# Show the gravitional field is conservative

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1. Nov 20, 2015

### Physics2341313

In my calculus textbook (section on vector calc) it is showing that the gravitational field is conservative. I followed fine except for the first part, defining the scalar function f.

Showing the field is conservative went something like this:

$f(x,y,z) = MM'G/\sqrt{x^2+y^2+z^2}$
$\nabla{f(x,y,z)} = \partial{f}/\partial{x}\hat{i}+\partial{f}/\partial{y}\hat{j}+\partial{f}/\partial{z}\hat{k}$
$= -MM'G/(x^2+y^2+z^2)^{3/2}\hat{i} + -MM'G/(x^2+y^2+z^2)^{3/2}\hat{j} + -MM'G/(x^2+y^2+z^2)^{3/2}\hat{k}$
= F(x, y, z)

Why, when defining the scalar function f is the $\sqrt{x^2 + y^2 + z^2}$ used?

2. Nov 20, 2015

### Staff: Mentor

It is the distance in cartesian coordinates.
Why the inverse distance and not its square or something else? Well, the inverse distance leads to the right force, other definitions do not.