I am taking a first year physics course, my calculus is really rusty and I am having a very hard time with (Potential) Gravitational Energy.

Let's consider a body of mass $m_1$ located at $r_1$ from another body with mass $m_2$. We move the body with mass $m_1$ to location $r_2$ such that $r_2 > r_1$. As far as I can tell, the work we do on $m_1$ is described by:

$$W_{r_1r_2} = \int_{r_1}^{r_2} F(r).dr=\int_{r_1}^{r_2}\frac {Gm_1m_2}{r^2}dr=-\frac {Gm_1m_2}{r} \biggr|_{r_1}^{r_2}$$

Which is:

$$\begin{equation}\tag{1}W_{r_1r_2} = Gm_1m_2 \left(\frac{-1}{r_2}-\frac{-1}{r_1}\right)=Gm_1m_2 \left(\frac{1}{r_1}-\frac{1}{r_2}\right)\end{equation}$$

Since $r_2>r_1$ then:

$$\begin{equation}\tag{2}W_{r_1r_2} > 0\end{equation}.$$

  1. The result (2) makes sense to me because we are doing work to move the body with mass $m_1$ away from the body with mass $m_2$. Is this correct?
  2. Is it correct to reason that the work done by the gravitational field would be $W_{g_{r_1r_2}}=-W_{r_1r_2}$ Since it is opposing the work we are doing?
  3. Thus the potential gravitational energy $\Delta U_g=-W_g=W_{r_1r_2}$. Is that why the gravitational potential energy is always defined as $-\frac {Gm_1m_2}{r} \biggr|_{r_1}^{r_2}$?
  4. If we move $m_1$ to $r_2$ such that $r_2$ is very far away (approaching infinity), then $(\frac{1}{r_2} \rightarrow 0)$ which would mean $W_{r_1\infty}=Gm_1m_2 \frac{1}{r_1}$. Is this right? I find this so counter-intuitive - although I can see from equation (1) that for any $r_2 > r_1 \rightarrow W_{r_1r_2} \leq W_{r_1\infty}$.

Before someone points it out I do not think this question is an exact duplicate of Concept of Gravitational potential energy or Is the Gravitational potential energy work done by Gravitational force or not?


1 Answer 1


Firstly, your initial calculation is wrong. And your conclusion is wrong.

You've just happened to label the path element "dr" instead of "dl". Your force is missing a multiple of "$-\hat r$" and in spherical coordinates, your path "dl" is $dr\hat r + r d\theta \hat \theta + rsin(\theta)\hat \phi $, whose dot product is the negative of what you've wrote down.

If you did it correctly, you'd find that when $r_{2} > r_{1}$, you would have a have a negative expression instead of positive. This formula is the amount of work done BY the gravitational field. When you throw something UPWARDS, gravity does NEGATIVE work, not positive.

The amount of work that I would have to do on the ball, against the gravitational field, would be the negative of this expression!

This is one of the reasons that potential is useful, because potential describes the amount of work that I would have to put in, in order to make an object move a certain distance, in the presence of a gravitational field.

The relationship between KE and potential, is that the change in potential difference is the negative of the change in KE

  • 1
    $\begingroup$ The potential energy has nothing to do with the work you will have to do. You can do any amount of work between two points in the field but the change in potential energy is still the same. Both potential and potential energy are related to the work done by the field only. $\endgroup$
    – nasu
    Mar 20, 2022 at 14:01
  • $\begingroup$ With zero change of Ke* $\endgroup$ Mar 20, 2022 at 14:09
  • $\begingroup$ Can you clarify what you are trying to say here? Aren't you arriving at the same conclusion as the OP? $\endgroup$
    – garyp
    Mar 20, 2022 at 14:10
  • $\begingroup$ Ignoring all other forces, I prefer to think about the change of potential energy as the amount of work that I would have to do on an object, in order to make that object move from A to B, in the presence of the field. Yes if I apply more work during those points it will also move from A to B, but it will exceed that distance $\endgroup$ Mar 20, 2022 at 14:16
  • $\begingroup$ And I was mainly trying to clarify OP's statement that int f.dr is the amount of work done by "me", instead of the work done by the field. $\endgroup$ Mar 20, 2022 at 14:17

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