Linking Gravitational Field Strength and Gravitational Potential confusion

Question

I want to ask a question about linking gravitational field strength g and gravitational potential V.

If I consider both quantities at a distance r from the centre of planet of mass M then I can see that they are closely related:

if $g = \frac{GM}{r^2}$ and $V = -\frac{GM}{r}$ then at a given point,

$$g = -\frac{-GM}{r}\times\frac{1}{r} = -\frac{V}{r}$$

so $$g = -\frac{V}{r}$$

Now, the book I am reading also states that I can link these variables through the graphs.

The rate of increase of the gravitational potential with distance r is known as the potential gradient, $\frac{\Delta V}{\Delta r}$ and that this value is equal to g, the gravitational field strength at that distance.

I see this is possible because using the equation we derived previously:

$$g = -\frac{V}{r}$$ $$-gr = V$$ $$-g = \frac{\Delta V}{\Delta R}$$ $$g = -\frac{\Delta V}{\Delta R}$$

but my question is this: using the graph, why can I not just take a point on the line and use the formula $g = -\frac{V}{r}$ instead of differentiation using $g = -\frac{\Delta V}{\Delta R}$ to find the value of g at a distance r?

Answer 1

In this case of gravitation force, you can indeed use $g=-\frac{V}{r}$ as well as differentiate V(r): $$ \frac{dV(r)}{dr}=-GM\frac{d}{dr}(\frac{1}{r})=GM(\frac{1}{r^2})=g $$ I guess the book is trying to establish the more general relationship between Force (F) and Energy (E). Broadly stating: $$ E=\int{}{}{F}dr ==> F=\frac{dE}{dr} $$

Answer 2

Post the discussion in comments, here is my answer:

It is not correct to say that $g = -\frac{\partial V}{\partial r}$, where $g$ is the magnitude of the gravitational field. The full statement for a field derived from a potential is

$$ \boldsymbol g = -\nabla V $$

The gravitational field of a point mass is $$ \boldsymbol g = -\frac{GM}{r^2}\hat r $$ for which $$ \begin{gather} V = -\frac{GM}{r} \\ -\nabla V = -\frac{\partial V}{\partial r}\hat r = \frac{V}{r}\hat r \end{gather} $$

Taking magnitudes, we see that $$ \begin{gather} g = \frac{\partial V}{\partial r} = -\frac{V}{r} \\ V = -gr \\ \end{gather} $$

Differentiating,

$$ \frac{\partial V}{\partial r} = -\frac{\partial (gr)}{\partial r} = -g - r\frac{\partial g}{\partial r} = -g + 2g = g $$

which is consistent.

It is important to note that $$ \frac{\partial V}{\partial r} = -\frac{V}{r} $$ is true only if $$ V = \frac{k}{r} $$ for some constant $k$.