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#11
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Kepler's 3rd with adjusted periods and radii
Peter Riedt wrote:
On Wednesday, February 17, 2016 at 1:17:00 AM UTC+8, Poutnik wrote: Dne úterý 16. února 2016 14:22:00 UTC+1 Peter Riedt napsal(a): On Tuesday, February 16, 2016 at 4:03:53 PM UTC+8, Mike Dworetsky wrote: Peter Riedt wrote: Kepler's 3rd with adjusted periods and radii Kepler's third law in the form t2/r3 gives a constant for solar objects such as planets and asteroids. It is a precise law but applying it with the measured values for the orbital periods in seconds for t and the measured semi major axis in metres for r produces results with small differences. Kepler's third law, in a more general form, involves the SUM of the masses of the two bodies: (M1+M2)P^2 = a^3 (using units of solar masses, years, and AUs) Hence for more massive planets like Jupiter you have to take this into account (Jupiter's mass is around 1/1000 of the Sun's mass). Just fiddling with numbers is not correct, and inadequate! I have used your formula (M1+M2)P^2 = a^3 and got the following results. Can you explain where I used the wrong data or arithmetic? - Missing gravitational constant and 4.pi^2 factor. G.(M1+M2). P^2 = 4.pi^2 . a^3 - Missing either conversion to meters and seconds, either conversion of the gravitational constant for AUs and years. - Persistent ignoring of rules of processing inaccurate data. The formula was posted by Mike. Read his post in this thread. Yes, I stated that the formulae I gave used units of solar mass, years, and AU. That is not what you are using. -- Mike Dworetsky (Remove pants sp*mbl*ck to reply) |
#12
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Kepler's 3rd with adjusted periods and radii
Dne úterý 16. února 2016 21:29:33 UTC+1 Martin Brown napsal(a):
On 16/02/2016 13:21, Peter Riedt wrote: (M1+M2)P^2 = a^3 (using units of solar masses, years, and AUs) *UNITS!!!* ^^^^^^^^^^^^ There is hidden implicit G, with the value approximately 1, and dimension of AU^3 / year^2 / Ms |
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