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Calculating the EXACT mass of the sun



 
 
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  #1  
Old November 5th 12, 01:54 AM posted to sci.astro
Peter Riedt
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Posts: 83
Default Calculating the EXACT mass of the sun

Calculating the EXACT mass of the sun

The relationship r*v^2 determines the mass of the sun by the distance of a planet from the sun (r in m) and the square of its speed (v in m/sec) exactly if the known velocity is adjusted for eccentric orbits. The adjusted velocities for nine planets are shown in the following table. All produce the standard value of GM with a deviation of 0.000000000000000000%. GM (1.327158270.E+20) is calculated from the Codata values G = 6.674E-11 and M = 1..988550E+30.
r/m v/msec adjusted v r*v^2=GM
MER 5.791000E+10 47,870 47,872 1.327158270.E+20
VEN 1.082100E+11 35,020 35,021 1.327158270.E+20
EAR 1.496000E+11 29,780 29,785 1.327158270.E+20
MAR 2.279200E+11 24,130 24,131 1.327158270.E+20
JUP 7.785700E+11 13,070 13,056 1.327158270.E+20
SAT 1.427000E+12 9,690 9,644 1.327158270.E+20
URA 2.871000E+12 6,810 6,799 1.327158270.E+20
NEP 4.497100E+12 5,430 5,432 1.327158270.E+20
PLU 5.913000E+12 4,720 4,738 1.327158270.E+20

Peter Riedt

  #2  
Old November 5th 12, 07:03 PM posted to sci.astro
Bill Owen
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Default Calculating the EXACT mass of the sun

Peter Riedt wrote:
Calculating the EXACT mass of the sun

The relationship r*v^2 determines the mass of the sun by the distance of a planet from the sun (r in m) and the square of its speed (v in m/sec) exactly if the known velocity is adjusted for eccentric orbits. The adjusted velocities for nine planets are shown in the following table. All produce the standard value of GM with a deviation of 0.000000000000000000%. GM (1.327158270.E+20) is calculated from the Codata values G = 6.674E-11 and M = 1.988550E+30.
r/m v/msec adjusted v r*v^2=GM
MER 5.791000E+10 47,870 47,872 1.327158270.E+20
VEN 1.082100E+11 35,020 35,021 1.327158270.E+20
EAR 1.496000E+11 29,780 29,785 1.327158270.E+20
MAR 2.279200E+11 24,130 24,131 1.327158270.E+20
JUP 7.785700E+11 13,070 13,056 1.327158270.E+20
SAT 1.427000E+12 9,690 9,644 1.327158270.E+20
URA 2.871000E+12 6,810 6,799 1.327158270.E+20
NEP 4.497100E+12 5,430 5,432 1.327158270.E+20
PLU 5.913000E+12 4,720 4,738 1.327158270.E+20

Peter Riedt


I see a few problems with this analysis.

1) The distance r is given to four or five significant digits, the
velocity v likewise. The quantity r*v^2 should therefore be given to no
more than five significant digits.

2) What is the adjustment for eccentricity?

3) What about perturbations from the other planets? Jupiter's mass is
about 1/1000 that of the Sun. For Uranus, Neptune and Pluto it almost
makes sense to add Jupiter's mass to the Sun's mass when working out
their r*v^2. (Certainly the inner four planets can be treated that way
under some circumstances.)

4) We can measure the product G*Msun quite well from radio tracking of
spacecraft. The most recent value (JPL planetary ephemeris DE425) is
1.3271244004E20 m^3/s^2, good to at least 10 digits. This differs from
the value above in the 6th significant digit.

5) Even though we know G*Msun very well, we don't know the sun's mass at
all well, because we don't know the value of G to more than four digits.
CODATA's currently recommended value is 6.67384E-11 +/- 0.00080E-11,
with a relative accuracy of 1.2E-4.

-- Bill Owen
  #3  
Old November 6th 12, 06:49 PM posted to sci.astro
Dr J R Stockton[_185_]
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Posts: 5
Default Calculating the EXACT mass of the sun

In sci.astro message 36f6dcb8-4f4e-4071-a649-3e61eaed1d32@googlegroups.
com, Sun, 4 Nov 2012 16:54:51, Peter Riedt posted:

Calculating the EXACT mass of the sun


Before you can calculate the EXACT mass of the Sun, you first need to
define what that actually *means*. In the Solar System, only bodies
smaller than the Moon can have a definite boundary.

You need to include the date as well, since the Sun loses around 1500
million tonnes per day in photons alone. It also gains occasional
comets.

You also need to consider the traceability of the mass standards used to
the BIPM kilogram.

