This presentation compares predictions of Kepler's 3rd Law to data from almost all of the orbital systems in the Solar System (including orbits of Pluto's moons), then predicts the height of a geostationary orbit.

1. Kepler and Newton vs.
Geocentrism, Flat Earth,
and the “Vortex”
JimSmithInChiapas
1 September 2013

2. Links to original sources for data
used here are given in the video
description.

3. In this video, we’ll learn
1. A relation among orbital
periods and sizes that was
first discovered by Kepler,
then explained by Newton;
and

4. 2. How to use that relation to
calculate the radius of the
orbit of a geostationary
satellite.

5. In my experience, geocentrists,
Flat-Earthers, Hollow-Earthers,
and believers in the “Vortex”
Solar System share many traits.
For example:

6. • All claim that “the evidence
proves they’re right”, even
though they come to wildly
different conclusions about
the Earth and Solar System.

7. • They have no explanation
for, nor do they
acknowledge, the data you’ll
see in this video.

8. As background to that data, I
need to mention that the orbits
of planets are ellipses.

9. So are the orbits of moons
around their planets.

10. Note: Although Pluto and
other trans-Neptunian
objects (TNOs) like Eris are
not considered planets, I’ll
refer to them as planets
here for simplicity.

11. The data you’re about to see for
each object refer to the orbital
period P (how long it takes the
object to complete one orbit), …

12. ..and to a, the orbit’s semi-
major axis.

13. An ellipse’s semi-major axis (a)
is half of its long dimension.

14. The next several slides will show
periods and lengths of the semi-
major axes for Solar orbits of
planets, …

15. and for orbits of up to 10 moons
of every planet that has them.
(Except the Earth’s. We’ll save
that for later.)

16. We’ll also see, for each of those
orbits, the value of the ratio
a3/P2 .
That is, the cube of a, divided by
the square of P.

17. If you check the original sources
from which the data were
taken, …

18. you’ll see that the orbits for the
objects shown vary immensely
in their a distances and
periods,…

19. and considerably in their
inclinations with respect to the
orbit of their planets.

20. Some of those orbits are even
retrograde.
That is, the orbits’ directions are
opposite to the direction of the
Earth’s orbit.

21. But despite all of those
differences in the nature of the
orbits, you’ll see that …

22. There are few cases where
an object’s a3/P2 differs by
more than a few % from
the average value of a3/P2
for other objects that orbit
the same body.

23. Now, here are the data:

24. Mars’s Moons
Name a (km) P (days) a3/P2 (km3/day2)
Phobos 9,378 0.31891 8.11E+12
Deimos 23,459 1.26244 8.10E+12

25. Jupiter’s Moons
Name a (km) P (days) a3/P2 (km3/day2)
Metis 128,000 0.294779 2.41E+16
Adrastea 129,000 0.29826 2.41E+16
Amalthea 181,400 0.498179 2.41E+16
Thebe 221,900 0.6745 2.40E+16
Io 421,600 1.769138 2.39E+16
Europa 670,900 3.551181 2.39E+16
Ganymede 1,070,400 7.154553 2.40E+16
Callisto 1,882,700 16.689018 2.40E+16
Themisto 7,507,000 130.02 2.50E+16
Leda 11,170,000 240.92 2.40E+16

26. Saturn’s Moons
Name a (km) P (days) a3/P2 (km3/day2)
Pan 133,583 0.575 7.21E+15
Daphnis 136,500 0.594 7.21E+15
Mimas 185,520 0.942421 7.19E+15
Enceladus 238,020 1.370218 7.18E+15
Tethys 294,660 1.887802 7.18E+15
Dione 377,400 2.73691 7.18E+15
Rhea 527,040 4.5175 7.17E+15
Titan 1,221,830 15.945421 7.17E+15
Hyperion 1,481,100 21.276609 7.18E+15
Iapetus 3,561,300 79.330183 7.18E+15

27. Uranus’s Moons
Name a (km) P (days) a3/P2 (km3/day2)
Cordelia 49,770 0.335034 1.10E+15
Ophelia 53,790 0.3764 1.10E+15
Bianca 59,170 0.434579 1.10E+15
Cressida 61,780 0.46357 1.10E+15
Desdemona 62,680 0.47365 1.10E+15
Miranda 129,390 1.413479 1.08E+15
Ariel 191,020 2.520379 1.10E+15
Umbriel 266,300 4.144177 1.10E+15
Titania 435,910 8.705872 1.09E+15
Oberon 583,520 13.463239 1.10E+15

