Submitted by LoreCriticizer t3_z53aqr in askscience
[deleted] t1_ixwaqzr wrote
Reply to comment by Humble_Indication798 in How would one calculate the gravity of a planet? by LoreCriticizer
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danceswithtree t1_ixwpoei wrote
I don't think that means what you think it means. The equation above gives only the first three digits for the gravitational constant-- because who wants to see it to the millionth decimal point? You can use as many decimal places as you want for the masses but the result (force in newtons) will only be good to 2 decimal places.
critterfluffy t1_ixwqome wrote
The gravitational constant has proven very difficult to measure. The recommended max significant figure is actually a bit more, at 6. More than what they had but it is still a strong limit to modern measurement. The memory I had of this was wrong on length but my statement still holds. Anything using G will be stuck to 6 digits.
Info: https://en.wikipedia.org/wiki/Gravitational_constant#Value_and_uncertainty
OlympusMons94 t1_ixxukz1 wrote
G and mass M can be measured with less precision individually, but the product G*M = mu, the standard gravitational parameter, can be measured with higher precision, and is more useful anyway, since most applications require or can use GM, not (necessarily) G and/or M separately.
But unless the distance from the mass is great enough to treat it as a single point mass, then well below even 6 digits of precision, the actual force of gravity at a given location cannot be treated as simply GM/r^2 . So a more precise value of G, M, or even GM, will not, by themselves, give a more accurate answer. Planets and other "spherical" celestial bodies are not perfectly spherical, and have topography, and an internal mass distribution that is not radially symmetric. Depending on the location and reference frame, you also have to consider centrifugal acceleration.
Because of the equatorial bulge and centrifugal acceleration, the effective force felt on Earth's surface is on average 1% lower at Earth's equator than at the poles. The non-uniform gravity field of Earth, and even more so the Moon, must be considered for spacecraft and artificial satellites orbiting these bodies.
danceswithtree t1_ixwrdpp wrote
So what does your statement even mean then? Your answer is only good to however many decimal places your least precise measurement/constant is. But what do you mean that mass is only good to three decimal places? You can have as accurate a mass as you want but the answer, a force in Newtons, will only be good to 3 or 6 or however many decimal places in your least precise constant or measurement.
critterfluffy t1_ixwrlcy wrote
The most accurate that the gravitational constant has been measured is 6 digits. After that the measurements drift. So anything being done using G will have a result no greater than 6 significant digits.
danceswithtree t1_ixwrytd wrote
This response has drifted at least 6 significant digits from your first response.
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