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Astrophysics/Is twice correct ?


QUESTION: Mass of earth in Kilograms = 5537831004648121688015772.9772114697
Mass of earth in Grams     = 5537831004648121688015772977.2114697

Our SOLARMath equations work with "Kilograms" ( )

But because Avogadro's number is atoms per mole which leads to atoms per "gram", prior to commencing the equation below, we need to convert the initial Mass of the earth to grams.

The Equation is;

The Mass of the earth divided by Gravity^4  x  Acceleration of earth in orbit  =  Avogadro's number.

5537831004648121688015772977.2114697 grams  /  (9.8)^4  =  600392689687827221617859.5082045622325038

600392689687827221617859.5082045622325038  x  .001003207246557  =  602318297074676474879.204

Now our question is;

Should this (above) "final value" be multiplied by 1,000 ?  Does the whole equation need a second x 1,000 in order to compensate for the very first conversion of kilograms to grams ?  It seems counter intuitive.  We figured the equation would need a "divided by 1,000" to compensated for the initial ( x 1,000 ) that was used to go from Kilograms to Grams.

When that final x 1,000 is done, the result gives us Avogadro's complete number = 602 318 297 074 676 474 879 204.

So that final x 1,000 gives us a correct value.

But ?  Is that final x 1,000 proper mathematics ?


ANSWER: Hi Brianna,

The link you gave me ( and the associated is pure rubbish. It is NOT science.

As a small example, just look at those 26 "equations". It appears that only one (#2) makes sense from a "units" standpoint (but not from a physical standpoint). In equation #1, you CANNOT get "speed of light" from "velocity squared". That's the first thing that a physicist learns. And one CANNOT measure the mass of the earth as 5537831004648121688015772977.2114697 grams - that's 36 SIGNIFICANT DIGITS!! That's impossible!

Please, Brianna, if you want to learn science, go to real legitimate sites - like or or There's lots more. But the site you're on is NOT scientific. It is rubbish.

By the way, the question you asked is meaningless, because the equation    
"The Mass of the earth divided by Gravity^4  x  Acceleration of earth in orbit = Avogadro's number" is meaningless. Sorry.

Prof. James Gort    

---------- FOLLOW-UP ----------


You say that this is just a . . . "coincidence" ?

According to Wikipedia, the recorded magnetic field of the earth is  .00003 Teslas   ( Wikipedia -  3.110−5 Teslas is the strength of earth's Magnetic Field on the equator, 0 latitude. )

From we see that the Tesla (Magnetic Field) is equal to  1/AZ  which is   1 / orbit Velocity.

The earth is moving through space at 30,689.1802374659 m/s.  When you divide 1 by that velocity (1/AZ) you get the exact

Magnetic Field of the earth  =  .00003 Teslas

Coincidence ? ? ?


Hi Bri,

"the Tesla (Magnetic Field) is equal to  1/AZ  which is   1 / orbit Velocity" is wrong. So if you get the "correct" magnetic field of the earth from an incorrect equation, then, yes, it's coincidence. Nothing more.

Here's a test - try it with Venus. Or Mars. It should work with any planet - correct?

Prof. James Gort  


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James Gort


Questions on observational astronomy, optics, and astrophysics. Specializing in the evolution of stars, variable stars, supernovae, neuton stars/pulsars, black holes, quasars, and cosmology.


I was a professional astronomer (University of Texas, McDonald Observatory), lecturer at the Adler Planetarium, professor of astrophysics, and amateur astronomer for 42 years. I have made numerous telescopes, and I am currently building one of the largest private observatories in Canada.

StarDate, University of Texas, numerous Journal Publications

B.A. Physics and Astronomy M.Sc. Physics Ph.D. Astrophysics

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