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# Advanced Math/Kilos to grams

Question
Hi,

All SOLARMath equations work with Kilograms ( http://members.shaw.ca/warmbeach/INDEX3.htm )

I don't want you to agree with SOLARMath, but please could you just tell me if multiplying this final value by 1,000 is correct.

Mass of earth in Kilograms  =  5537831004648121688015772.9772114697
Mass of earth in grams        =  5537831004648121688015772977.2114697

Note:  All SOLARMath equations are Kilograms.  Hence, because Avogadro's number is atoms / mole which leads to atoms / gram, the initial Mass of the earth must begin with grams.

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

5537831004648121688015772977.2114697   /   (9.8)^4   =   600392689687827221617859.5082045622325038

600392689687827221617859.5082045622325038   x   .001003207246557   =   602318297074676474879.204

That final value must be multiplied by 1,000  due to the initial conversion of kilograms to grams, thus giving Avogadro's complete number;   602 318 297 074 676 474 879 204.

Thanks,
Bri
;0)

Kilograms (kg) to grams (g) is an easy conversion. We have by definition, 1 kg = 1000 g = 10^3 g.

Using scientific notation on the number you provided, the mass of the earth in kilograms is

5537831004648121688015772.9772114697 ≈ 5.537x10^24 kg.

Converting this to gram just mean multiplying by 1000 which is equivalent to adding 3 to the exponent of 10, so that

Mass of earth in grams  ≈ 5.537x10^27 g.

This conversion is equivalent to moving the decimal point over 3 positions, so it looks like your first expression is correct.

5537831004648121688015772.9772114697 kg = 5537831004648121688015772977.2114697 g.

The Avogodro's Number calculation also looks correct since the mass of the earth comes in as a proportionality constant (it doesn't come in as squared or 1/mass or any other more complicated way). This means you can just multiply by 1000 in the final answer to go from kg to g, as you say.

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#### randy patton

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