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Humble_Indication798 t1_ixuejf8 wrote

They were probably talking about the net gravitational force felt from the spaceship or some object from multiple planetary bodies.

The equation for force from gravity is (Gm1m2)/r^2 G is a constant 6.67*10^-11 M1 is the planet M2 is the object you are interested in R is the distance BETWEEN THE TWO CENTERS OF THE OBJECTS

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sawkonmaicok t1_ixw2u4m wrote

Actually not the centers of the objects, but their centers of mass. Usually this is around the center but not always since things can have an uneven disteibution of density. Also that formula is only an approximation.

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Kered13 t1_ixy5uc7 wrote

For a planet sized object you can safely assume that the center of mass will effectively be it's geometric center.

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_Jaquen_Hgar_ t1_ixwrspc wrote

The question asked was about the strength of gravity exerted by one planet, so a slightly more apposite formula is the one for the gravitational acceleration felt by an object of any mass in free fall:

g=G.M/r²

i.e. it’s independent of the mass of the other object.

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Humble_Indication798 t1_ixwub3p wrote

Yeah but if you consider force=massacceleration, m2A=gm1m2/r^2 then A=g*m1/r^2 and A is acceleration from gravity which is the same as the formula you posted. Which is independent of the mass of the second object

I assumed the person who posed the question initially could get to this point algebraically.

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lacgibra t1_ixuglp0 wrote

Yeah right, the only change would depends whether spaceship or the object is bound to the system or not ( planet's atmosphere then the distance between them would be radius of the planet ), if it's extra planetary then the r would be distance between them, pretty much the same.

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[deleted] t1_ixwaqzr wrote

[removed]

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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.

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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

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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.

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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.

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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.

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danceswithtree t1_ixwrytd wrote

This response has drifted at least 6 significant digits from your first response.

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