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I am not much clear regarding the defintion of "gravitational potential":

enter image description here

Is the work done for bringing the unit mass from infinity to that point by, gravitaional force or external force? (As you can see in the image, one source say external force and the other source says gravitational force)

And for the negative sign,

Firstly,by formula Gravitational potential is equal to:

**"work done/mass"

Some things i think that are fixed:

  1. mass is always positive.
  2. object is moving from 'infinity' to the 'gravitational field producing source'.
  3. sign of gravitational potential depends on the sign of work done only as mass is always positive.**

Am I correct till here?

Case 1) work done by gravitational force:

as the object moves from 'infinity' to the 'gravitational field source mass' and the gravitational force is also in the same direction that is from 'infinity' to the 'gravitational field source mass'. And work done is (force.displacement) (dot product) and since both are in the same direction, work done by gravitational force is positive. And since mass is also positive. Thus making gravitational potential positive in this case.

Case 2) Work done by external force:
if the work done is done by external force, then work can be both:
a) positive (if the external force is also directed toward the 'gravitational field producing source mass' thus same as the object distance direction and making work done positive.

b) negative (if the external force is directed from infinity to further infinity (i.e away from the 'gravitational producing source mass' and thus also opposite to the direction of the object moving direction and thus making work done negative)

Case 3) if gravitational force, then applying work energy theorem:
Work done is also (change in kinetic Energy) or (negative of change in potential energy) and as a fact I know that potential energy is comparitively lesser when the object is closer to the 'gravitational field producing source mass' and comparitively higher when a bit further away from the 'gravitational field producing source mass' and as our object final position is closer to the 'gravitational field producing source mass'
Thus change in potential energy is therefore:
lesser value - higher value = negative value.

Now work done is negative of change in potential energy:
Negative×(negative value) = positive value.
Thus making work done in this case postive.

Now, the sign of gravitational potential in formula in every book and online is negative(fixed) and for that work done has to be negative

and only (case 2.b) is giving work done negative and all other giving positive work done.

I am confused how the negative sign is coming.

And if the force doing the work done, is gravitational force, but then the value of work done by it is positive and not negative and as gravitational force is conservative, can you also explain how is it consistent with work energy theorem? Can you please and explain in detail. Thank you so much.

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  • $\begingroup$ Please do not post images. Use MathJax for writing formulas. $\endgroup$
    – Hyperon
    Mar 1 at 8:11
  • $\begingroup$ The absolute value of the potential energy has little meaning (at least definitely so for classical mechanics), and in this case, the only thing that is truly observable is the difference in the potential energy. Imagine: You will need to use your energy to get father away from a star. that corresponds to having a higher potential energy there. So having a minus sign makes sense. You can also understand that the minus sign must come out, as you take the derivative of $1/r^2$ (force) which gives you $-1/r$ (potential energy). $\endgroup$ Mar 1 at 8:56
  • $\begingroup$ The concept of 'work' appears completely useless to me and only leads to confusion. It is probably a marketing term from the early days of mechanisation. $\endgroup$
    – my2cts
    Mar 1 at 10:31
  • $\begingroup$ @Hyperon ohk, got it. Thank you $\endgroup$ Mar 4 at 15:52

2 Answers 2

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The second definition is incorrect and so that answers a number of your questions.

First start by defining the system which in this case is the "unit mass" and the masses causing the gravitational field.

The definition should contain the words work done on the unit mass by an external force or minus the work done on the unit mass by the gravitational force on it (which is an internal force).

It does not matter at what velocity the unit mass moves at as long as the kinetic energy of the unit mass at the start and finish is the same.

To make the calculation easier it is assumed that the external force is equal in magnitude and opposite in direction to the gravitational force, ie there is no change in the kinetic energy of the unit mass all the way along its path.


For consistency when using an external force or the gravitational force to evaluate the potential it should make no difference to the value of the potential (including sign) which is found.

Using the external force $\vec F_{\rm ext}$ the potential at some position $A$ is given by $\displaystyle \int_\infty^A \vec F_{\rm ext}\cdot d\vec r = {\color{red} +} a$.
Using the gravitational force $\vec F_{\rm grav}$ one evaluates the integral, $\displaystyle \int_\infty^A \vec F_{\rm grav}\cdot d\vec r$, and finds that the value comes out to be ${\color{red} -}a$.
To get the same value using either method it has decided that when using the gravitational force to find the potential one needs to use $\displaystyle {\color{red} -}\int_\infty^A \vec F_{\rm grav}\cdot d\vec r$ which will returns the value $-(-a)=+a$, the same value as when using the external force.

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  • $\begingroup$ Thank you so much for answering. I understood that both the force can do work; but you say with gravitational force there will be minus sign. I cant understand why? Can you please explain in detail. Thank you so much. Because the object displacement is in the direction of source mass and so is gravitational force, doesnt that make the work done positive? (Thetha=0). Thank you $\endgroup$ Mar 4 at 15:49
  • $\begingroup$ There are two forces on the unit mass, gravitational and external. With the definition of potential in terms of the external force the work done by the gravitational force will have the opposite sign to the work done by the external force. $\endgroup$
    – Farcher
    Mar 4 at 20:38
  • $\begingroup$ ohk, so i also understood that work done will be opposites because the forces are oppposites too. The angle between gravitational force and displacement vector is zero° and with external force 180°, correct? And i was taught work done to be dot product of force vector and displacement vector. So work done by gravitational force will be positive (cos 0 = +1 ) and by external force be negativen (cos 180 = -1 ); as the direction of obect displacement is from infinity to the point i.e towards the source mass and gravitational force is also toward source mass from infinity. $\endgroup$ Mar 5 at 9:33
  • $\begingroup$ as you can see we have opposites work done by those forces but the minus sign is coming with gravitational force and not the external force (-1) ; but the dot product is positive. That's why i am still confuse regarding the minus sign. Sorry. And thank you. $\endgroup$ Mar 5 at 9:38
  • $\begingroup$ @Cerebralcortex I have added to my answer. $\endgroup$
    – Farcher
    Mar 5 at 10:07
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You should know that physicists defined the potential to have a minus sign.

A potential of a force $\vec{F}(\vec{r})$ is defined to satisfy: $\vec{F}=-\nabla {U}$. See here if you don't know what a gradient($\nabla$) is.

Gravitational force always points to the point mass; therefore the potential has to have a negative slope towards the point mass.

Is the work done for bringing the unit mass from infinity to that point by, gravitaional force or external force?

It is the work done by an external force. Imagine a particle infinitely far away from a point mass. It will be slowly dragged by gravity. You apply external force to the particle so that it cancels out the gravitational force. You slightly weaken the external force by $\epsilon$ so that the particle moves towards the point mass by distance $d(\epsilon)$. Then you change the external force so that it cancels gravitational force again. By repeating this process for an infinitely long time, you can move this particle to a position that has a finite distance from the point mass. This process was done to maintain velocity $v\approx 0$, so the kinetic energy is equal to 0.

Therefore the work done to the particle is completely converted to potential energy. The direction of the external force is the opposite of the gravitational force, giving the minus sign.

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