Starting off, I first want to know the relation between work and potential energy.

$\Delta\mathbf U = - W $

How was this expression formulated?

Moving on,

My second doubt was in the derivation of the expression:

$\mathbf U = \frac{-GMm}{r}$


  1. Why are we bringing the point mass from infinity to $\mathbf (x = r)$
  2. What's with the negative sign?
  3. Who's work done are we talking about here?

Another thing, Why are we supposed to move the object without an acceleration? Is that even possible?!

Last but not the least, $$\mathbf V_B - V_A = \frac{U_B - U_A }{m}$$ Where, A mass $\mathbf m$ has been brought from point A to B under a gravitational field and,

$\mathbf V_B - V_A$ is the change in potential

$\mathbf U_B$ and $\mathbf U_A$ denote the gravitational potential energy when the mass m is at point B and A respectively.

Can you explain this equation to me? and what $\mathbf V_B - V_A$ mean?

I know these questions are pretty obvious and dumb but I am a fresh mind to physics and the deep concept of potential energy. Also, it's my first time posting here. Forgive me if my questions weren't concise enough. My physics teacher just skimmed through this topic!!!


1 Answer 1


Let's start by the definition of work. For simplicity I use,

$W = F \cdot \Delta x$

$W = m a_x \Delta x$

recall, $v^2 -v_o^2 = 2 a_x \Delta x $, so

$W = \frac{1}{2} m (v^2 -v_o^2 )$

$W = \frac{1}{2} m v^2 - \frac{1}{2} m v_o^2$

$W = K_f - K_i$

where $K$ is the kinetic energy.

Now, let's look at the conservation of energy.

$E_i = E_f$

$K_i + U_i = K_f + U_f$

$U_i - U_f = K_f - K_i$

The RHS is work. To get a $\Delta U$ we must have final minus initial so,

$- (U_f - U_i) = W$

$W = -\Delta U$.

Note that work is proportional to the change in potential energy. that means that to calculate work you do not need the exact value of the potential you just need to know how much it changes.

Let's try to find ( and define) gravitational potential.

$W = F \cdot \Delta x$

$W = F \cdot (x_f - x_i)$

$W = \frac{GMm}{r^2} \cdot (x_f - x_i)$

Let's replace ($W = -\Delta U= U_i - U_f$)

$U_i - U_f = \frac{GMm}{r^2} \cdot x_f - \frac{GMm}{r^2}\cdot x_i$

Let's look at the meaning of each term, $r$ is the distance between the two objects, $x_i$ is the initial distance between the objects, $x_f$ is the final distance between the objects. If you want to calculate the current gravitational potential between two planets, i.e. $r = x_f$, by comparing both sides of the equation we get (notice what is happening to the negative sign)

$U_i = - \frac{GMm}{x_i^2} x_i $

$U_f = - \frac{GMm}{x_f^2} x_f = - \frac{GMm}{x_f} $

We are only interested in the final position because it corresponds to the current positions of the planets. The initial one is not interesting to us because it does not tell us anything about what is happening between them now. Because it does not change our final answer we can assume that $x_i$ was very large and $U_i$ was practically zero. So in general it is correct to say.

$U = - \frac{GMm}{x} $

Who moved the planets together? gravity. The negative sign (which came out of the calculations) makes sure that you end up with a correct sign for your work.

Finally $\Delta V$, There are two masses in this equation $M$ and $m$. $\Delta V$ tells us what is the gravitational potential energy of M on 1 kg object at different locations ($r$).

  • $\begingroup$ How did we end up with positive initial potential? Isn't it supposed to be < 0? And thank you very much for clarifying in such a simple manner. Physics is fun when you get it. $\endgroup$
    – Jeeshaan
    Dec 16, 2020 at 3:44
  • $\begingroup$ You got me. I fixed it. :) I agree physics is fun when we get it. $\endgroup$ Dec 16, 2020 at 4:42

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