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Total mechanical energy of a point mass in an elliptical orbit

  1. Jul 25, 2012 #1
    Hey,

    I was trying to prove to myself an expression for the total energy, and I got stuck

    Here is the picture (it's transparent, so I won't embed it here).

    The problem I seem to be having is when I observe these two points where the speed is in its highest and its lowest. Since in these points is the force perpendicular to the velocity vector, I conclude that the acceleration in these points must only have its centripetal component, and therefore the centripetal force and the gravity force must be the same
    [itex]\frac{mv_1 ^2}{r_1} = \frac{GMm}{r_1 ^2 },[/itex]

    and similarly

    [itex]\frac{mv_2 ^2}{r_2} = \frac{GMm}{r_2 ^2 }.[/itex]

    The total energy must remain constant

    [itex]E = K + U = \frac{mv_1 ^2}{2} - \frac{GMm}{r_1} = \frac{mv_2 ^2}{2} - \frac{GMm}{r_2},[/itex]

    and using the results above, I get

    [itex]E = - \frac{GMm}{2r_1 } = - \frac{GMm}{2r_2 },[/itex] which is obviously wrong.

    I can't seem to see where my reasoning went wrong? Maybe someone could help me? Any help is appreciated!

    Thanks.
     
  2. jcsd
  3. Jul 25, 2012 #2

    CWatters

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  4. Jul 25, 2012 #3
    Hi,

    Thank you for your response, and I am aware of google and wikipedia, however I am trying to find a mistake in my reasoning above.
     
  5. Jul 25, 2012 #4

    Drakkith

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    My math is a bit rusty, but where did the two MV2/2 terms go?
     
  6. Jul 25, 2012 #5
    Hm, if you're referring to the speed of the other body, I just assumed it's zero.

    As in, if I am looking at the motion of the satellite orbiting the Sun, I assume that the Sun is stationary in its position.

    Edit: Oh, and if you are talking about my equations above, then

    [itex]\frac{mv_1 ^2}{r_1} = \frac{GMm}{r_1 ^2 },[/itex]
    you divide by two and multiply by [itex]r_1[/itex]
    [itex]\frac{mv_1 ^2}{2} = \frac{GMm}{2r_1 },[/itex]
    and then
    [itex]E = K + U = \frac{mv_1 ^2}{2} - \frac{GMm}{r_1} = \frac{GMm}{2r_1 } - \frac{GMm}{r_1} = - \frac{GMm}{2r_1}.[/itex]

    The procedure is identical for the other point.
     
    Last edited: Jul 25, 2012
  7. Jul 25, 2012 #6

    Drakkith

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    I could be mistaken, but I don't think you can multiply mv2/2 and GMm/r1 together, you need to subtract.
    Edit: Perhaps I am misreading the math. If so, just tell me to shut up and I'll go away lol.
     
  8. Jul 25, 2012 #7

    mfb

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    The last conclusion is wrong. If those forces would be the same, the object would move in a circular orbit, which is wrong in the general case. In the point of largest velocity, the gravitational force is smaller (and therefore the object begins to increase the distance), at the other point the gravitational force is larger.


    1000. Post :D
     
  9. Jul 25, 2012 #8

    Drakkith

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    How is the force smaller at it's closest approach?
     
  10. Jul 25, 2012 #9
    I think the very first line is wrong. The acceleration is not equal to mv^2/r, that formula applies to circular motion. If you like you can imagine the motion is instantaneously circular when the acceleration is perpendicular to the velocity, which is the argument you seem to be making, but the centre of this circle will not be the central mass, it could be anywhere. The mv^2/r applies only if you take r to be the radius of curvature at that point, which is different in general from the distance to the central mass. In fact they will only be equal when the orbit is a circle I think, so you've probably derived the total energy for a circular orbit.
     
  11. Jul 25, 2012 #10
    @mfb Well, they are not the same all the time, they are only the same in these particular points, in the case of an ellipse, that is.

    If they were not the same in those points, wouldn't that mean that there would be a tangential acceleration as well, and therefore a tangential force? We know there isn't one.

    @Drakkith I am sorry, I am not following you at the moment, but it could be because I am tired :)

    @TobyC Yeah, that I was the argument I was making. Thank you, that haven't crossed my mind! I shall ponder about it!
     
    Last edited: Jul 25, 2012
  12. Jul 25, 2012 #11

    mfb

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    The gravitational force at the point of closest approach is smaller than the force required for a circular orbit (centrifugal force in a rotation coordinate system).


