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Velocity when falling into planet

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi! I would need a little help with the following problem:
    We have found a new planet with density ρ and radius R, and drill a hole to its center. Then accidentally, one person falls into the hole. What is his velocity when reaching the bottom (the center of the planet)?

    2. Relevant equations
    Force of gravity: F = G *M *m/22
    Gravitational portential: E = (-) GMm/r
    When inside a homogen sphere with mass, the gravitational forces of the sphere cancel --> only the mass inside the spehere will accelerate an object (if the object is outside that mass).
    Kinetic translational energy: E = mv2/2
    acceleration of simple harmonic oscillator: y''=(-) ω2*y
    velocity of simpple harmonic oscillator: y'=ω*y


    3. The attempt at a solution
    I attempted to solve the problem with energy conservation. I thought that in point A on the surface the energy is E = GMm/r , where M is the mass of the planet and m of the person falling. Then, at the bottom the whole gravitational energy has been cocnverted to kinetic energy, thus:
    v = (2GM/R)^(1/2)

    However, this answer showed to be one factor 2^(1/2) wrong. My book proposes the following solution:

    The acceleration of the object (person) is:
    a= F/m = G M(r)/r2 = G/rr * 4πr3*ρ/3 = 4πGρ r/3
    (because the mass affecting the object decreases as it falls)
    From this equation it is visible that the force and acceleration of the object is proportional to its distance from the middle of the planet, where it is in equilibrium. This is just as in the case of simple harmonic motion (in this case with amplitude R). Thus, using y''=(-) ω2*y , we get:
    ω = (4πGρ/3)^(1/2)
    therefore the velocity at the middle of the planet (the maximal velocity in the equilibrium position) is:
    v = ω*A = (4πGρ/3)^(1/2)*R = (4πR3Gρ/3R)^(1/2) = (MG/R)^(1/2)

    This answer does, as previously mentioned, resemble the one obtained with energy conservation, but it differs a factor 2^(1/2). My question is: Why do I miss a factor of 2^(1/2) when using energy conservation? How should I modify my solution to obtain the correct answer with energy conservation?

    Thank you! :)
     

    Attached Files:

  2. jcsd
  3. Nov 2, 2015 #2

    PeroK

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    What is the potential energy of something at the centre of the planet?
     
  4. Nov 2, 2015 #3
    Well, because it can be seen as an object inside a sphere with thickness/radius R, and because the gravitational forces on it cancel, I would say that the potential energy is zero. Am I wrong?
     
  5. Nov 2, 2015 #4

    PeroK

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    Potential is only 0 at a very large distance from a massive body. Potential does not cancel out like force. If you are, say, half way between two identical masses, then the force is 0, but the potential is twice what it is for each mass.

    At the centre of a sphere, the potential is the integral over all points of the sphere. And that won't be 0.
     
  6. Nov 2, 2015 #5
    Oh, that was a useful example. Thank you! :)

    So how can I take the integral over all points of the planet? Should I divide it into thin spheres?
     
  7. Nov 2, 2015 #6

    PeroK

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    Yes.
     
  8. Nov 2, 2015 #7

    haruspex

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    @Alettix , note also that you ignored the sign of the potential at the surface. That should have been a clue.
     
  9. Nov 2, 2015 #8
    Does EP = 2πgρMR2 sound right? If I have calculated properly, it gives the right value for v.
     
  10. Nov 2, 2015 #9
    I thought about it like: Change in energy = final potential - potential at the surface = 0 - (-GmM/R) = GmM/R
    so that's why the sign dropped. Now I do however know that the final potential in the middle is not zero.

    But do you have any other suggestion of how the sign should be handled? I do struggel a bit with it sometimes.
     
  11. Nov 2, 2015 #10

    haruspex

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    That would give you the increase in GPE, so a negative change in KE.
     
  12. Nov 3, 2015 #11

    PeroK

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    Also, you know that the potential is 0 at "infinity" and that is greater than at the surface. So, you don't fall off the planet!

    If potential were 0 at the centre, then likewise you wouldn't fall into the planet.
     
  13. Nov 3, 2015 #12
    1. Set '0' potential energy at the center of the planet.
    2. Calculate the energy needed to raise mass 'm' from the center to the surface. This is your available potential energy.
    3. Equate this potential energy to kinetic energy of 'm' and you'll get the correct answer.
     
  14. Nov 3, 2015 #13
    Thank you very much everybody, I think I got it now! :)

    The only thing I still wonder is if any of you know if -2πgρMR2 is the correct potantial in the middle of the planet (if the potential at the surface is -GMm/R)?
     
  15. Nov 3, 2015 #14

    haruspex

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    Nearly right, but to be clear these are gravitational potential energies, not potentials. Potential is independent of the test mass, like voltage is independent of the test charge.
     
  16. Nov 4, 2015 #15

    PeroK

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    Since you know the answer to the original problem (KE at the centre of the planet), you can deduce the PE at the centre from that. In any case, you should be able to simplify the expression you have so that it looks similar to the one for the surface.
     
  17. Nov 4, 2015 #16
    Yes, I am able to to that. My orginal expression was E = -2πGmR2ρ (there were some typos with the capital letters), which can be simplified to:
    E = - 3/2 * MmG/R if I'm not wrong. :)
     
  18. Nov 4, 2015 #17

    PeroK

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    You're not wrong!
     
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