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Time-Averaging the Potential Energy

  1. Oct 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Screenshot (25).png

    2. Relevant equations
    Gravitational potential energy
    3. The attempt at a solution
    Isn't this the solution? Screenshot (26).png
    So Is <U> = U(r) where a = r? How do I incorporate the ##\frac {1}{\tau}## and the given "useful definite integral" from zero to 2π? Thanks
     
  2. jcsd
  3. Oct 16, 2015 #2
    I tried taking the integral but I got undefined anwer
    Gravitational potential energy = ##U = \frac {-GMm}{r^2}##
    ##-GMm \int_{0}^{a} \frac{1}{r^2}dr##
    ##-GMm (\frac {-1}{a} + \frac{-1}{0})##
     
  4. Oct 16, 2015 #3
    I thought the limits are final on top and initial on the bottom. We're going from the surface (0) to a distance a away from the planet. Why should the bottom limit be infinity?
     
  5. Oct 16, 2015 #4
    I think I understand now, in the second picture of the first post, they're bringing an object from infinitely far away to a radius of r. To clarify, I think ##F = \frac {GMm}{r^2}## is negative because it is an attractive force. Can this reasoning be justified? So for post #2, I'll just replace 0 with a , and the top with infinity. ##\int_{a}^{\infty}##. That would give me a positive value because I must put in work to move the object from a distance of a to infinity.
     
  6. Oct 16, 2015 #5
    As a side note this seems to indicate that the formula for potential energy breaks down if I cross the Earth's core?
     
  7. Oct 16, 2015 #6
    More þan one way to do it?
     
  8. Oct 16, 2015 #7
    Now I have the solutions guide but it confuses me even more. Where do the marked definitions come from? Screenshot (27).png

    This is an online solutions guide, so I don't know what the teacher's lecture is.
     
  9. Oct 16, 2015 #8

    Nathanael

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    Yes, this is because that is how gravitational potential energy is (typically) defined.
    Yes that is right. It's truly a vector equation. The force (from A on B) points along the radius vector (from A to B), but in the opposite direction (because it's attractive), and so there is the negative sign.
    You could do that. Now your distance of zero potential energy is "a." The reason it is typically defined the other way around (from infinity to a) is so that the distance of zero potential energy is infinity.
    It breaks down before then, namely once you go below the surface of Earth. The reason is that the force equation assumes the Earth to be a point-object. Once you go below the surface, some of the mass will be above you, and so the force equation must be modified.



    No it's not. That is the way to get the potential energy function from the law of gravity, but that is not what the problem asked for. Re-read the definition of the time average <f> of a function f. You want to apply this operation <U> to the potential energy U.
     
  10. Oct 16, 2015 #9
    Here is the second page to the solution
    Screenshot (29).png
     
  11. Oct 16, 2015 #10
    It's frustrating when I don't understand the solutions guide
     
  12. Oct 16, 2015 #11

    Nathanael

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    I missed your post #7 at first.

    The first equation you marked can be understood algebraically. ##\frac{1}{\frac{d\theta}{dt}}=\frac{dt}{d\theta}##.
    Then they just used the dot notation (it's just shorthand convention for writing time derivatives) ##\dot \theta \equiv \frac{d\theta}{dt}##

    The second equation you marked can be understood geometrically. h is (twice) the speed at which area is swept out by the position vector (sometimes called "areal velocity").
    The understanding is geometrical, but I will try to give you the picture with words. In a short amount of time dt, the position vector sweeps out a thin triangle. The length of the long side will be R and the length of the short side will be Rdθ. The formula for the area of a triangle is half the product of these sides. So then the area per time dt is 0.5R2dθ/dt=0.5h
    The fact that h is constant is one of Kepler's laws. It can also be understood in terms of angular momentum. (That law of Kepler's is equivalent to saying angular momentum is conserved.)
    I hope you have been taught one of these two things (Kepler's law or that angular momentum is conserved in orbit). That is where that equation comes from.

    It's more fun to not check the solution and be confused than it is to check the solution and be frustrated :wink:
     
  13. Oct 16, 2015 #12
    Just to clarify, it seems that time average <f(x)> is very similar to the average value of an integral avg-val.jpg
     
  14. Oct 16, 2015 #13
    I can't find anything about h in my textbook
     
  15. Oct 16, 2015 #14
    Yes, I have been taught both, but I can't apply it to this context.
    I know that the second law says "...a line between the sun and the planet sweeps equal areas in equal times."
    Angular momentum is conserved if there is no net torque and it is ##L = I \times \omega ##
     
  16. Oct 16, 2015 #15
    So it looks like h is defined as #h = R^2 d\theta ## But if we let more time pass, then the angle theta will be greater, so shouldn't h be greater to? Thus h changes and isn't constant.

    Kepler's second law is just another way to express angular momentum?
     
  17. Oct 16, 2015 #16

    Nathanael

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    Yes, they are exactly the same in logic. The only difference of course, is that the time average uses time as the independent variable.
    Is the problem from your textbook? Is the solution from your textbook? Does your text talk about the polar equation of ellipses?
    You forgot the dt in there, ##h=R^2 \frac{d\theta}{dt}##
    Yes, Kepler's law of areas implies conservation of angular momentum in orbit. The speed at which area is swept out is ##\frac{1}{2}R^2\frac{d\theta}{dt}=\text{constant by Kepler's law}## (I explain this formula in post #11). The equation for the angular momentum is ##mR^2\frac{d\theta}{dt}## which you can now see is 2m times the speed at which areas are swept out. Since this areal-speed is constant, and m is constant, that means that the angular momentum is constant.

    Kepler's law is from observations, though. The theoretical reason angular momentum is conserved is because the force of gravity always acts along the radius vector, and so ##\text{torque}=\vec R \times \vec F= 0##
     
  18. Oct 16, 2015 #17
    I thought the further away the object is from Earth, the more potential energy it has. For example, a rock held 10 feet above the ground has more potential energy that one held only one foot.
     
  19. Oct 16, 2015 #18
    No it is not
     
  20. Oct 16, 2015 #19

    Nathanael

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    How much potential energy something has is really arbitrary and meaningless. How much potential energy something has is just an artifact of where you choose the zero to be. If you're dealing with earthly problems, you might choose the surface to be the location of zero potential energy. Typically with astronomical problems (like this one) you will choose the zero to be at infinity.

    If it doesn't matter the magnitude of the potential energy, then what does matter? The important detail about potential energy functions is how they change. Changes in potential energy are independent of where you choose the zero to be.
    Choosing a different zero just shifts the potential energy function, say from U to U+C where C is constant, but you see ##(U_f+C)-(U_i+C)=U_f-U_i##
     
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