Deriving gravitational potential energy -- mistake

[jl12]

Homework Statement

Hi I'm attempting to derive the gravitational potential energy of a point mass ($m$) that's moving from infinity to a point r' inside a gravitational field produced by a another mass $M$. For simplicity I treated it as a one dimensional case. The problem I get is that the final answer is out by a factor of $-1$.

Homework Equations

$U=-W$ (from Wikipedia about the relation between potential energy and work for a conservative force.)
$F=GMm/(r^2)$

The Attempt at a Solution

My incorrect derivation is attached. Any help would be very much appreciated.

 

[Orodruin]

The force and the displacement vector with your parametrisation using r do not point in the same direction.

 

[Charles Link]

@jl12 If you write $\vec F=-\frac{GMm}{r^2} \hat{a}_r$ , then $\vec{F} \cdot d \vec{r} =-\frac{GMm}{r^2} \, dr$, because $dr=\hat{a}_r \cdot d \vec{r}$. (Note: $d \vec{r}=dr \,\hat{a}_r+r \, d \theta \, \hat{a}_{\theta}+r \sin{\theta} \, d \phi \, \hat{a}_{\phi}$ in a spherical coordinate system, so that $\hat{a}_r \cdot \, d \vec{r}=dr$).

 

[jl12]

The force and the displacement vector with your parametrisation using r do not point in the same direction.

Why is the dr vector in the opposite direction to if I'm taking it from infinity to r'. Surely $m$ is moving to the left? Btw thanks both for the quick reply both of you.

 

[Charles Link]

Why is the dr vector in the opposite direction to if I'm taking it from infinity to r'. Surely $m$ is moving to the left? Btw thanks both for the quick reply both of you.

This one is a little tricky. One thing to notice is that the integral of $\int \vec{F} \cdot \, d \vec{r}$ needs to be positive because $\vec{F}$ points along the direction of your path. As it turns out, because you are following a path of decreasing $r$ in the limits of your integral, the differential $dr=\Delta r_i$ in your integral $\int F \, dr=\sum\limits_{i} F_i \, \Delta r_i$ becomes negative, giving your integral $\int F \, dr$ a negative result. $\\$ Alternatively, if you want a very mechanical/methodical approach, see post 3 again.

 

[jl12]

This one is a little tricky. One thing to notice is that the integral of $\int \vec{F} \cdot \, d \vec{r}$ needs to be positive because $\vec{F}$ points along the direction of your path. As it turns out, because you are following a path of decreasing $r$ in the limits of your integral, the $dr=\Delta r_i$ in your integral $\int F \, dr=\sum\limits_{i} F_i \, \Delta r_i$ becomes negative, giving your integral a negative result.

So when I integrate it I should replace dr with -dr.

This one is a little tricky. One thing to notice is that the integral of $\int \vec{F} \cdot \, d \vec{r}$ needs to be positive because $\vec{F}$ points along the direction of your path. As it turns out, because you are following a path of decreasing $r$ in the limits of your integral, the differential $dr=\Delta r_i$ in your integral $\int F \, dr=\sum\limits_{i} F_i \, \Delta r_i$ becomes negative, giving your integral $\int F \, dr$ a negative result.

This one is a little tricky. One thing to notice is that the integral of $\int \vec{F} \cdot \, d \vec{r}$ needs to be positive because $\vec{F}$ points along the direction of your path. As it turns out, because you are following a path of decreasing $r$ in the limits of your integral, the differential $dr=\Delta r_i$ in your integral $\int F \, dr=\sum\limits_{i} F_i \, \Delta r_i$ becomes negative, giving your integral $\int F \, dr$ a negative result.

So is the dr always assumed to be positive so when I take the dot product dr is always in the radial direction?

 

[Charles Link]

So when I integrate it I should replace dr with -dr.

That seems to be a very necessary step here. $\\$The alternative is to use simple vectors in one dimension. Instead of calling it $r$, let's call it $x$, and $\vec{F}=-\frac{GMm}{r^2} \hat{a}_x$ for $x>0$. (The force points to the left). Then, you will notice that $\vec{F} \cdot d \vec{x}=-\frac{GMm}{x^2} \, dx$, ( $d \vec{x}=dx \, \hat{a}_x$), and you can get the correct result very mechanically. By this second method, $dx$ is automatically negative in the integral with the way the limits are input, making $\int \vec{F} \cdot d \vec{x}$ positive. (You don't need to put in any artificial minus signs).

 

[jl12]

That seems to be a very necessary step here. $\\$The alternative is to use simple vectors in one dimension. Instead of calling it $r$, let's call it $x$, and $\vec{F}=-\frac{GMm}{r^2} \hat{a}_x$ for $x>0$. (The force points to the left). Then, you will notice that $\vec{F} \cdot d \vec{x}=-\frac{GMm}{x^2} \, dx$, and you can get the correct result very mechanically.

Sorry to keep spamming but just to be clear so when you define dr or dx or whatever as vectors they don't depend on the limits of integration they are just always defined such that they go in the positive (or radial for dr) direction.

 

[Charles Link]

That is correct. See also the addition in post 7 that $d \vec{x}=dx \, \hat{a}_x$.

 

[jl12]

That is correct. See also the addition in post 7 that $d \vec{x}=dx \, \hat{a}_x$.

Oh ok that clears up so much thanks for your support.