# Gauss's Law vs Newton's Law

This is thought experiment. I couldn't get a good answer because I keep getting negative mass.

Gauss's Law say that eletric field is proportional to charge, how much charged is enclosed. Newton's gravitational pretty much says the same thing. More mass, strong gravitational strength.

So to put it mathematically

$$\vec{E} \propto \sum Q_{en}$$

But charge can be negative, a negative sum of charge means the electric field is going inwards.

$$\vec{g} \propto \sum M_{en}$$

But mass can only be positive, but g is then propotional to the mass enclosed, which means it will go radially outwards. That doesn't make sense because we know gravity is radially inward for a spherical shape (like earth)

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Constants of proportionality can be negative... – dmckee Jan 1 '12 at 22:02

In the equations as you've written them, the constant of proportionality is an outward-pointing vector for the electric field and an inward-pointing vector for the gravitational field. Or in other words, if you take the radial component only: it's a positive constant for the electric force and a negative constant for the gravitational force.

The details: Gauss's law for electrostatics actually says

$$\iint\vec{E}\cdot\mathrm{d}\vec{A} = \frac{Q_\text{enc}}{\epsilon_0}$$

and for Newtonian gravity, you can write

$$\iint\vec{g}\cdot\mathrm{d}\vec{A} = -4\pi G M_\text{enc}$$

For a spherically symmetric surface and mass/charge distribution, letting $\hat{n}$ represent the outward-pointing normal vector at each point on the surface, these simplify to

$$\vec{E} = \frac{Q_\text{enc}}{4\pi\epsilon_0 r^2}\hat{n}$$

and

$$\vec{g} = -\frac{GM_\text{enc}}{r^2}\hat{n}$$

Note that the constant of proportionality in the first case is $\hat{n}/4\pi\epsilon_0 r^2$, which points outward, and in the second case is $-G\hat{n}/r^2$, which points inwards.

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Nice answer, but you should be drinking for new years. Get off stackexchange! – Colin K Jan 1 '12 at 5:30
For all you know I could be drinking now ;-) (Not that I am, but that's beside the point) – David Z Jan 1 '12 at 5:33
So does that mean that it should actually be $\vec{g} \propto -\sum M_{en}$? I seriously thought that proportional symbol was alpha. Thanks! – Hawk Jan 1 '12 at 19:13
You could write $\vec{g}\propto -\sum M_{en}$ or $\vec{g}\propto\sum M_{en}$. They mean the same thing, same as $\vec{g}\propto 2\sum M_{en}$ or $\vec{g}\propto -5\sum M_{en}$ or so on. The proportionality relation $\propto$ says nothing about whether the constant is positive or negative. – David Z Jan 1 '12 at 22:18