# Missing mass in gravitational force

1. Apr 19, 2013

### PLuz

This is a really dumb question but I can't seem to make sense out of this...

For a conservative force, we have $\vec{F}=-\nabla \phi$, where $\phi$ stands for the potential. So let's take the gravitational potential, given by:
$$\phi=-G_N \frac{M}{r}.$$
Then, by the previous formula: $\vec{F_g}=-G_N \frac{M}{r^2}\hat{e_r}$...but this is the expression for the gravitational field (force per unit mass) not the gravitational force... What am I doing wrong?

Thank you very much.

2. Apr 19, 2013

### WannabeNewton

No, $a = -\nabla \phi$. $F = -\nabla \phi$ holds for a unit test mass. You are confusing the potential $\phi$ with the potential energy $U$ for which it is true that $F = -\nabla U$. Whenever you have doubts like this, just check units. $\phi$ has units of Joules/kg so $\nabla \phi$ has units of Joules/(kg*m) = N*m/(kg*m) = N/kg = m/s^2