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Dot product in the Gravitational Potential Energy formula

  1. May 22, 2015 #1
    This is the gravitational potential energy formula
    $$U = -\int_\infty^r\vec{F}_\text{field}\cdot d\vec{r}$$
    If r vector's direction is form infinity to r, then it means it has same direction as Gravitational Force. So cos0=1
    But after multiplication there is a negative sign here: "-GMm"

    $$U = -\int_\infty^r \frac{-GMm}{r^2} dr$$

    Why is there a negative sign here(before GMm)? Doesn't gravitational force vector and r vector(from infinity to r) have same direction?
     
  2. jcsd
  3. May 22, 2015 #2
    Let name a force F, we can express it a gradient of some potential Φ
    Mathematically : F = ∇Φ
    Gravity is an attractive force, it's tends pull and bring you to itself, but that mean F = - ∇Φ, if you consider a force field that is pushing you away, then you can say that F = ∇Φ,
    The minus sign just tells that the force increase whenever the potential decreases .
     
  4. May 22, 2015 #3
    Thanks but I am not talking about the negative sign before the integral.

    I am asking the negative sign that was added after that operation: $$\vec{F}_\text{field}\cdot d\vec{r}$$

    Here F is gravitational force, right? So it must be toward the origin, so is r vector, right? So the angle is 0. And there mustn't be a negative sign there. But there is.

    Why is negative sign added there?
     
  5. May 22, 2015 #4
    The r vector is not "from infinity to r". (whatever that means).
    It is from the origin to the given point. If you write the gravitational force as you did, the origin is at the center of a spherical body or at a point mass.
    The r vector has a direction opposite to that of the force.
    .
     
  6. May 22, 2015 #5
    Because F = -GMm/r^2, the direction of force is opposite to the direction the angle isn't 0 it's pi (180 degree )
     
  7. May 22, 2015 #6
    I am confused here.

    dr is a differential displacement vector and it shows direction of movement right?

    So we set infinity to zero potential, make calculations from infinity to point r. Here we are bringing the object from infinity to point r. So r vector is from infinity to point r. Gravitational force vector is also toward origin. So displacement and force vectors have same direction.

    Why did you say: "The r vector has a direction opposite to that of the force"?
     
  8. May 22, 2015 #7
    Because it does. The r vector is from origin to the mass. The force is from the mass to the origin, the mass is attracted to the origin. This is for the vector r.

    .
     
  9. May 22, 2015 #8
    Look at what nasu just said !
     
  10. May 22, 2015 #9

    Dale

    Staff: Mentor

    F points towards the origin, and the unit vector r points away. So F is some negative scalar multiple of the unit vector r. It has nothing to do with the dot product. The sign is in the force law itself.

    This doesn't make sense.
     
    Last edited: May 22, 2015
  11. May 22, 2015 #10
    As you answered my other thread, there mustn't be any sign there. Because it is "magnitude" of the gravitational force. I think there is no unit vector there. It must be displacement vector.

    And I still don't understand, if we integrate from infinity to r point which means bringing object from infinity to point r, how does displacement vector r points from origin to point r, not from infinity to point r???
     
  12. May 23, 2015 #11
    Then why is r called "displacement vector"?
     
  13. May 23, 2015 #12
    You are right but displacement mean:
    $$ \delta{\mathbf{r}} = \mathbf{r}_f - \mathbf{r}_i $$
    rf and ri are parallel for direct central movement.
     
    Last edited: May 23, 2015
  14. May 23, 2015 #13
    I looked up Halliday's Fundamentals of Physics book, in section 13-6, he called r as a displacement vector.

    He gave an example about shooting a baseball directly away from Earth and integrate from R to infinity(not infinity to R) and said: "The integral contains the scalar (or dot) product of the force F(r) and the differential displacement vector dr along the ball’s path" and integrate
    $$\vec{F}_\text{r}\cdot d\vec{r}$$= F(r)drcosϕ

    where ϕ is the angle between the directions of F(r) and dr. When we substitute 180° for ϕ"

    Why did he say this?

    Because F is toward earth and dr is directed away from earth. So F and r vectors have opposite direction.

    But here, F is gravitational force so, it is directed to earth and the ball is bringing from infinity to point R but according to examples dr is still pointing infinity from earth.
     
  15. May 23, 2015 #14
    See what does vector minus vector mean:
    $$ \mathbf{A}-\mathbf{B} = \mathbf{A} + (-\mathbf{B}) $$ is another vector than starts from the end of B and end to the end of A.
     
  16. May 23, 2015 #15

    robphy

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    I think there is a confusion in the use and/or interpretation of ##d\vec r## as an infinitesimal element of a directed path
    and as an increment of the ##r##-coordinate. Let's use ##d\vec s## for the infinitesimal element of a path.

    $$U = -\int_{P}^{Q}\vec{F}_\text{field}\cdot d\vec{s},$$
    where ##P## is the start-point (which will be taken at infinite-radius from the origin) and ##Q## is the end-point (which will be taken at finite-radius ##R## from the origin.