    No. Velocity and acceleration are always perpendicular at those points, independent of their magnitude. The magnitude of gravity determines the radial force. And this is not zero.
     
  13. Jul 25, 2012 #12

    Drakkith

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    I'm sorry, I'm trying to understand it myself. Now, in the equation E=K+U, can I ask why you are using GMm/r as the potential energy, the U term? And is MV2/2 the kinetic energy part?
     
  14. Jul 25, 2012 #13

    Drakkith

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    I don't think I'm following you here. Are you saying that the magnitude of the force is less during closest approach than when further away? That doesn't make sense to me as the force drops off with distance. Perhaps I am misunderstanding terminology or something.
     
  15. Jul 25, 2012 #14
    The second Newton's law tells us that the force vector has the same direction as the acceleration vector.
    Therefore, since the acceleration is normal to the direction of the velocity, as you say, it cannot have tangential acceleration, only centripetal.
    Therefore, the centripetal force is equal to the gravity force in that point.

    So we agree?

    I think the answer to the puzzle is what TobyC suggested.

    We are talking about gravitational force here. -GMm/r is the expression for the gravitational energy - it is the negative gradient of the gravitional force. Does this make sense to you?

    And yes, that is the kinetic energy part :)
     
  16. Jul 25, 2012 #15

    Drakkith

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    Ah, now it makes sense. Sorry about all that!
     
  17. Jul 25, 2012 #16
    The error consist in assuming the same "radius" r1 on both sides.
    The centripetal force is given by
    [itex]\frac{mv_1 ^2}{R} [/itex]
    where R is the radius of curvature at that point.
    For ellipse this is R=b^2/a at the points situated on the opposite ends of the large axis.

    To see that the energy is conserved, you may express the r1 in the gravitational force in terms of the parameters of the ellipse as
    r1=a(1-e)
    where
    [itex]e=\sqrt{1-\frac{b^2}{a^2}}[/itex]
    At the other end of the orbit, r2=a(1+e).
    R and r1,r2 are not the same.
    If you express the energy in terms of the parameters of the ellipse you will see that is independent of the position on the trajectory.
     
  18. Jul 26, 2012 #17

    mfb

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    nasu's explanation is a different way to say the same thing. Maybe it is clearer if stated like this, but I'll try to rephrase my point:

    At the point of closest approach, the gravitational force is [itex]\frac{GMm}{r_p^2}[/itex].
    If the object would move on a circular orbit, this would be the same as the centrifugal force (in rotating coordinates) or centripetal force for a circular motion (in an inertial system) [itex]\frac{mv_p^2}{r_p}[/itex]. But it not in a circular orbit! Clearly the kinetic energy and velocity is higher, as it can increase its distance afterwards:
    [tex]\frac{mv_p^2}{r_p} > \frac{GMm}{r_p^2}[/tex]

    At the opposite point, the gravitational force is [itex]\frac{GMm}{r_a^2}[/itex]. If the object would move on a circular orbit, this would be the same as the centrifugal force (in rotating coordinates) or centripetal force for a circular motion (in an inertial system) [itex]\frac{mv_a^2}{r_a}[/itex]. But it not in a circular orbit! Clearly the kinetic energy and velocity is lower, as it comes closer afterwards:
    [tex]\frac{mv_a^2}{r_a} < \frac{GMm}{r_a^2}[/tex]

    Now, write down the total energy:
    [tex]E =\frac{mv_p^2}{2} - \frac{GMm}{r_p^2}[/tex]
    [tex]E = \frac{mv_a^2}{2} - \frac{GMm}{r_a^2}[/tex]
    You cannot replace the gravitational potential by a kinetic energy here, as we only have inequalities. A closer look at the details of the ellipse might give equations for them, but you would get a prefactor depending on the excentricity.
     
  19. Jul 26, 2012 #18
    It is rather a more quantitative. I like your explanation, which explain why the OP approach does not work for ellipse.

    However I feel that it may convey the idea that the centripetal acceleration and force are associated only with a circular orbit, which is not the case.
    Newton's second law F=m*a_cp still is applies for these points, only use the right expression for a_cp, by using the radius of curvature at that point and not distance to the focal point.

    And you can still relate the kinetic and potential energies, if you write the right formulas, with the correct radius of curvature. So the OP approach is OK in principle, just use the right parameters, for the specific curve.
     
  20. Jul 26, 2012 #19

    CWatters

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    I don't think that's correct either. Otherwise if you set them going at one of those points it would continue in a circle. There needs to be a second order term?

    Try..

    http://en.wikipedia.org/wiki/Vis-viva_equation#Derivation
     
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