# Linking Gravitational Field Strength and Gravitational Potential confusion

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?

• The derivative of $-gr$ wrt $r$ is not $-g$, it is $-g - r\frac{\partial g}{\partial r}$
– Styg
Dec 24, 2017 at 14:17
• @Samarth I don't believe so, are you sure? Dec 24, 2017 at 14:24
• Yes, apply the product rule
– Styg
Dec 24, 2017 at 14:26
• @Samarth so how is the equation $g = -\frac{\Delta V}{\Delta R}$ then achieved? Is it as $\Delta R$ approaches 0, the expression is reduced to $-g$? Dec 24, 2017 at 14:31
• I just noticed, your equation $g = -\frac{\partial V}{\partial r}$ is wrong too. Posting an answer in a bit.
– Styg
Dec 24, 2017 at 15:31

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: $$E=\int{}{}{F}dr ==> F=\frac{dE}{dr}$$

• so both methods are valid? Dec 24, 2017 at 14:55
• Yes, for the gravitation case. Dec 24, 2017 at 15:01

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 $k$.

• Agree with the corrections but the problem is stated in scalars and I guess should be answered accordingly. Dec 24, 2017 at 15:45
• If the problem stated in scalars you're talking about is whether it is okay to use either differentiating or dividing by $r$ (with the correct signs) to find $g$, then yes, both give you the same answer. If the problem stated in scalars is your "derivation" of $g = -\frac{\partial V}{\partial r}$ from $-gr = V$, I have already talked about how it is wrong in the comments below your question and in my answer.
– Styg
Dec 24, 2017 at 15:58