    Now let us evaluate this symbolic line-integral in polar-coordinates. Keeping only the radial-components (since ##\vec F## is radial),
    $$U = -\int_{P}^{Q}\left(-\frac{GMm}{r}\hat r\right) \cdot \left( dr\ \hat r\right),$$
    where I have left the endpoints of the path symbolic. Note the sense of the path to be taken hasn't been used at this point. Note: ##dr\ \hat r## is the direction of increasing-##r##.
    Simplifying,
    $$U = -\int_{P}^{Q}\left(-\frac{GMm}{r} dr\ \right),$$
    Now, choosing ##P## and ##Q## according to the definition:
    $$U = -\int_{r=\infty}^{r=R}\left(-\frac{GMm}{r} dr\ \right).$$
     
  17. May 23, 2015 #16
    Thank you.
    If we integrate from Q to P, will ##dr##'s direction change?

    Can you please give more comment about transforming from ds to dr.
     
  18. May 23, 2015 #17

    robphy

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    No, ##dr\ \hat r## (the direction of increasing r) always points radially outwards.
    In polar coordinates, an infinitesimal displacement is ##d\vec s= dr\ \hat r+rd\theta\ \hat\theta##.
     
  19. May 23, 2015 #18
    And Gravitational Force always points inwards. So for gravitational force, the result of dot multiplication is always negative. Then what is the point of dot multiplication.

    How can we determine gravitational force make positive or negative work???

    Thanks for help...
     
  20. May 24, 2015 #19
    Gravitational potential energy is defined the way it is so that it can have real values. Radius = 0 cannot be used because at r=0, by the force formula, the force of gravity is infinite, and any object at a radius greater than zero would then have infinite potential energy relative to r=0. But at r=infinity, the force of gravity is zero and any object at a radius less than infinity can easily compute a real potential energy relative to infinite radius. This potential energy then has to be defined as negative so that when computing the potential energy difference between two radii, the higher radius will have a less negative potential energy and the difference in energy is positive when looking upward. So, it takes positive energy to climb upward.
     
  21. May 24, 2015 #20

    jtbell

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    Clearly the the work has to have the same magnitude but opposite signs when we switch the direction of travel (i.e. from P to Q versus from Q to P). In the work integral, we can accomplish this either by switching the limits of integration or by changing the sign on ##\hat r dr##. But if we do both of these, there is no net effect. The convention is to switch the limits of integration, because that is consistent with what we expect from ordinary single-variable integration. (At least, I think it's simply a convention in line integration. If someone can give a more substantial reason, I'd be happy to see it!)
     
  22. May 24, 2015 #21

    Dale

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    I think that you are getting confused here. To be clear, I will use ##\vec{x}## to denote a vector and denote its magnitude ##|\vec{x}|## and I use ##\hat{x}## to denote a unit vector defined such that ##|\hat{x}|=1##.

    Now, when you write a vector in coordinates you write it as a sum of some basis vectors, like ##\vec{a} = a_x \hat{x} + a_y \hat{y} + a_z \hat{z}##, where the basis vectors are defined at each point as the unit vector in the direction of increasing coordinate. So ##\hat{x}## is the unit vector which points in the direction that the x coordinate increases (other coordinates held constant). Therefore in spherical coordinates the ##\hat{r}## vector points in the direction of increasing r coordinate, which is outwards, away from the origin.

    Up to this point all of this is purely math, with no physics content. Now, suppose we have a central particle of mass ##M## located at the origin and a test particle of mass ##m## located at coordinates ##(r,\theta,\phi)## in a spherical coordinate system. From Newton's law of gravitation we know that the gravitational force is proportional to both ##M## and ##m## and inversely proportional to ##r^2##. We also know that the gravitational force is attractive, so the force on the test particle points inwards, towards the origin which is the opposite direction as ##\hat{r}##.

    So, if we write the gravitational force on the test particle as ##\vec{F}=-G\frac{Mm}{r^2} \hat{r}## then we get a force that points in the correct direction, i.e. it attracts towards the origin. If we omit the negative sign then we would erroneously get ##\vec{F}## pointing in the same direction as ##\hat{r}## which is away from the origin, i.e. a repulsive force on the test mass.

    The dot product doesn't even enter in yet. There is a negative in the gravitational force law itself, regardless of what you might calculate afterwards. You have to start with the correct force law, and only then can you begin to calculate correct quantities based on it. Is that clear?
     
    Last edited: May 24, 2015
  23. May 28, 2015 #22

    Mark Harder

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    Consider what would happen if you brought a test object from infinity to r and r had it's origin at infinity. First, let us dispense with the vector dot product business. Modelling the earth as a sphere, (Which admittedly it isn't; but I believe you're are making the same assumption.), the central symmetry of the problem allows us to treat both as scalars. You must also realize that if you subtract any number (scalars now) from infinity the result is still infinity. That being the case, the displacement, - r = ∞ , and the integral over the displacement must be taken from infinity to infinity, so the potential is zero! Reductio ad absurdum, it makes no sense to employ the convention that vector r has its origin at infinity. But what does it mean when the origin is at the center of the force and the object is brought from infinity in to a point at distance r from the center? In this case r is finite, so the value of the antiderivative of the force function at r is trivial to find. Technically, the value of the antiderivative at r= infinity is actually the limit of the antiderivative as r approaches infinity. That's not difficult in the case where the potential has r in its denominator; the potential goes to zero when r goes to infinity. So the problem can only be done, and done handily, when r = zero at the center of the sphere.
     
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