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Find total mechanical energy

  1. Jul 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A satellite of mass m is orbiting the earth. The radius of the orbit is [tex]r_{0}[/tex] and the mass of earth is [tex]M_{e}[/tex].

    a. Find total mechanical energy.
    b. Now suppose that satellite encounters constant frictional force f which retards its motion. The satellite will spiral to the earth. Assume the radius changes so slowly that you can treat the satellite as being in circular orbit of average radius r. Find approximate change in radius per revolution of satellite.

    I found the first a. I think it's -[tex]\frac{GmM_{e}}{r_{0}}[/tex].
    But then for part b, I assume that the frictional force is in the direction of its velocity. And so, I think I have [tex]\frac{\partial^2\theta}{dt^2}[/tex] = f. But then I am not sure what to do after that? I also know that the work done by f is equal to the change in energy? Thanks.

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jul 17, 2009
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  3. Jul 17, 2009 #2
    Re: Gravitation

    Um, I meant d^2theta/dt^2 = f. BUt somehow Latex came out the other way.
     
  4. Jul 17, 2009 #3

    alphysicist

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    Re: Gravitation

    Hi bodensee9,

    No, I don't think that is correct. Remember that they ask for the total mechanical energy.

    That cannot be the right relation, since it does not have the same units on each side of the equation.

    Instead, I would suggest thinking about how much work is done by the frictional force every time the satellite goes around once. How will that work affect the total energy that you found in part a?
     
  5. Jul 18, 2009 #4
    Re: Gravitation

    Hi

    Oh, right. So the total mechanical energy E would be then [tex]\frac{l^2}{mr^2} - \frac{GmM_{e}}{r}[/tex], where l is the angular momentum of the satellite. And you can find l because for a circular orbit, [tex]\frac{dE}{dr}= 0[/tex].

    Okay, I think the amount of work done by friction as the satellite goes around once would be [tex]2fr\pi[/tex]? Hm .. as the satellite gets closer, the speed increases but potential energy decreases. So then the change in energy is then [tex]2fr\pi[/tex]? Then would I solve for r using the energy equation above?

    Thanks.
     
    Last edited: Jul 18, 2009
  6. Jul 18, 2009 #5
    Re: Gravitation

    sorry, i meant dE/dr = 0, but somehow the latex came out the other way.
     
  7. Jul 18, 2009 #6

    alphysicist

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    Re: Gravitation

    I don't think that's right; it looks like you're missing a numerical factor.

    But in the end I think you want to find a single term for part a. Just use translational kinetic energy:

    [tex]
    E_{\rm total}=\frac{1}{2}mv^2-\frac{GMm}{r}
    [/tex]

    Since it is a circular orbit, you can also find v in terms of r. Once you plug that back into this energy expression, you'll get a much simpler form that makes solving part b easier.

    Okay, but don't forget the sign. Friction is causing the satellite to lose energy.

    Yes, I think you would want to use the work-energy formula for this, since energy is not conserved.
     
  8. Jul 19, 2009 #7
    Re: Gravitation

    Thanks!
    You're right; I meant to have a 1/2 in front of the l^2/mr^2 term. For some reason I'm having trouble with Latex.
     
  9. Jul 19, 2009 #8

    alphysicist

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    Re: Gravitation

    That's right. And here, angular momentum is not conserved, so you should eliminate L, and get an expression for the total energy that has r (but not L or v).
     
  10. Jul 20, 2009 #9
    Re: Gravitation

    Okay, since I will have
    [tex]\frac{mv^2}{r}[/tex] = [tex]\frac{GmM}{r^2}[/tex],
    so then my [tex]v^{2}[/tex] = [tex]\frac{GM}{r}[/tex].
    Then plugging in this means that my total energy at any given r is
    [tex]\frac{-GmM}{2r}[/tex].
    Then I would have:
    [tex]\frac{-GmM}{2r_{i}}[/tex] = [tex]\frac{-GmM}{2r_{f}}[/tex] - [tex]2r_{i}f\pi[/tex]?

    Where [tex]r_{i}[/tex] is the initial radius and [tex]r_{f}[/tex] is the new radius after one revolution has passed.
    Then I will have

    [tex]GmM(\frac{r_{f}-r_{i}}{2r_{i}r_{f}})[/tex] = [tex]2r_{i}f\pi[/tex].

    So then can I make the approximation that since r_final and r_initial won't differ very much from revolution to revolution, so effectively in the denominator I have [tex]2r_{i}^2[/tex]? And then I can simply the expression to get

    [tex](r_{f}-r_{i})[/tex] = [tex]\frac{4r_{i}^3f\pi}{GmM}[/tex]

    Thanks! Sorry this is taking so long.
     
  11. Jul 20, 2009 #10

    Redbelly98

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    Re: Gravitation

    You're very nearly right.

    A couple of questions:

    Do you expect that:

    • the initial energy is less than or greater than the final energy?
    • the final radius is smaller or larger than the initial radius?

    Think about the answers to those questions, and whether these equations are consistant with or contradict them:

     
  12. Jul 20, 2009 #11
    Re: Gravitation

    Um, I thought that the potential energy will be less than the initial energy (because the done by friction will actually go into the increase in kinetic energy?) and I thought the final radius will be smaller than the initial radius...? Thanks.
     
  13. Jul 20, 2009 #12
    Re: Gravitation

    oh right, I should have -GmM/r_initial - 2(pi)r_initial*f = -GmM/r_final. Thanks!!
     
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