Wiki says it is the product of G and M, G=6.7E-11, and for Earth, M=~6E24 Kg.
That multiplied together is about 4E14. The thing I don't understand is why they list that as being a bit under 400,000 for Earth. The moon is 4902 and change.
How do they derive that number? It is a billion times lower than just multiplying G and M directly.
Originally posted by sonhouse Wiki says it is the product of G and M, G=6.7E-11, and for Earth, M=~6E24 Kg.
That multiplied together is about 4E14. The thing I don't understand is why they list that as being a bit under 400,000 for Earth. The moon is 4902 and change.
How do they derive that number? It is a billion times lower than just multiplying G and M directly.
Probably a unit conversion problem. Change the units of G to cgs and you get E-8 instead of E-11. Using hours instead of seconds you get 8.650E-1.
The SI units of the standard gravitational parameter are m^3/s^2...The value for the Earth is called the geocentric gravitational constant and equals [~400,000] km^3/s^2
Originally posted by AThousandYoung Probably a unit conversion problem. Change the units of G to cgs and you get E-8 instead of E-11. Using hours instead of seconds you get 8.650E-1.
The SI units of the standard gravitational parameter are m^3/s^2...The value for the Earth is called the geocentric gravitational constant and equals [~400,000] km^3/s^2
6.67 E-11 is the # I always use for G and the mass of Earth is in kilograms, the units are correct. I don't get it.
Originally posted by AThousandYoung Your calculated answer uses meters. Wiki expresses the same answer using kilometers. km^3 is 1 billion times larger than m^3.
If I used Km instead of meters that would only be 3 orders of magnitude difference, a thousand to one. Instead I have a thousand MILLION to one difference.
But you are right, G can be expressed as 6.67 E-8 in Km. That would not change the mass which is in Kg.
Question in physics, "Std. gravitational parameter':
28 Aug '12 18:40
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