# Homework Help: Gravitational potential energy from ∞ to r

1. May 12, 2013

### Saix7

Hello, I'm a high school student studying for the AP Physics test tomorrow. I've been having trouble proving the negative value of gravitational potential energy through working out the work integral. I will greatly appreciate any help and clarification.

1. The problem statement, all variables and given/known data

Find the change in gravitational potential energy $\Delta U$ by calculating the work done by the gravitational force to bring a particle of mass $m_2$ from infinity, to a position at a distance $r$ from another particle of mass $m_1$

2. Relevant equations

\begin{align}\Delta U&=-W\\ U_r-U_\infty&=-W_G \\ W_F &=\int\vec{F}\cdot d\vec{r}\\ W_G &=\int^r_\infty \vec{F_G}\cdot d\vec{r} \end{align}

3. The attempt at a solution

\begin{align}\vec{F_G}\cdot d\vec{r} &=-\frac{Gm_1 m_2}{r^2}\hat r\cdot -d\vec r\\ &=\frac{Gm_1 m_2}{r^2} dr\\ W_G &=Gm_1 m_2 \int^r_\infty \frac{dr}{r^2} \\&=Gm_1 m_2 \Big ( \frac1 \infty - \frac1 r\Big )\\ &=-\frac {Gm_1 m_2}{r}\\ U_\infty&=0 \\ U_r&=\frac {GMm}{r}\end{align}

Of course the answer should be negative, since $\vec F_G$ does positive work in moving a particle inwards from an infinite distance. The calculation works out fine when I move a particle from an arbitrary distance $r$ to $\infty$, but not the other way around. I suspect that I may be working out the dot product wrong in the work equation. The way I see it is that both $d\vec r$ and $\vec F_G$ are in the negative radial direction, so their dot product is positive. Is the infinitesimal $d\vec r$ intrinsically always positive even if the displacement is in the negative radial direction?

2. May 12, 2013

### voko

First of all, note that you abused notation: you can't equate $\vec{F}_G \cdot d\vec{r}$ with $\vec{A} \cdot -d\vec{r}$ because you can't have $d\vec{r} = - d\vec{r}$.

Then you ditched $d\vec{r}$ and introduced $dr$, which apparently denotes the magnitude of $d\vec{r}$, which is always positive by definition. But then the subsequent integral does not make sense, because it ranges from a greater bound to a lower one, thus implying $dr$ is negative.

That's how you got the wrong sign.

3. May 12, 2013

### Saix7

I understand now, thank you very much voko!