Derive Newtonian gravity from symmetry?

[Ans]

Is it possible to derive laws of Newtonian gravity from some symmetries?

 

[Meir Achuz]

It can be derived from Gauss's law and spherical symmetry, if that's what you mean.

 

[etotheipi]

It can be derived from Gauss's law and spherical symmetry, if that's what you mean.

Is one more fundamental than the other? You can certainly consider a sphere $S$ centred on a point mass and look at $\iint_S \vec{g} \cdot d\vec{A}= \iint_S -\frac{Gm}{r^2} \hat{r} \cdot d\vec{A} = -4\pi G m$, but can the constant on the right be obtained without referring to Newton's law of gravitation in the first place?

 

[Meir Achuz]

You just have to assume spherical symmetry, including that $\vec g$ is in the radial direction.
Then, $\int\int{\vec g}\cdot{\vec{dA}}=-g\int\int dA=-4\pi R^2=-4\pi Gm$.

 

[etotheipi]

You just have to assume spherical symmetry, including that $\vec g$ is in the radial direction.
Then, $\int\int{\vec g}\cdot{\vec{dA}}=-g\int\int dA=-4\pi R^2=-4\pi Gm$.

But don't we still need to use that $g = \frac{Gm}{r^2}$, Newton's law, in order to get that $-g\iint dA = -4\pi R^2 \frac{Gm}{r^2} = -4\pi G m$? In other words, is there a way to derive that the gravitational flux is $-4\pi G m$ without referring to Newton's law of gravitation?

 

[Meir Achuz]

Why do you say that? Gauss's law is $\int\int{\vec g}\cdot{\vec{dA}}=-4\pi Gm$, with no mention of Newton's law.

 

[etotheipi]

Why do you say that? Gauss's law is $\int\int{\vec g}\cdot{\vec{dA}}=-4\pi Gm$, with no mention of Newton's law.

Fair enough, my point was just that the two seem to be equivalent statements of the same law, and it is not so much deriving one from the other as re-stating it in a different form.