can you tell me another formula which involves G?you cannot define G on the back of g,you big savants...
and please don`t babble formulas imaginated by other `big savants`...
then G is used to calculate g on other planets...based on what?
on the ratio earth/water meant by cavendish?
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the gravitational force between two masses is proportional to the product of their masses divided by the square of the distance between their centres of mass.
G is simply the constant of proportionality. It has units of N m^2 / kg^2
so that the result of F = G m1m2/r^2 comes out in Newtons.
I see no g!
Perhaps you are thinking Newons are defined in terms of g. They are not. (see link 1)
Perhaps you are thinking kilograms are defined in terms of g. They are not. (see link 2)
I see no g!
E
The value of G is not calculated based on g. And even if it were, what's the problem? g is an experimental value, there's nothing wrong with calculating things from it.
The value of G was first measured by measuring the force between lead spheres. No "g" involved. Once G was known, then g was used to calculate the mass of the Earth. It's not as if we had some other way to know the mass of the Earth, and then used g to calculate G.
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Edit: I'm starting to doubt you are serious, but G can be used to calculate g on other planets based on the mass of those planets (which can be determined from G, and the orbital parameters of any natural or artificial satellite the planet has) and their planetary radius as measured through a telescope.
By Cavendish's work on the ratio of earth to water, I assume you mean his determination of the specific gravity of the Earth, or the ratio of the Earth's overall density to that of water. Because the volume of the Earth was known, his determination of the specific gravity is equivalent to a determination of the mass of the Earth. His work with the lead spheres was, to him, just a step along the way to find the specific gravity of the Earth. But to us, it was all that was needed for a determination of G, independent of g.
As far as babbling formulas is concerned, you do realize you asked us to tell you another formula involving G, right?
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Later Edit (for dawgdays): g isn't found from G by using
F/m = g = GM/r²
g is measured with falling objects, pendula, etc. When you use g = GM/r², you are calculating M, the mass of the Earth. g and r are measured directly, G is measured with experiments similar to Cavendish's, leaving M to be calculated.
dawgdays
There may be a problem of notation. "g" is used for the unit "gram". "g" is also used for the acceleration caused by gravity at the surface of the Earth, which can be written in terms of "G", the universal gravitational constant.
"G", the universal gravitional constant is not derived from "g". It isn't derived at all. It is measured. It is a conversion factor. It is a property of the behavior of two masses upon each other in Newtonian gravitation.
"g" for the Earth is actually derived using "G". g = G*M/R^2, where M is the mass of the Earth, and R is the radius of the Earth. In this instance, "g" is shorthand for part of the universal gravitational formula. Because of this, each planet could have a value of "g" that was used as a shorthand representation of the expression, g(planet) = GM(planet)/R(planet)^2, where M(planet) and R(planet) are the mass and radius of that planet, respectively.
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Newton's gravitational constant (G) is just that, constant, the acceleration due to gravity (g) varies with mass and density. Thereby if mentioned in the same sentence they are not the same thing, this is known through dimensional analysis. Big G's units are N (m/s)^2, whereas little g's units are m/s^2. This disproves your qualm of G being defined by g as a tautology. Besides no where in any definition of either do they reference the other.