# Defining gravitational potential

I recently came across the definition of gravitational potential where .....

Suppose a particle of mass $m$ is taken from a point $A$ to $B$. Let $U(A)$ and $U(B)$ denote the gravitational potential energy when the mass $m$ is at point $A$ and point $B$ respectively ..Also $V(B) - V(A)$ is the change in potential, then $$V(B) - V(A) =\frac{U(B) - U(A)}{m}$$

My teacher then said that if we take $A$ to be the reference point then $V(A)$ becomes equal to $0$ while I have no problem with this but why can't we also say that $U(A)$ is also equal to zero? After all we are taking $A$ as the origin? But my teacher says that we cant take $U(A)$ as equal to zero. Can anyone please explain why?

My book(Concepts of Physics by HC verma) also says the same thing.

• @Freelander: You can check the edit history yourself by clicking the 'edited' button. – Qmechanic Nov 9 '15 at 19:02

## 3 Answers

Your teacher is wrong. The gravitational potential $V(x)$ is generally defined as potential energy per unit mass i.e. $V(x) \equiv \dfrac{U(x)}{m}$. So for the points where $U(x)$ is zero, $V(x)$ is zero and vice-versa by definition.

EDIT: After you added the comment and a snapshot of the book, I realized your book has defined Gravitational Potential in a different manner as $V(B)-V(A)=\dfrac{U(B)-U(A)}{m}$ than the standard $V(r)=U(r)/m$. This allows the room to choose different references for both $V(r)$ and $U(r)$. So to solve your confusion, the reference of $U(r)$ might be different than that of $V(r)$ so $U(A)$ need not be equal to $mV(A)$.

• So ..that means if $V(A)=0$ then $U(A)=0$...?? – Freelancer Nov 10 '15 at 2:01
• But my book concepts of physics by HC Verma also says the same thing as my teacher did...I have Added the picture of the page also...please look at it.. – Freelancer Nov 10 '15 at 2:13
• @Freelancer Your book does not say anything about why $U(x)$ should not be zero. – user40330 Nov 10 '15 at 7:01
• Yes!! That's what I am saying .!!.and I want to know why the book or. My teacher isn't also saying that!! – Freelancer Nov 10 '15 at 7:50
• Ok..what you are trying to say is that ..my book is taking one reference of Gravitational potential energy $U(r)$ and another(different reference) for Gravitational potential $V(r)$ ..right?? – Freelancer Nov 10 '15 at 11:05

Your copy of Verma has already defined gravitational potential energy previously in (11.3)

The gravitational potential energy of a two particle system is $$U(r) = -\frac{Gm_1m_2}{r} \tag{11.3}$$ where $m_1$ and $m_2$ are the masses of the particles, $r$ is the separation between the particles and the potential energy is chosen to be zero when the separation is infinite.

This explicitly states that the $GPE$ is negative (non-zero) at any finite separation.

Now given that definition, you can find the potential difference between two points. Since only the potential difference is defined, any point can represent zero potential, even a point that has non-zero potential energy.

Defining a point as the zero potential does not change the potential energy of that point.

• OK..so as the book had already defined and said that the $GPE$ is taken as zero at infinity ..so I can't now possibly take any other point having zero potential energy.. right?? – Freelancer Nov 10 '15 at 11:03
• Correct. While other conventions are possible, $U(\infty)=0$ is standard. The important thing is that defining the $GPE$ for certain points does not limit your choice of any arbitrary point to be zero potential. The two are not linked. – BowlOfRed Nov 10 '15 at 11:16
• The two( GPE and Gravitational potential) are not linked..!!.well..that's quite weird...they have the same name..at least....!! Also in the definition of potential I gave above ..they certainly are related by a neat looking equation...!! – Freelancer Nov 10 '15 at 11:32
• But let's leave that for now...as you also said that defining the GPE for certain points does not limit your choice of any arbitrary point to be zero potential ..but ..its not clear to me (intuitively) ..taking two different refernces for two things which are related by a fine (given)equation... – Freelancer Nov 10 '15 at 11:39
• No one is saying you take two different origins for initial and final. It's saying that you can pick an arbitrary origin (that is valid for both) and it doesn't affect the acceleration or the displacement because on the difference is used. – BowlOfRed Nov 10 '15 at 15:54

To answer the question, let me write the relation between potential energy and force.

\begin{align}\vec{E} &= -\dfrac{\mathrm d V}{\mathrm d \vec r}\\ \implies \mathrm dV &= -\ \vec E . \mathrm d \vec r \\ \implies \displaystyle \int \mathrm dV &= -\displaystyle \int \vec E . \mathrm d \vec r \\ \implies V + c &= -\displaystyle \int \vec E . \mathrm d \vec r\end{align}

Where $c$ is the constant of integration. Note that $c$ has the same dimensions as of $V$.

Now, this $c$ makes sure that you can choose any point as reference point where the gravitational potential energy is zero.

Take the general concept you have been using since your childhood. What is the gravitational potential energy of a mass $m$ at a height $h$ from the surface of the Earth ? The answer is $mgh$ and hence the gravitational potential is $gh$. Have you ever noticed that the potential at the surface of the Earth is taken to be zero in this case (and hence as the reference point). The question is how ? The $c$ does the magic. Take $c$ such that the gravitational potential on the surface of the Earth becomes zero (In this case, you have to take $c =\dfrac{GM}{R}$, $R$ being the radius of the Earth). You are free to take any value of $c$ making your reference point as your wish.

The potential difference between 2 points won't change because the both of the $c$ will cancel out (as the difference is taken).

Since $V$ and $U$ differ by a factor $m$, if you can $V$ to be zero, you can also take $U$ to be zero at your reference point.

(Do the calculations for force and potential energy. You'll get that $U +c' = -\displaystyle \int \vec F . \mathrm d \vec r$. See, here too, you can choose any point as the reference point)

• This answer is based on all the comments and answers and the question itself. In case you notice any error, please point it out – user174490 Dec 10 '17 at 5:20
• Just to be picky: one can't divide by a vector so $dV/d\vec r$ is technically incorrect. It should be $\hat x\frac{\partial V}{\partial x}+\hat y\frac{\partial V}{\partial y}+\hat z\frac{\partial V}{\partial z}$. Not sure why your answer was downvoted but you did ask if we noticed errors. – ZeroTheHero Dec 17 '17 at 18:34