28. Neptune’s Moons
Name a (km) P (days) a3/P2 (km3/day2)
Naiad 48,227 0.294396 1.29E+15
Thalassa 50,075 0.311485 1.29E+15
Despina 52,526 0.334655 1.29E+15
Galatea 61,953 0.428745 1.29E+15
Larissa 73,548 0.554654 1.29E+15
S/2004 105,300 0.95 1.29E+15
Proteus 117,647 1.122315 1.29E+15
Triton 354,760 5.876854 1.29E+15
Nereid 5,513,400 360.13619 1.29E+15
Halimede 15,730,000 1879.7 1.10E+15

29. Pluto’s Moons
Name a (km) P (days) a3/P2 (km3/day2)
Charon 19,600 6.3872 1.85E+11
Styx 42,000 20.2 1.82E+11
Nix 48,700 24.9 1.86E+11
Kerberos 59,000 32.1 1.99E+11
Hydra 64,750 38.2 1.86E+11

30. Solar Orbits of Planets and TNOs
Name a (km) P (days) a3/P2 (km3/day2)
Mercury 57,910,000 87.968 2.51E+19
Venus 108,210,000 224.695 2.51E+19
Earth 149,600,000 365.242 2.51E+19
Mars 227,920,000 686.973 2.51E+19
Jupiter 778,570,000 4,330.60 2.52E+19
Saturn 1,433,530,000 10,746.94 2.55E+19
Uranus 2,872,460,000 30,588.74 2.53E+19
Neptune 4,495,060,000 59,799.90 2.54E+19
Pluto 5,906,380,000 90,588 2.51E+19
Eris 10,120,000,000 204,870 2.47E+19

31. You may have noticed that
although there was little
variation among the values of
a3/P2 on any given slide, …

32. there was a great difference from
one slide to another. In fact, the
value for orbits around the Sun
was more than 100,000,000
times greater than for orbits
around Pluto.

33. Kepler was the first to discover
that a3/P2 was (almost) constant
for the orbits of all of the planets
then known (1619). He
discovered this through analysis
of Tycho Brahe’s observational
data.

34. Kepler later found (1622) that
the a3/P2 valuesfor orbits of
Jupiter’s moons were constant,
but not equal to the value of
a3/P2 for orbits of planets around
the Sun.*
http://en.wikipedia.org/wiki/Godefroy_Wendelin#cite_n
ote-2

35. Many years later (1687),
Newton’s Principia Mechanica
presented a theory of forces—
including gravity—that explained
Kepler’s results.

36. This triumph gained Newton
lasting fame. Those who are
interested can read the details in
the link given below.*
http://hyperphysics.phy-
astr.gsu.edu/hbase/kepler.html

37. What we would like to do now, is
use Newton’s results—as
substantiated by the data we
examined—to predict the value
of a needed for a satellite to be
geostationary.

38. That is, for the satellite to remain
above the same point on the
Earth.

39. That is, for the satellite to remain
above the same point on the
Earth.
To do so, the satellite’s orbital
period must be exactly one day.

40. Basing our calculation upon the
work of Kepler and Newton, we’ll
assume that all objects that orbit
the Earth have the same value of
a3/P2.

41. So, we assume that
a3/P2 of the satellite
will be equal to
a3/P2 of the Earth’s Moon.

42. The data for Earth’s Moon are:
a = 384,400 km,
P = 27.3217 days.

43. P for the satellite must be 1 day.
Therefore,
a3/P2 of satellite = a3/P2 of Moon
becomes
(a of satellite)3/(1 day)2
= (384,400 km)3/(27.3217 days)2.

44. Solving for a,
a (km) =
cube root of (384,4003/27.32172)
= 42,330 km. ←

45. No satellite can be geostationary
unless its orbit is circular, and
directly over the Equator.

46. This is another fact that cannot
be explained by geocentrists, Flat
Earthers, and believers in the
“Vortex” Solar System.

47. For a circular orbit, a is equal to
the orbit’s radius.

48. The Earth’s equatorial radius is
6,378 km, so an orbit whose
radius is 42,330 km (that’s the
number we calculated a few
slides ago) …

49. will have a height (above the
Equator) of
42,330 km - 6,378 km
= 35,950 km. ←

50. That answer (35,950 km) is
within 1% of that given by the
NOAA (35,800 km).

51. Summary
• Our analysis of data from
seven orbital systems (Mars,
Jupiter, Saturn, Uranus,
Neptune, Pluto, and Sun) …

52. Summary (cont’d)
confirmed Kepler’s and
Newton’s observations and
predictions about values of
a3/P2.

53. Summary (cont’d)
• Using those predictions, we
made an accurate calculation
of the radius needed for a
geostationary orbit.

54. Summary (cont’d)
• To my knowledge, no
geocentrist, Flat Earther, or
supporter of the “Vortex”
Solar System has even
attempted to calculate that
radius using their respective
theories.

55. END