# Relationship between gravitational potential and gravitational field strength

So I was trying to think of a reasonable relationship between gravitational potential and gravitational field strength. However, I'm not sure whether this is correct:

$$g=\frac{GM}{r^2}$$ where $$g$$ is the gravitational field strength at a point which is $$r$$ meters away from a point mass.

$$\phi = -\frac{GM}{r}$$ where $$\phi$$ is the gravitational potential at a point which is also $$r$$ meters away from a point mass.

Now, since the gravitational potential is defined as:

Gravitational potential at a point in a gravitational field is defined as the work done per unit mass in bringing a small test mass from infinity to the point.

So therefore:

$$\phi=\frac{GM}{r^2}\times(\infty-r)$$
$$\phi = \infty - \frac{GMr}{r^2}$$
$$\phi = \infty - \frac{GM}{r}$$

Here, I noticed that this is not equal to the formula $$\phi = -\frac{GM}{r}$$.

Can someone please explain to me why this is the case? Or is this way of deriving the formula for gravitational potential completely wrong?

• Completely wrong. The correct formula is $\phi = \int g(r) dr$, with $\phi(\infty)=0$! – Kostas Nov 21 '19 at 10:45

The potential at a point is derived with calculus,you cannot multiply $$\frac{GM}{r^2}$$ with the difference in positions because the gravitational field varies with distance from centre of mass of the body.Change in potential energy is defined as negative of the work done by conservative force so $$-F_{conservative} \Delta x = \Delta U$$$$F_{conservative}=- \frac{dU}{dx}$$(for infinitesimal changes).You can verify this as$$\frac{d}{dr}(\frac{-GM}{r})= \frac{GM}{r^2}$$