Show the gravitional field is conservative

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
Physics2341313
Messages
53
Reaction score
0
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:

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

Why, when defining the scalar function f is the [itex]\sqrt{x^2 + y^2 + z^2}[/itex] used?
 
Physics news on Phys.org
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.