Mechanical Engineering/Two times ?
Mass of earth in Kilograms = 5537831004648121688015772.9772114697
Mass of earth in Grams = 5537831004648121688015772977.2114697
Our SOLARMath equations work with "Kilograms" ( http://members.shaw.ca/warmbeach/INDEX3.htm
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 ?
All you need to do is to stack up the units to see if you need to do the last multiplication or division or not, rember to end up with no unit though.
But why do you even need to find avogadro's constant for a singular and unique mass like earth, you can't use that number anywhere else, you can't find a system that contains milions of earths.
in my opinion finding avogadro's number for earth is waist of time.