# Newton's law of gravitation

1. Apr 9, 2015

### Calpalned

1. The problem statement, all variables and given/known
according
to my calculus III textbook, the gravitational force acting on an object at $x = <x,y,z>$ is $F(x)=-\frac{mMG}{|x|^3} x$. What's the point of having a cube on the bottom. Why shouldn't I memorize it as $F(x)=-\frac{mMG}{|x|^2}$

2. Relevant equations
See above

3. The attempt at a solution

2. Apr 9, 2015

3. Apr 9, 2015

### Staff: Mentor

x is a vector. $G\frac{M m}{|\mathbf{x}|^2}$ is the magnitude of the force, while $\frac{\mathbf{x}}{|\mathbf{x}|}$ is a unit vector in the direction of the force.