1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential energy and gravitation

  1. May 18, 2006 #1
    Sir,
    A body is projected in space from the earth’s surface with escape velocity. It is said that at the time of projection its total energy is Kinetic. Is it true? If so, doesn’t it have gravitational P.E at the time of projection?
     
  2. jcsd
  3. May 18, 2006 #2
    The thing about potential energy is that the total amount of potential energy is often undefined. What really matters is the change in potential energy. Thus, the problem is only asking you to arbitrarily set the V(R) = 0 where R is the distance from the center of the earth when you are at its surface.

    This is in the same way as you are setting kinetic energy to zero directly before the body is projected into space. You are picking zero arbitarily, in reference to the earth. The body could be said from the view of the sun, to have a bunch of kinetic energy as it is revolving around the sun at a very fast pace, but as only changes in kinetic and potential energy matter in newtonian physics, we ignore those effects.

    ~Lyuokdea
     
  4. May 19, 2006 #3
    At the time of release to make calculations easier , we set the P.E at the surface of earth =0 , so that , we can easily calculate its K.E at some higher point as a function of distance from earth's surface.

    BJ.
     
  5. May 19, 2006 #4
    I don't see how the energy at the time of projection is purely kinetic!!!

    The total energy of a body of mass [itex]m[/itex] (moving with speed [itex]v[/itex]) in the central field is always the sum of its kinetic energy,

    [tex]T = \frac{1}{2}mv^2[/tex]

    and the potential energy,

    [tex]V = -\frac{GmM_{e}}{r}[/tex]

    ([itex]M_{e}[/itex] = mass of the earth)

    where [itex]r[/itex] is the radial distance from the center of the earth. If you impart escape velocity to the body at the time of projection, then at [itex]r=\infty[/itex] it has zero energy. At any intermediate stage, the body obviously has some kinetic energy and some potential energy.

    The potential energy at infinite separation is always zero but escape velocity guarantees that the kinetic energy at [itex]r=\infty[/itex] is zero too.

    An energy balance at [itex]r=R_{e}[/itex] (where [itex]R_{e}[/itex] is the radius of the earth) and [itex]r=\infty[/itex] gives

    [tex]T + V = 0[/itex]

    or

    [tex]\frac{1}{2}mv_{esc}^2-\frac{GmM_{e}}{R_{e}} = 0[/tex]

    [itex]v_{esc}[/itex] is the (required) escape speed.

    Hope this helps.
     
    Last edited: May 19, 2006
  6. May 19, 2006 #5
    Via your definition of V=0 at infinity, then of course there will be a non zero v at r=radius of the earth, but try this conversiont:

    [tex] V = \frac{GmM_e}{r} [/tex]

    set:

    [tex] V(R_e) = 0 [/tex]

    R_e = Radius of the Earth

    this gives you:

    [tex] V(\infty) = + \frac{GmM_e}{R_e} [/tex]

    Now, try to find a way that this changes the physics of the situation, it can't.

    The only thing that matters is this change in the potential energy. If you are familiar with calculus this might help:

    [tex] V = \int E * dl [/tex]

    where * is the dot product, thus V has a degree of freedom of being able to be shifted by an arbitrary constant C, as do all integrals. If I try to find the electric field from the potential.

    [tex] E = grad(V) [/tex]

    for any arbitarary shift V' = V + C

    [tex] E = grad(V') = grad(V+C) = grad(V) + grad(C) = grad(V) [/tex]

    Thus adding any constant to C is completely arbitrary and does not change the physics of the problem. Sorry I can't figure out how to do gradients in tex all of a sudden.

    ~Lyuokdea
     
    Last edited: May 19, 2006
  7. May 19, 2006 #6

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I agree except that here I think you meant
    "consider adding at all points the value
    [tex] V = \frac{GmM_e}{R_e} [/tex]
    and this gives
    [tex] V(\infty) = + \frac{GmM_e}{R_e} [/tex]"


    Just to make things clear to the other posters (using the variable r instead of R_e might have confused them).

    Patrick
     
  8. May 19, 2006 #7
    yes, that's what I meant, I changed it in the original to be clearer.

    Thanks,

    ~Lyuokdea
     
  9. May 20, 2006 #8
    I see what you mean but I don't think he is comfortable with the idea of reference in potentials because the question as it stands then has no "unique" answer. The fact that the body is on the earth gives it a potential energy but if that happens to be your datum, you say its zero. Thats not mathematically incorrect either.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Potential energy and gravitation
Loading...