3
$\begingroup$

This may just be a simple Misconception Question, here goes:

Definition for Gravitational Potential Energy:

  1. The work done by gravity to pull an object to the ground.

    $E=-(\frac{GMm}{r}-\frac{GMm}{R})$
    Where, $r =$ Distance from Centre of Mass , $R =$ Radius of Earth

    Example:
    Object of 1 kg released from 1 meter on the surface of Earth
    $E=-(\frac{GMm}{r}-\frac{GMm}{R})$
    $E = -(\frac{( 6.67 x 10^{-11}Nm^2kg^{-2} )(6.0x10^{24}kg)(1kg)}{6.4x10^6m + 1m}-\frac{( 6.67 x 10^{-11}Nm^2kg^{-2} )(6.0x10^{24}kg)(1kg)}{6.4x10^6m})$
    $E = 9.77J$

    Which is quite consistent from the formula unless $h$ is too big
    $E = mgh$
    $E = (1kg)(9.81ms^{-1})(1m)$
    $E = 9.81J$

  2. The work done by gravity to pull an object from infinity to point 'r'.

    $E=-\frac{GMm}{r}$
    Where, $r =$ Distance from Centre of Mass

So,
Question is why are they using the same name if they don't represent the same thing?

$\endgroup$
  • $\begingroup$ Keep in mind that when falling from infinity, the radius of the Earth is negligible, as indicated by its omission from 2. You're looking at only a few percent difference in calculated potential for something even as close as the moon due to ignoring the radii of either object, and the moon doesn't even begin to compare to "at infinity." $\endgroup$ – Asher Oct 3 '15 at 15:35
1
$\begingroup$

As per the formal definition: the potential energy of a field $\mathbf{v}$ is any function $f$ such that $\mathbf{v} = - \textrm{grad} f$ anywhere in the domain of definition of $\mathbf{v}$ (or wherever it makes sense). According to the above if $f$ is a potential for the field $\mathbf{v}$ so is $f + c, c$ being any constant; therefore if a field admits one potential energy, it admits infinite thereof and you can span some of them with at least an additional constant. To directly answer your question:

Question is why are they using the same name if they don't represent the same thing?

The definition of potential energy is unique but the collection of functions satisfying it is infinite.

Furthermore the potential energies are such that the work done by the field between two points can be expressed as the difference in potential energy between the two, i. e. $$ W_{A\to B}(\textbf{v}) = f(A) - f(B). $$ One can now easily see that no matter which of the functions in the collection you have chosen, the additional constant always drops off when subtracting the contributions on the right hand side. Consequently, it does not really matter which one you choose as long as you look at differences and work done (which is what one does in physics).

This said, it is very often conventional to choose the additional constant to be zero just for the sake of avoiding to append one more term that by no means contributes. Since, given the fields at hand, the potential energy scales as $1/r$, it reaches zero when $r\to\infty$, so that one reads that as to be the work done by the field to move a unit (mass) charge from where it stands to infinity.

$\endgroup$
  • $\begingroup$ Thanks for the answer. But if someone ask me what is the Gravitational Potential Energy of an Object. Should I use the First Formula or Second Formula? Is there a way to be more specific on which one to use? $\endgroup$ – MrYellow Oct 4 '15 at 5:37
  • $\begingroup$ @MrYellow The second formula is fine enough, if you just want to know what to answer. $\endgroup$ – gented Oct 4 '15 at 10:49
1
$\begingroup$

When working with potential energy in classical mechanics we always compare the difference in energy between different points. It's only this difference that matters and not the absolute value of the energy (which is not observable). We are therefore free to take the zero-point wherever we want. This is the difference between the two formulas: in the second formula the zero point is at infinity and in the first the zero-point is at $r=R$.

Since the choice of zero point does not matter for the dynamics both formulas give rise to the same physics. That being said the second formula is the most natural choice to make (as is does not have the unneccesary reference to the Earth in it).

$\endgroup$
  • $\begingroup$ In the second version, the zero of P.E. is taken to be at infinity. (you have it the other way around). Also (small point), the zero in the first version is immaterial. It could be at infinity, or earth surface or anywhere else. $\endgroup$ – garyp Oct 3 '15 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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