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Kepler’s Validation of G
Kepler’s Validation of G
Cavendish discovered the universal gravitational constant G in his laboratory in 1798 with a value of 6.74E11. Today the recognized value is 6.67259E11. The formula P=sqrt((4pi2r3)/(G(msun+mplanet)), which was derived from Kepler’s planetary laws, provides a link between G and the orbital periods of free fall bodies in the solar system. If we apply this formula with an adjusted value of G=6.66662E11 (for a better fit) we get the following differences in days between observed (in Pluto’s case predicted) and calculated orbital periods of the selected planets and moons: OBS DAYS CALC DAYS DAYS DIFF MERCURY 87.96 87.97 0.0087428 VENUS 224.68 224.70 0.0177057 EARTH 365.26 365.25 0.0061423 MARS 686.98 686.87 0.1145676 JUPITER 4332.71 4334.48 1.7681635 SATURN 10759.10 10759.00 0.0982464 URANUS 30707.41 30707.07 0.3376758 NEPTUNE 60198.50 60198.51 0.0112854 PLUTO 90474.90 90763.22 288.3172342 MOON 27.32 27.27 0.0513968 IO 1.77 1.77 0.0000000 EUROPA 3.55 3.55 0.0000000 GANYMEDE 7.15 7.15 0.0000000 CALLISTO 16.69 16.69 0.0000000 The results from the formula derived from Kepler’s planetary laws with the adjusted value of G validate the constant G as a universal constant in laboratory, heliocentric, geocentric and joviancentric environments. In all cases except Pluto’s and Jupiter’s, the difference is less than +0.4 days. The large difference for Pluto may indicate an error in the predicted orbital period or the mean distance from the sun or the assumed mass. For Jupiter it is remarkable that while the planet has a difference of 1.7 days in 4334 days, his four moons have no difference at all. Peter Riedt 
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#2




Kepler’s Validation of G
On Mar 15, 10:32*am, Peter Riedt wrote:
Kepler’s Validation of G Cavendish discovered the universal gravitational constant G in his laboratory in 1798 with a value of 6.74E11. Today the recognized value is 6.67259E11. The formula P=sqrt((4pi2r3)/(G(msun+mplanet)), which was derived from Kepler’s planetary laws, provides a link between G and the orbital periods of free fall bodies in the solar system. If we apply this formula with an adjusted value of G=6.66662E11 (for a better fit) we get the following differences in days between observed (in Pluto’s case predicted) and calculated orbital periods of the selected planets and moons: * * * * * * * * OBS DAYS * * * *CALC DAYS * * * DAYS DIFF MERCURY * * * * 87.96 * 87.97 * 0.0087428 VENUS * * * * * 224.68 *224.70 *0.0177057 EARTH * * * * * 365.26 *365.25 *0.0061423 MARS * * * * * *686.98 *686.87 *0.1145676 JUPITER * * * * 4332.71 4334.48 1.7681635 SATURN * * * * *10759.10 * * * *10759.00 * * * *0.0982464 URANUS * * * * *30707.41 * * * *30707.07 * * * *0.3376758 NEPTUNE * * * * 60198.50 * * * *60198.51 * * * *0..0112854 PLUTO * * * * * 90474.90 * * * *90763.22 * * * *288.3172342 MOON * * * * * *27.32 * 27.27 * 0.0513968 IO * * * * * * *1.77 * *1.77 * *0.0000000 EUROPA * * * * *3.55 * *3.55 * *0.0000000 GANYMEDE * * * * * * * *7.15 * *7.15 * *0.0000000 CALLISTO * * * * * * * *16.69 * 16.69 * 0.0000000 The results from the formula derived from Kepler’s planetary laws with the adjusted value of G validate the constant G as a universal constant in laboratory, heliocentric, geocentric and joviancentric environments. In all cases except Pluto’s and Jupiter’s, the difference is less than +0.4 days. The large difference for Pluto may indicate an error in the predicted orbital period or the mean distance from the sun or the assumed mass. For Jupiter it is remarkable that while the planet has a difference of 1.7 days in 4334 days, his four moons have no difference at all. Peter Riedt Correction, the adjusted value of G should be 6.6662E11 not 6.66662E11. 
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