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Gravitational potential energy from ∞ to r

  1. May 12, 2013 #1
    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. jcsd
  3. May 12, 2013 #2
    You suspicion is correct. In fact, many people get confused about this.

    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.
     
  4. May 12, 2013 #3
    I understand now, thank you very much voko!
     
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