderly creation. There remain six bodies, the sphere and the five regular polyhedra. To the sphere corresponds the heaven. On the other hand, the dynamic world is represented by the flat-faces solids. Of these there are five: when viewed as boundaries, however, these five determine six distinct things: hence the six planets that revolve about the sun. This is also the reason why there are but six planets.

I have further shown that the regular solids fall into two groups: three in one, and two in the other. To the larger group belongs, first of all, the Cube, then the Pyramid, and finally the Dodecahedron. To the second group belongs, first, the Octahedron, and second, the Icosahedron. That is why the most important portion of the universe, the Earth -- where God's image is reflected in man -- separates the two groups. For, as I have proved next, the solids of the first group must lie beyond the earth's orbit, and those of the second group within… Thus I was led to assign the Cube to Saturn, the Tetrahedron to Jupiter, the Dodecahedron to Mars, the Icosahedron to Venus, and the Octahedron to Mercury.


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outer solar system   inner solar system
[magnify]

 

      radius of sphere ratio R/r
  planet platonic solid outer (R) inner (r) kepler modern
  saturn r = 9.53876          
        cube
√3
2
1
2
√3 =  1.732 1.833
  jupiter r = 5.20283          
        tetrahedron
√6
4
√6
12
3 3.415
  mars r = 1.52369          
        dodecahedron
√15 + √3
4
√(250 + 110√5)
20
1.258 1.524
  earth r = 1          
        icosahedron
√(10 + 2√5)
4
3√3 + √15
12
1.258 1.383
  venus r = 0.72333          
        octahedron
1
√2
1
√6
√3 =  1.732 1.869
  mercury  r = 0.38710          

Kepler decided the observations of the planets were wrong, not his model of nested platonic solids.

This model was rendered useless when Uranus (the seventh planet) was discovered in 1781, Neptune (the eighth planet) in 1846, Ceres (the first asteroid) in 1801, and Pluto (the first Kuiper belt object) in 1930. More than 10,000 objects orbiting the sun have been identified.

The counter reformation steps in and Kepler is driven out of Graz.

rather make a desert of the country than rule over heretics Philip III (1598-1621) Son of Philip II becomes leader of Spanish house of Hapsburg says to Pope: "I would rather lose a hundred lives, if I had them, than consent to rule over heretics." This rebellion would drag on until 1648, become part of the wider European struggle known as the Thirty Years War (1618-1648)

Kepler's Good Stuff

Additional Quotes

Ubi materia, ibi geometria.   Where there is matter, there is geometry.
     
Geometria una et æterna est in mente Dei refulgens: cuius consortium hominibus tributum inter causas est, cur homo sit imago Dei.   Geometry is one and eternal shining in the mind of God: that share in it accorded to men is one of the reasons that Man is the image of God.
     
Mensus eram cœlos, nunc Terræ metior umbras. Mens cœlestis erat, corporis umbra jacet.   I used to measure the Heavens, now I measure the shadows of Earth. The mind belonged to Heaven, the body's shadow lies here.
     
    Geometry existed before the Creation. It is co-eternal with the mind of God. Geometry provided God with a model for the Creation. Geometry is God Himself.
     
    Geometry, which before the origin of things was coeternal with the divine mind and is God himself (for what could there be in God which would not be God himself?), supplied God with patterns for the creation of the world, and passed over to Man along with the image of God.
     
    My brain gets tired when I try to understand what I wrote, and I find it hard to rediscover the connexion between the figures and the text, that I established myself.

Kepler wrote the first work of science fiction -- the Somnium (1634) -- published by Kepler's son Ludwig, four years after his death.

Summary

Problems

practice

  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.)

    Solutions …

    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 in the section called Orbital Mechanics I.)
      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 ⇒ r3satellite  =  r3moon
        T2satellite T2moon (1 day)2 (27 days)2
                     
        rsatellite ≈  1  earth-moon distance
        9
           
      2. The earth-moon distance is about sixty times the radius of the earth.
             
        rsatellite ≈  60  ≈ 7 earth radii
        9
           
      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 ≈   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 3600 = 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.
  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.
     
