Maybe what I'm going to ask sounds stupid but it means a lot to me that's why I'm asking. I've problem with just a single step in derivation.

Let's start. I'm keeping in mind the direction of each vector and that's why I'm using direction of vector.Didn't get readymade pic that's why made it by myself

Now, the force on mass $m$ is: $$\vec{F}=-\frac{GMm}{|\vec{r}_{12}|^2}\vec{1}_{12}$$ Hope you're seeing the negative sign.(Now, the line below is the line I have problem with.I think this is what am I doing i correct but it's not.)Now,to displace $m$ through small displacement $dr$ small work $dw$ will be $$dw=-\frac{GMm}{|\vec{r}_{12}|^2}\vec{1}_{12}\cdot d\vec{r}$$ Since $\vec{1}_{12}$ and $d\vec{r}$ has opposite direction thus dot product yield one more negative sign then $dw$ $$dw=-(-\frac{GMm}{|\vec{r}_{12}|^2}) d\vec{r}$$ It give$$dw=\frac{GMm}{|\vec{r}_{12}|^2} d\vec{r}$$ On solving i.e. on doing integration value of $w$ will be $$dw=-\frac{GMm}{|\vec{r}_{12}|}$$ (But this value is worng isn't it.)Thus potential energy $U$ will be $$U=-w=\frac{GMm}{|\vec{r}_{12}|}$$ But every one know it's wrong.

Help me. I know that's wrong but can't find any thing that proves that it is worng. If I'm doing anything wrong except that (what I've explained) you can explain that. And please help me with statement 2 which I've told earlier. Is that wrong? If it is, why? Can you explain it?


Force exerted by $M$ on $m$ is given by

$$\mathbf F= G\frac{Mm}{r_{12}^2}\, (-\mathbf 1_{12})\;.\tag 1$$

Now, work done it by $\mathbf F$ displacing $m$ infinitesimally from $ r_{12}\mathbf 1_{12}$ to $r_{12}\mathbf 1_{12}+ \mathrm dr_{12}\mathbf{1_{12}}$ is given by

\begin{align}\mathrm dw &=\mathbf F\cdot \mathrm dr_{12}\mathbf{1_{12}}\\ &=G\frac{Mm}{r_{12}^2}\, (-\mathbf 1_{12})\cdot \mathrm dr_{12}\mathbf{1_{12}}\\ &= - G\frac{Mm}{r_{12}^2}\mathrm dr_{12}\;.\tag 2\end{align}

Note in $(2)\,,$ there is only one negative sign; it didn't disappear due to dot product.

You can carry on by integrating with proper limits to get the potential energy function.

  • $\begingroup$ Cmon as I can see in figure dr is in opposite direction to position vector then why have you taken the same as to direction of position vector. $\endgroup$ – Brett Leigh May 28 '16 at 1:39

Actually, all of your derivation is correct. But, potential energy is the energy due to gravitational field with respect to point at infinity.

By your derivation, u at infinity is zero. u at (r,0) is GMm/r . Thus, potential energy at (r,0) is

U(r,0) = u(0,inf) - u(0, r)

       = 0 - GMm/r

       = -GMm/r

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