# Deriving gravitational potential energy - why is $r_0 = \infty$?

I'm wondering about some assumptions I have to make in deriving the gravitational potential energy. This arises from the following exercise: -

Since the net force acting on the satellite is conservative, $$W=\oint_\mathcal{C}\vec F\cdot\mathrm{d}\vec r=0$$

Although I don't think that bit of information is necessary.

$$W = -GMm \ \int_{r_0}^{r_1} 1/r^2 dr$$

$$W = -GMm \left[-\frac{1}{r}\right]_{r_0}^{r_1}$$

Now, to arrive at the equation I'm looking to express, $r_0$ must equal $\infty$. However, I don't why this must be the case, to derive the potential energy. I know it's conventional to take the reference point of potential energy from an infinitely far away point, which is why $GPE$ is always negative, but it doesn't logically follow in the integral for me to do this, as it seems to imply the object was brought from $\infty$ to $r_1$ which doesn't make sense to me. So why is it that we set $r_0 = \infty$?

• physics.stackexchange.com/q/326797 is in the context of electrostatics rather than gravitation, but the same reasoning applies. Oct 28, 2017 at 16:48

Because the integral you write down is work, which is change in potential energy. If you want the potential at a point $r_1$, you need to calculate the work from the position $r_0$ where $V=0$. You can choose this arbitrarily by shifting the potential with a constant, but the problem specifies to take it at infinity.
In a practical problem that, that is, a problem that asks you to predict the value of some quantity that you can measure, one always deals with differences or changes in the energy of a system. Knowing "the energy" of a particular system at a particular time is only useful if you know something about "the energy" at a later time. $\Delta E$. Or you may be given information about power or work and be asked to find $\Delta E$ (which could be used to answer questions about still-future events).