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    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

    Solutions …

    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 all the extra 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 …

       
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    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.

     
    Features of the Asteroid Belt Shaped by Orbital Resonance with Jupiter
    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
     
  3. Write something different.
    • Answer it.
  4. Write something different.
    • Answer it.

conceptual

  1. Kepler's first two laws of planetary motion state that "the path of each planet about the sun is an ellipse with the sun at one focus" and "each planet moves so that a line drawn from the sun to the planet sweeps out equal areas in equal periods of time." Explain in one paragraph how these two laws together agree with the more fundamental laws of …
    1. conservation of angular momentum and
    2. conservation of energy.

numerical

  1. In 1984 Marc Davis, Piet Hut, and Richard A. Muller presented the following highly speculative hypothesis in a letter to the British science journal Nature.
    A 26-Myr periodicity has recently been seen in the fossil record of extinction in the geological past. At least two of these extinctions are known to be associated with the impact on the Earth of a comet or asteroid with a diameter of a few kilometres. We propose that the periodic events are triggered by an unseen companion to the Sun, travelling in a moderately eccentric orbit, which at its closest approach (perihelion) passes through the "Oort cloud" of comets which surrounds the Sun. During each passage this unseen solar companion perturbs the orbits of these comets, sending a large number of them (over 1 × 109) into paths which reach the inner Solar System. Several of these hit the Earth, on average, in the following million years….
    This "unseen solar companion" later acquired the name Nemesis, after the Greek goddess of divine retribution (Νέμεσις).

    Determine the average distance from Nemesis to the sun. Give your answer in …
    1. astronomical units and
    2. light years and
    3. compare it to the distance to the nearest star (4.3 light years).

statistical

  1. de-revolutionibus.txt
    Read the following passage from the English translation of De Revolutionibus. Determine the period of each of the five planets known to Copernicus in the Sixteenth Century using his measurements. State your answers in years or days as appropriate. Compare them to the Twenty-first Century values. Compile your results in a table like the one below.

    planet period
    (copernicus)    		 period
    (contemporary)    		 per cent
    deviation
    saturn   29.4580 years  
    jupiter   11.8625 years  
    mars   686.980 days  
    venus   224.701 days  
    mercury   87.9708 days  
        average deviation →  

  2. The asteroids in the following table are reported to be in resonant orbits with a planet. Determine
    1. the ratio of the number of periods of the asteroid to the number of periods of the planet,
    2. the most likely ideal ratio that relates them, and
    3. the per cent deviation between the observed and ideal ratios.

    asteroid planet nasteroid : nplanet deviation
    name r (AU) name r (AU) observed ideal (%)
    1685 toro 1.367 venus 0.723      
    1685 toro 1.367 earth 1.000      
    1221 amor 1.920 earth 1.000      
    3753 cruithne 0.998 earth 1.000      
    87 alinda 2.485 jupiter 5.204      
    8 flora 2.201 jupiter 5.204      
      pluto 39.482 neptune 30.047      

  3. Write some sort of resonant orbit problem to go with this nice illustration.
 
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  janus mimas enceladus tethys dione rhea titan
1.120 rs ringlet              
1.190 rs ringlet              
guerin division
d-c ring interface		              
titan ringlet              
maxwell ringlet?
maxwell division?
lyot division? french division?		              
1.470 rs ringlet              
1.495 rs ringlet              
bond-dawes gap
c-b ring interface		              
huygens ringlet              
end of b ring              
cassini division   2:1 3:1 4:1 5:1 9:1  
start of a ring              
encke gap?
encke minima?		   5:3          
keeler gap              
outer edge of a ring 7:6            
f ring              
start of g ring              
end of g ring              

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