How does gravitational potential relate to gravitational acceleration?

I've been wondering about what the gravitational potential V actually tells us, and how it relates to gravitational force and/or acceleration. The formula is $$V = -\frac{GM}{r}$$. I did some calculations for V at the surface of the Earth, and got $$-6.256 * 10^7m^2s^{-2}$$. At the surface of the Earth, $$g = 9.82ms^{-2}$$

I then took that same value for V but changed M to be the mass of the sun, and solved for r. I got $$r=2.122*10^{12}m$$. Naively, I used those numbers to compare the gravitational force at those radii, but the numbers did not match.

What is the relationship between gravitational potential and gravitational acceleration? I read somewhere that $$g = - \nabla V$$,but why does the same value for V give different g?

As you wrote, $$\vec{g}=\vec{F}/m=-\nabla V$$ $$\vec{g} = -(-GM)\frac{\partial (1/r)}{\partial r}\hat{r}$$ $$\vec{g}= -\frac{GM}{r^2}\hat{r}$$ This equation is only valid if all the mass is concentrated at a point, or if you are outside a sphere of (angularly-uniform density) mass.

• Regarding "why does the same value for V give different g" ... This is the correct relationship between the scalar gravitational potential and the gravitational field $\vec{g}$ around a massive object $M$. You see that it does not depend on the value of $V(r)$, but only on the derivative (gradient) of $V(\vec{r}$). Commented Nov 24, 2022 at 14:23

Yes $$g=-\nabla V$$ if $$V=\frac{-GM}{r}$$. They are as related as force with potential energy.

Reason is that value of $$r$$ is different for earth and sun. So $$V_e=V_s$$,$$\frac{M}{R}=\frac{m}{r}$$where $$M, R$$ are mass and distance for sun, and $$m, r$$ for earth.

As $$R$$ is big, so square of $$R$$ is much larger for acceleration due to gravity. Thus value of $$g$$ for same potential is,$$\frac{g_s}{g_e}=\frac{r}{R}$$For a unit test mass, forces for sun and earth at same potential is given by,$$\frac{F_s}{F_m}=\frac{r}{R}$$ Thus forces are unequal and smaller for sun compared to earth

Starting from the second Newton's law: $$\mathbf F = m\mathbf a$$. For a test mass $$m$$ outside a spherical celestial body: $$-\frac{GMm\hat {\mathbf r}}{r^2} = -\frac{GMm \mathbf r}{r^3} = m \frac{\mathbf {dv}}{dt}$$

Here we have how to calculate the acceleration at each $$r$$. The meaning and utility of $$V$$ has to be understood after some manipulation of that equation:

Making the dot product at both sides by the infinitesimal displacement vector $$\mathbf {dr}$$ and dividing by $$m$$:

$$-\frac{GM \mathbf {r.dr}}{r^3} = \frac{\mathbf {dv.dr}}{dt} = \mathbf {v.dv} = d(\frac{1}{2}v^2)$$

For the the left side: $$-\frac{GM \mathbf {r.dr}}{r^3} = \nabla \left(\frac{GM}{r}\right)\mathbf{.dr} = d\left(\frac{GM}{r}\right)$$

Calling $$V = -\frac{GM}{r}$$ we get: $$V + \frac{1}{2}v^2 = cte$$

As $$V$$ is only function of $$r$$, and has that amazing relation with $$v^2$$, it allow us to get easily the escape velocity for instance: $$V_0 + \frac{1}{2}v_e^2 = V_{\infty} + 0 = 0$$