- #1
Saix7
- 2
- 0
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.
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##
$$\begin{align}\Delta U&=-W\\ U_r-U_\infty&=-W_G \\ W_F &=\int\vec{F}\cdot d\vec{r}\\ W_G &=\int^r_\infty \vec{F_G}\cdot d\vec{r} \end{align}$$
$$\begin{align}\vec{F_G}\cdot d\vec{r} &=-\frac{Gm_1 m_2}{r^2}\hat r\cdot -d\vec r\\ &=\frac{Gm_1 m_2}{r^2} dr\\ W_G &=Gm_1 m_2 \int^r_\infty \frac{dr}{r^2} \\&=Gm_1 m_2 \Big ( \frac1 \infty - \frac1 r\Big )\\ &=-\frac {Gm_1 m_2}{r}\\ U_\infty&=0 \\ U_r&=\frac {GMm}{r}\end{align}$$
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?
Homework Statement
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##
Homework Equations
$$\begin{align}\Delta U&=-W\\ U_r-U_\infty&=-W_G \\ W_F &=\int\vec{F}\cdot d\vec{r}\\ W_G &=\int^r_\infty \vec{F_G}\cdot d\vec{r} \end{align}$$
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?
