# What is the zero of potential?

A satellite of mass $m$ is travelling at a speed $v_0$ in a circular orbit of radius $r_0$ around a fixed mass at $O$. Taking the zero of potential at $r=\infty$, show that the total energy of the satellite is $\frac{-mv^2_0}{2}$.

I'm having trouble understanding the question. What does the zero of potential mean?

• the point at which the value of the potential is zero. In this example, at infinity. – Wolphram jonny Nov 30 '14 at 14:07
• @wolprhramjonny that should be an answer. – David Z Nov 30 '14 at 14:20
• George, you're asking a good question, but what did you do to try to figure out the answer yourself? – David Z Nov 30 '14 at 14:20
• @DavidZ Thanks! but I am sure it will be downvoted by people saying that it should be a comment (it happened to me in the past with longer answers) – Wolphram jonny Nov 30 '14 at 14:28
• @wolprhramjonny I didn't mean that as a compliment, I meant that you made your post in the wrong place. It shouldn't be a comment, because it answers the question (and anyone who tells you that that sort of thing should be a comment is wrong :-P). But it's true that an answer that short is not a very good one. To keep everyone happy, I would say, either expand on that a bit and post it as an answer (doesn't have to be much, a few sentences is often enough to make a decent answer), or don't post it at all. – David Z Nov 30 '14 at 14:42

What they are basically telling is that the gravitational potential at an infinite distance from the fixed mass is zero. The total energy of the satellite would be K.E. + P.E.

Since, $$V(r)=-\int{E(r)}$$ $$V(r)=\frac{-GM}{r}$$ With M being the mass of the fixed object. As r tends to infinity, it is clear that the potential of the system rapidly decreases, so it is assumed to be zero. (extremely negligible.)

The orbital velocity of the object is $v_o$. and therefore its kinetic energy is $\frac{mv_o^2}{2}$. In this type of simple orbit scenario, it is always true that $P.E=-2(K.E)$.

Therefore the total energy is $K.E-2(K.E)=-K.E.$ which is $\frac {-mv_o^2}{2}$

In a general satellite planet system, the satellite travels with an orbital velocity. Since the orbital velocity in this case makes the object travel in a circular orbit, it must have centripetal acceleration.

Therefore, the centripetal acceleration $\frac{mv_o^2}{r}=\frac{GMm}{r^2}$, $v_o=(\frac{GM}{r})^{1/2}$. Substituting in the K.E. equation, you have $K.E=\frac{GMm}{2r}$ and the gravitational potential energy at the point, by definition is $\frac{-GMm}{r}$ and hence P.E.=-2(K.E)

The potential $V(r)=\int \frac{F(r)}{m} dr$ as integral of gravitational force $F(r)$ divided by mass $m$ is only determined up to an arbitrary constant.

A satellite coming from infinity will get kinetic energy from falling to the gravity center. So it is a convenient convention to say the potential $V(r) = 0$ and therefore also the potential energy $E_{p}=mV(r)= 0$ at $r\to\infty$.

Chosen this way

• the kinetic energy of a mass falling from $r=\infty$ (with zero speed) is $E_k=-E_p=-mV(r)$ and
• a satellite is bound to a center of gravity, if its total energy (kinetic + potential) is below 0.