0
$\begingroup$

Change in potential energy is $$\Delta U = -\text{Work Done} \, .$$ While deriving the equation for potential energy we just equate the work done by gravitational force with the potential energy and do not consider $\Delta U$, i.e. $U(r) - U(\infty)$, where $r$ is the distance of the object from the gravitating source. If we do that then the gravitational potential energy would come out positive. Why is that we don't take $\Delta U$ in this case?

$\endgroup$
  • 1
    $\begingroup$ I'm not sure what you mean by $U(r)$. Potential energy isn't absolute, i.e. you can only define it relative to a reference point. Often, we take a point infinitely far away from the system of interest as the reference point. Then we define the potential energy at point $p$ as the work needed to move a particle from infinitely far away to the point $p$. Is this what you're asking about? $\endgroup$ – DanielSank Jan 12 '17 at 3:25
  • $\begingroup$ Yes exactly. Why is the work done equated with potential energy. What's the logic behind. If it's something related to "storing energy" then what about the change in kinetic energy due to that work? $\endgroup$ – Jasgeet Singh Jan 12 '17 at 3:39
2
$\begingroup$

Change in potential energy is $\Delta U =$ -Work Done

is to be read as

Change in potential energy is $\Delta U = -$Work Done by gravitational forces

So imagine that you have two point masses as the system that you are considering and define the gravitational potential energy to be zero when their separation is infinite, $U\infty) = 0$

The separation of the masses is then decrease to $r$ and so the gravitational potential energy of the masses is now $U(r)$.

In going from infinity to separation $r$ the change in gravitational potential energy of the system of two masses is $\Delta U = U(r) - U(\infty) = U(r)$

So that is the left hand side of your equation dealt with.

The gravitational force is attractive and along the line joining the two point masses.

Imagine that something (you?) is holding the masses apart but allows them to start at rest when at an infinite separation and finish at rest when the separation is $r$.
Whilst the separation is decreasing the gravitational forces are moving and doing work.
This work is positive because the direction of the gravitational forces and their direction of movement are the same.

So minus the work done by the gravitational forces is negative which means that the change in gravitational potential energy of the system is negative ie the gravitational potential energy has decreased.


Another way of defining the change in gravitational potential energy is $\Delta U = $Work Done by external forces

Now the external forces are opposite in direction to the gravitational forces so the direction of the external forces is opposite to the direction of movement of the external forces.
Hence the work done by the external forces is negative as is the change in gravitational potential energy of the system.


If the system os two masses is isolated then the centre of mass of the system cannot move.
This means that if the masses are allowed to move closer together due to their mutual attraction their centre of mass does not move.
This in turn means that both masses move towards the centre of mass and the gravitational forces (equal in magnitude by Newton's third law) on both bodies do work.

However in the case of a system like the Earth and an object on the Earth with the mass of the object being very much less than that of the Earth the movement of the Earth towards the centre of mass of the system is negligible compared with the movement of the object.
The Earth is assumed not to move at all.

$\endgroup$
2
$\begingroup$

I think you're a little bit confused. $$\Delta U = \text{-work done BY the object} = \text{+work done ON the object}$$ This is critical! So, if you push a block, you're losing your "potential" to do work, and the block is GAINING said ability. (Friction aside, of course). You need to differentiate between work done BY and work done ON. In fact, in some sense, potential is the total amount of work done ON an object; once the object "gains" that energy, it can release it. So the potential energy is kind of the work done by "God" (or your teacher, or some angry really bored person who's been around since the dawn of time) on your object to get it the way it is.

As you know: $$\text{Work done BY object}=\int_a^b F dx$$ As I mentioned earlier, you can see potential as the work done by the universe to get an object into the state that it's currently in. Since it's stupid to say "let's measure gravity relative to a distance exactly 37.6 AU from some random object", we might as well just say "let's measure compared to being infinitely far away", because it's way more useful and easier to work with.

Well, now everything is easy:

$$\text{Force exerted by gravity} = -\frac{GMm}{r^2}$$ Now we shove that into our integral:

$$\text{- Work done ON object}=\text{Work done BY object} = \int_\infty^{r_{0}}{-\frac{GMm}{r^2}dr}=[\frac{GMm}{r}]^{r_0}_\infty= \frac{GMm}{r_{0}}$$

And we're done! Since potential energy is work done ON object ("by the universe" is implied), and that's negative work done BY the object, so our potential is negative, which it intuitively should be!

$\endgroup$
1
$\begingroup$

I don't know where you're getting your signs from, but the procedure you're considering is exactly what is done in practice.

If you let a particle of mass $m$ fall in from rest at infinity to some radius $r$, it will pick up some kinetic energy and speed: this means that it goes from kinetic energy 0 to positive kinetic energy, so positive work was done on it.

Since the change in potential energy is the negative of the work, the potential energy must have decreased, not increased. So the change is negative, not positive.

$\endgroup$
  • $\begingroup$ So the work done $W=U(r)-U(\infty)=\int_{\infty}^r\frac{GmM}{r^2}dr=-\frac{mMG}{r}$ is the work done by the gravity force, to bring $m$ from infinity to $r$. The work done on the mass $m$ is the negative of this. Therefore the potential energy of $m$ has decreased. In other terms, $m$ gains kinetic energy on the expense of potential, and the balance is zero. The total energy remains unaltered. Right? $\endgroup$ – Alexander Cska Mar 31 '18 at 12:50
  • $\begingroup$ @AlexanderCska almost; I would take issue with saying that the first expression is "the work done by the gravity force to bring $m$ from infinity to $r$," that is the exact opposite of what your next sentence says, which I would regard as correct. A work is a force on an object integrated over distance, which by the work-energy theorem is also a change in the kinetic energy of that object. So there is no difference between "the work done on the mass $m$" and "the work done by the gravity force". $\endgroup$ – CR Drost Apr 2 '18 at 3:13

protected by Qmechanic Jan 12 '17 at 8:19

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.