Why gauss's law of gravity has negative sign?

Question

Why Gauss's law of gravitational field have negative sign but gauss's law of electric field is positive sign?

$$\nabla\cdot{\bf{g}}=-4\pi \rho G$$ $$\nabla\cdot{\bf{E}}=\frac{\rho}{\epsilon_0}$$

Answer 1

It's the same sign that differs between Newton's universal law of gravitation, on one side,

$\displaystyle \vec{F}_g = -G \frac{m_1 m_2}{r^2} \vec{r}_{12}$,

and Coulomb's law of electrostatics, on the other side,

$\displaystyle \vec{F}_q = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \vec{r}_{12}$.

Answer 2

It's because the gravitational force between two objects of positive mass is in the direction from one object toward the other, while the electrical force between two objects of positive charge is in the opposite direction, pushing the objects apart.

Answer 3

Two objects with the "gravitational charge" (let it be mass) of the same sign attract each other, but two electric charges of the same sign push each other.
That means then when we write some vector laws we have to use different signs for it.

Let $\vec{r}_{12}$ be the vector from the first point to the second. Then gravity force has the opposite direction (it tries to make the distance between the objects smaller) so we write $\displaystyle \vec{F}_g \propto-\frac{q_{g1} q_{g2}}{r^2} \vec{r}_{12}$, where $q_g$ is just the mass.
And electric force does the opposite thing: if $q_{e1} q_{e2}>0$ it pushes, if $q_{e1} q_{e2}<0$ it pulls. So we write $\displaystyle \vec{F}_g \propto\frac{q_{e1} q_{e2}}{r^2} \vec{r}_{12}$.