Relation Between Gravitational Potential and Acceleration due to gravity

I have always had this question in mind. I know expression for gravitational potential is given as

$$V=-\frac{GM}{r}$$

While acceleration due to gravity is given as

$$g=\frac{GM}{r}$$

So i assume

$$V=-g$$

Is my assumption correct? Even if it is, I am not able to understand this relation intuitively. Initially I thought that this meant that thier (acceleration due to gravity and gravitational potential) directions were different. But Gravitational potential is scalar and cannot have direction. If my assumption is not correct, is there any intuitive relation between $$V$$ and $$g$$?

• Acceleration is proportional to the inverse of the distance squared. V and g take different units (V is measured in $J/kg$, or $m^2/s^2$, while acceleration is measured in $m/s^2$), they cannot become equal by simple negation. Jun 2 at 15:40
• What made you think $g\propto 1/r$ rather than $g\propto 1/r^2$?
– J.G.
Jun 2 at 15:45

$$V \ne -g$$. For one thing, the gravitational potential is a scalar while $$\vec g$$ is a vector field, so they can't be equal. They aren't equal in magnitude either: $$V = -\frac{GM}{r}$$ $$\vec g = -\frac{GM}{r^2}\hat r \ \ \Rightarrow \ \ g = |\vec g| = \frac{GM}{r^2}$$ The relation between $$V$$ and $$\vec g$$ is $$\vec g = -\vec\nabla V,$$ i.e. $$\vec g$$ is the negative gradient of $$V$$.