Gravitational Potential Derivation

Question

The definition of Gravitational Potential at a point is the work done per unit mass in moving it from infinity to that point. However the work is positive and if you perform the integral you get a positive value as you should. $$W=\int_\infty^r\left(-\frac{GMm}{r^2}\right)dr=\frac{GMm}{r}$$ $$\frac{W}{m}=\frac{GM}{r}=-V$$ $F=-\frac{GMm}{r^2}\hat{r}$ as the unit vector, $r$, is pointing towards the mass $M$. So I know why the Gravitational potential has to be negative so shouldn't the definition be "The magnitude of the gravitational potential at a point is the work done per unit mass in moving it from infinity to that point."

Answer 1

First, the definition should be

The work done per unit mass in moving it slowly from infinity to that point

since we don't want to include kinetic energy here (for example, I could do more work than needed to get the mass to position $r$, so if we don't include slowly here then this definition does not give a unique potential energy).

Second, you are mixing up your signs and/or who/what should be doing the work we consider. You have calculated the work done by gravity. Therefore, you have correctly found that $W_\text{grav}=-\Delta U$, which is why you are getting a positive sign.

To use your definition, you have to consider the work done by you to move the mass from infinity. This force points radially outwards because we want to move the mass in a way that its very slow velocity remains constant, so your integral should be $$\Delta U=W_\text{you}=\int_\infty^r\frac{GMm}{r^2}dr=-\frac{GMm}{r}$$

Since $U(\infty)=0$, we can just say that $\Delta U=U(r)=-\frac{GMm}{r}$, and this is probably what you were expecting.

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Honestly, I prefer the definition of potential energy that just relies on the conservative force, and not this idea of the work an external agent does against this conservative force. $$\mathbf F=-\nabla U$$ $$U(\mathbf b)-U(\mathbf a)=-\int_{\mathbf a}^{\mathbf b}\mathbf F\cdot\text d\mathbf r$$