The connection between potential energy and force

[NODARman]

Homework Statement

Is force the derivative of potential energy?

Relevant Equations

F=-du/dh=-d/dh(mgh)

Hi, if the force is the derivative of potential energy, does it mean that the force is equal to mg and with a constant gravity, it will be the same at any height?
But in real life, F (or mg) would be different on the Earth's surface and 400 km above it (~8 m/s^2).
So, this formula is used to calculate the gravity field force that is applied to the object. right? Or do I understand differently?

 

[Orodruin]

Gravitational potential being equal to mgh is an approximation that is only valid in a small enough region that the gravitational field is approximately constant. In general, outside a spherically symmetric object, the gravitational potential energy of a test mass is given by
$$
U = -\frac{GmM}{r}
$$
where $G$ is Newton’s gravitational constant, $M$ the mass of the gravitating body, $m$ the test mass, and $r$ the distance to the test mass from the gravitational center.

 

[Delta2]

In more accurate words, and in the language of vector calculus, if $U(x,y,z)$ or $U(r,\theta,\phi)$ is the potential energy of a test "charge" m , inside the field at position $(x,y,z)$ (or at position $(r,\theta,\phi)$), then the force from this field to the test charge is $$\mathbf{F}=-\nabla U$$ , that is the force F is the negative of the gradient of the potential energy of the test charge m at position (x,y,z). When we have one dimension , say x, the gradient reduces to the derivative $dU/dx$. Check the definition of gradient at Wikipedia or elsewhere in the web, as I said in one dimension is the same as the derivative.
https://en.wikipedia.org/wiki/Gradient

What do you mean by "So this formula is used to calculate the gravity field force..." but the gravitational field force is the weight. It is just that the weight, being the gradient of a not necessarily constant function $U$(depends on the approximation we make for the potential energy $U$) is not necessarily constant but it depends on the position of the test "charge" $(x,y,z)$ (or $(r,\theta,\phi)$.