The meaning of "exact" in metrology is definitely not of itself exact.

--
(c) John Stockton, nr London, UK. Mail via homepage. Turnpike v6.05 MIME.
Web http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms and links;
Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.
No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.
  #4  
Old November 10th 12, 02:32 PM posted to sci.astro
Peter Riedt
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Posts: 83
Default Calculating the EXACT mass of the sun

On Wednesday, November 7, 2012 1:49:27 AM UTC+8, Dr J R Stockton wrote:
In sci.astro message 36f6dcb8-4f4e-4071-a649-3e61eaed1d32@googlegroups.

com, Sun, 4 Nov 2012 16:54:51, Peter Riedt posted:



Calculating the EXACT mass of the sun




Before you can calculate the EXACT mass of the Sun, you first need to

define what that actually *means*. In the Solar System, only bodies

smaller than the Moon can have a definite boundary.



You need to include the date as well, since the Sun loses around 1500

million tonnes per day in photons alone. It also gains occasional

comets.



You also need to consider the traceability of the mass standards used to

the BIPM kilogram.



The meaning of "exact" in metrology is definitely not of itself exact.



--

(c) John Stockton, nr London, UK. Mail via homepage. Turnpike v6.05 MIME.

Web http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms and links;

Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc.

No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News.


Dr Stockton, I agree with your comments. However, I believe my calculations
confirmed two facts:

1. The value of G is certain.
2. Applying r and the adjusted v of nine different planets and using
a variant of Kepler's third planetary law produced one identical result
for GM.

  #5  
Old November 10th 12, 02:44 PM posted to sci.astro
Peter Riedt
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Posts: 83
Default Calculating the EXACT mass of the sun

On Tuesday, November 6, 2012 2:03:14 AM UTC+8, Bill Owen wrote:
Peter Riedt wrote:

Calculating the EXACT mass of the sun




The relationship r*v^2 determines the mass of the sun by the distance of a planet from the sun (r in m) and the square of its speed (v in m/sec) exactly if the known velocity is adjusted for eccentric orbits. The adjusted velocities for nine planets are shown in the following table. All produce the standard value of GM with a deviation of 0.000000000000000000%. GM (1.327158270.E+20) is calculated from the Codata values G = 6.674E-11 and M = 1.988550E+30.


r/m v/msec adjusted v r*v^2=GM


MER 5.791000E+10 47,870 47,872 1.327158270.E+20


VEN 1.082100E+11 35,020 35,021 1.327158270.E+20


EAR 1.496000E+11 29,780 29,785 1.327158270.E+20


MAR 2.279200E+11 24,130 24,131 1.327158270.E+20


JUP 7.785700E+11 13,070 13,056 1.327158270.E+20


SAT 1.427000E+12 9,690 9,644 1.327158270.E+20


URA 2.871000E+12 6,810 6,799 1.327158270.E+20


NEP 4.497100E+12 5,430 5,432 1.327158270.E+20


PLU 5.913000E+12 4,720 4,738 1.327158270.E+20




Peter Riedt






I see a few problems with this analysis.



1) The distance r is given to four or five significant digits, the

velocity v likewise. The quantity r*v^2 should therefore be given to no

more than five significant digits.



2) What is the adjustment for eccentricity?



3) What about perturbations from the other planets? Jupiter's mass is

about 1/1000 that of the Sun. For Uranus, Neptune and Pluto it almost

makes sense to add Jupiter's mass to the Sun's mass when working out

their r*v^2. (Certainly the inner four planets can be treated that way

under some circumstances.)



4) We can measure the product G*Msun quite well from radio tracking of

spacecraft. The most recent value (JPL planetary ephemeris DE425) is

1.3271244004E20 m^3/s^2, good to at least 10 digits. This differs from

the value above in the 6th significant digit.



5) Even though we know G*Msun very well, we don't know the sun's mass at

all well, because we don't know the value of G to more than four digits.

CODATA's currently recommended value is 6.67384E-11 +/- 0.00080E-11,

with a relative accuracy of 1.2E-4.



-- Bill Owen


Bill, it is not an analysis; it is a calculation using a variant of Kepler's third planetary law. The remarkable thing about it is that for nine planets each result is identical in terms of GM and therefore the values of G and M are validated regardless of the number of significant digits. The adjustments to v are minimal, they range from 1m/sec to 46m/sec. They were necessary because rv^2is for circular orbits and all planets are ellipses due to perturbations. I will never doubt again that G represents a true constant. G will be part of my theory of the cause of gravity. Kepler and Cavendish were geniuses.

Peter Riedt
 




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