Derivation of the gravitation potential energy and gravitation potential

Question

I have some slight confusing in deriving the gravitation potential energy. In the image below, it explains that the gravitation potential energy is equal to the work done from infinity to a distance r between two masses. And that the work done here will be stored as gravitational potential energy at point r.

How is this possible? I would have imagined that the gravitational potential energy at r would be the summation of work done from the closest distance of two masses to r. Only this energy can be converted equivalently to kinetic energy... Taking the summation of work done from infinity to r would not be equivalent to the converted energy at r into kinetic energy.

Any ideas anyone?

Answer 1

Only changes in potential energy have physical significance, the absolute values that we assign to potential energies are a matter of convention – as to where in space we choose to have zero potential energy.

For example, close to the Earth, where the gravitational field is almost uniform, we may use the formula: PE=$mgh$. We could choose $h=0$ to be at the local ground level, but if we had an underground laboratory we might put $h=0$ at laboratory floor level ��� and so on.

You are concerned with the field, both close and distant, due to a spherically symmetrical body, of radius $R$, say. The work that must be supplied by an external agency to move $m$ at constant speed from $r$ to $r+dr$ is $(GMm/r^2)dr.$ If we choose to put our zero of PE on the body's surface, then the PE at distance $r_1$ from the body's centre will be $$\text{PE}=\int_R^{r_1} \frac{GMm}{r^2} dr=GMm\left(\frac 1R - \frac 1{r_1}\right)$$ For someone based on the body's surface this does have the merit of being positive: the further we take $m$ from the body (mass $M$) the more positive the PE. And putting $r_1=R+h$ we find that for $h\ll R$, $\text{PE}= mgh$ in which $g=GM/R^2$.

For an astronomical body, we don't usually think of ourselves as being on its surface, and it's more convenient to take the zero of PE as at infinity, giving the simple formula

$$\text{PE}=\int_\infty^{r_1} \frac{GMm}{r^2} dr=-GMm\frac 1{r_1}.$$ This means that the PE as we come closer to $M$ becomes increasingly negative, rather than becoming less positive, but both formulae above give the same value for the change in PE as we take $m$ from $r_1=a$ to $r_1=b$.

What about choosing to put the zero of PE at the centre of $M$? If $M$ is a homogeneous sphere, the PE of $m$ at $r=r_1$, for $r_1>R$ is
$$\text{PE}=\int_0^{R} GM\frac{r^3}{R^3}m\frac 1{r^2} dr+\int_R^{r_1} \frac{GMm}{r^2} dr=GMm\left(\frac 3{2R} - \frac 1{r_1}\right)$$ This somewhat messy formula seems to have no advantages – especially since a homogeneous sphere is a poor model for any large astronomical body. If $M$ is a point mass, that is we consider $R$ approaching zero, the formula becomes unusable.

Answer 2

It would seem natural to start at zero distance between the objects and going from there to $r$. In most cases, as in yours, the objects are treated as point masses and the distance $r$ is measured to be the separation between those points of masses. So if you would want to start at the closest separation between them, that would be at $r=0$. In other words, you would be attempting to measure potential energy of two objects that are located at the same point which is not feasible to do.

Instead, we choose to start out at $r= \infty$ due to inverse relation between potential energy and $r$. So for a practical purpose, $V=0$ at $r=\infty$. Then shifting the object to some finite distance $r$ would give the potential $V=-\frac{GMm}{r}$ at the distance $r$ from the other mass.

In the same way, choosing to go from $r=\infty$ to $r=r_1$ and $r=\infty$ to $=r_2$ and taking their difference would give the $\Delta V = -GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$