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Time period of oscillation and gravitation

  1. Sep 30, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=62332&stc=1&d=1380567544.png


    2. Relevant equations



    3. The attempt at a solution
    I really don't know how to start with this problem. The four point masses of mass m oscillate together so I am confused as to how should I begin making the equations. Just a guess, should I write down the expression for potential energy of the system?

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Sep 30, 2013 #2

    mfb

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    Staff: Mentor

    Sure. Add the kinetic energy for the mentioned oscillation afterwards.
    This is just a harmonic oscillator (for small excitations).
     
  4. Oct 1, 2013 #3
    This is what I get:
    [tex]E=\frac{2GMm}{a}+\frac{2GMm}{b}+\frac{4Gm^2}{\sqrt{a^2+b^2}}+\frac{Gm^2}{2a}+\frac{Gm^2}{2b}+\frac{G(M+4m)M_e}{R}+2mv^2[/tex]
    where ##M_e## is the mass of earth and ##R## is the radius of orbit.
    Is this correct? What should be my next step?
     
  5. Oct 1, 2013 #4
    Do not compute the potential energy of the masses with respect to each other. The satellite is considered rigid, so it is a constant. Compute the the difference in potential energy of the masses with respect to the Earth if the masses are rotated a small angle from the stable position about the central mass. Use the fact that the distances are very much smaller than the radius of the orbit. Then compute the kinetic energy of this small motion.
     
  6. Oct 1, 2013 #5
    Please see the attachment.
    [tex]U_i=\frac{GM_em}{R-a}+\frac{GM_em}{R+a}+\frac{2GM_em}{R}+C[/tex]
    where C is the potential energy of central mass with respect to earth and the potential energy of point masses with respect to each other and the central mass.
    [tex]U_f=\frac{GM_em}{R-a\cos\theta}+\frac{GM_em}{R+a\cos\theta}+\frac{GM_em}{R-b\sin\theta}+\frac{GM_em}{R+b\sin\theta}+C[/tex]
    I am thinking of using the following approximations:
    [tex]\frac{1}{R-a}=\frac{1}{R}\left(1+\frac{a}{R}\right)[/tex]
    and
    [tex]\frac{1}{R-a\cos\theta}=\frac{1}{R}\left(1+\frac{a\cos\theta}{R}\right)[/tex]
    Should I replace ##\cos\theta## with ##1-\theta^2/2## for small angle approximation?

    How to find the kinetic energy? Do I have to work in the frame of CM? Is it simply ##2mv^2## where v is the velocity of each point mass?
     

    Attached Files:

  7. Oct 1, 2013 #6
    Your approximation should work, but, personally, I would just find the second derivative with regard to the angle directly, and form ##k\theta^2## as the potential energy due to the small displacement. That should eliminate any concerns about the validity of approximations.

    Regarding the kinetic energy, we only care about the motion due to the small displacement, which makes CM the frame of choice.
     
  8. Oct 1, 2013 #7
    [tex]\frac{dU}{d\theta}=GM_em\left(\frac{-a\sin\theta}{(R-a\cos\theta)^2}+\frac{a\sin\theta}{(R+a\cos\theta)^2}+\frac{a\cos\theta}{(R-a\sin\theta)^2}+\frac{-a\cos\theta}{(R+a\sin\theta)^2}\right)[/tex]
    Should I simplify this using approximation to find the second derivative or should I go on differentiating this? This seems a bit dirty.

    Should I use the approximation:
    [tex]\frac{1}{(R-a\cos\theta)^2}=\frac{1}{R}\left(1+\frac{2a\cos\theta}{R}\right)=\frac{1}{R}\left(1+\frac{2a}{R}-\frac{a\theta^2}{R}\right)[/tex]
     
  9. Oct 1, 2013 #8
    Actually, upon thinking some more, the successive approximations approach results in four terms that cancel one another.

    $$ \frac {1} {R - a \cos \theta} + \frac {1} {R + a \cos \theta} \approx \frac 1 R \left( 1 + \frac {a \cos \theta} {R} \right) + \frac 1 R \left( 1 - \frac {a \cos \theta} {R} \right) = \frac 1 R $$

    Secondly, the gravitational potential energy must be negative.
     
  10. Oct 1, 2013 #9
    The two final terms should have ##b##, not ##a##. See me previous message as well.

    There is no need to be smart here. Just differentiate the four terms treating each as a ##f(\theta)g(\theta)## product; that will give you eight terms, then you let ##\theta = 0## and find ## U \approx U_0 + U''(0) \theta^2 / 2## (throw ##U_0## away then).
     
  11. Oct 1, 2013 #10
    Ok I found
    $$U''(0)=GM_em\left(\frac{-a}{(R-a)^2}+\frac{a}{(R+a)^2}+\frac{4ab}{R^3}\right)$$

    How do you get this: ## U \approx U_0 + U''(0) \theta^2 / 2##? :confused:
     
  12. Oct 1, 2013 #11
    This is Taylor's expansion, which makes use of the fact that ##U'(0) = 0## (why is that so?).
     
  13. Oct 1, 2013 #12
    ##U'(0)=0## because at ##\theta=0##, the potential energy is minimum according to the given question.

    I used the approximations and got
    $$U=U_0+\frac{4ab-2a^2}{R^3}\frac{\theta^2}{2}$$
    What should I do now?
     
  14. Oct 1, 2013 #13
    Have you found the kinetic energy of the small oscillations? Conservation of total energy should do the rest.
     
  15. Oct 1, 2013 #14
    Is it ##2mv^2##?
     
  16. Oct 1, 2013 #15
    What is ##v## here?
     
  17. Oct 1, 2013 #16
    Velocity of point masses.
     
  18. Oct 1, 2013 #17

    mfb

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    What is the relation between v and θ?
    Can you express the kinetic energy with θ? Afterwards, compare it to the potential and kinetic energy of a simple harmonic oscillator.
     
  19. Oct 1, 2013 #18
    From conservation of energy,
    $$2mv^2=\frac{2a^2-4ab}{R^3}\frac{\theta^2}{2}$$
    $$\Rightarrow v=\frac{\theta}{2R}\sqrt{\frac{2a^2-4ab}{m}}$$
    Do you ask me the kinetic energy of a single point mass? I don't really understand where this is heading. :(
     
  20. Oct 1, 2013 #19
    How come the velocities of all the masses are equal?

    Re-read the second paragraph in #6.
     
  21. Oct 1, 2013 #20
    Do I have to use conservation of angular momentum?
     
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