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Homework Help: Momentum Planetary Motion Problem

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Two spheres of mass m and radius r, are released from rest in empty space. The centers of the spheres are separated by a distance R. They end up colliding due to gravitation attraction. Find the magnitude of the impulse just before they collide.

    2. Relevant equations

    Eg= -Gmm/r
    ek= 1/2mv^2

    3. The attempt at a solution

    I first drew my problem out with two frames.

    The first frame of the two masses separated by a distance R. The second frame, where they are just about to touch separated by a distance of 2r.

    I calculated Change in momentum for both objects.



    i found that the change in momentum was reliant upon the final velocity..

    Initially when i observe the energy of the two spheres, They are seperated by a distance of R, and have no kinetic energy. So if i calculate the Initial energy total for 1 mass, (Which is the same as the other) i get.

    Et1= Eg1
    Et1= -G*(m^2)/R

    In my second frame when i calculate total energy the two spheres are now separated by a distance of 2r. The spheres now have kinetic energy after having moved toward each other..

    Et2= Eg2 + Ek2

    Et2= -G*((m^2)/2r

    Now when i look at this question, i was thinking that it is similar to an orbital question where you have to move an object on the surface to an orbital level and it will have a certain speed... The kinetic energy in a question like that is based on the difference of the First Gravitational Potential Energy Subtract The Second.

    I came up with this relation using conservation of energy (i'm assuming no energy is lost cause they are in space and no external forces have acted upon the objects only the conservative force of gravity.)

    Eg1 = Eg2 + Ek2
    Ek2= Eg1-Eg2
    Ek2 = -G(m^2)/R - (-G(m^2)/2r)
    Ek2= G(m^2)/2r - Gm^2/R

    so 1/2mv^2 = G(m^2)/2r - Gm^2/R

    this leaves me with

    V = √(Gm(1/r - 2/R))

    so J = ΔP
    so J = mV
    so J = m*√(Gm(1/r - 2/R))

    so j = √(Gm^3(1/r - 2/R))

    This is my dilemma... I found a textbook online which has this same question stated, with the answer to the question i'm solving.... There answer is different than mine so i'm wondering if some one could point out to me what i'm doing wrong. There answer seems to have 1/2r where i have 1/r and 1/R where i have 2/R...

    There answer is:

    J = √(Gm^3(1/2r - 1/R))

    http://books.google.ca/books?id=Kuh...r centers separated by the distance R&f=false

    Question #48 P382 (SHOULD BE A DIRECT LINK)

    Anyways i would really appreciate some help cause i can't seem to find where i'm going wrong, and i'm not exactly positive if the textbook answer is even correct, because it is just stated in the question itself... I don't even really have a solid "Answer" to check my answer with other than this one... ^.^
    Last edited: Mar 4, 2013
  2. jcsd
  3. Mar 4, 2013 #2


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    2017 Award

    Staff: Mentor

    The gravitational energy you calculated is the energy of the whole system, not the energy of a single object.

    You have so many equations for something which can be done in a few lines...
    Initial energy of the system: Ei=-Gm^2/R
    Final energy of the system: Ef=-Gm^2/(2r) + 2*(1/2)mv^2

    Energy conservation gives Ei = Ef or
    => ##m^2v^2=Gm^3\left(\frac{1}{2r}-\frac{1}{R}\right)##
    => ##mv=\sqrt{Gm^3\left(\frac{1}{2r}-\frac{1}{R}\right)}##
  4. Mar 4, 2013 #3
    So taking Eg1 = -Gm^2/R is not taking the the energy of a single object? Cause i thought you did the same thing in the next line...? I don't understand can you explain what you mean please?

    And here for the EFinal

    You are taking -Gm^2/2r + (1/2mv^2 + 1/2mv^2) <=== this is where you get 2MV^2 i'm assuming(please explain)?
    but again isn't this taking the energy of the whole system? So which energy am i taking, of the entire system or of a single object?

    Thank you for your time and i would really appreciate a response <3.
  5. Mar 4, 2013 #4


    User Avatar
    2017 Award

    Staff: Mentor

    It is the energy you need to separate those objects to "infinite" distance. Once you moved one object to "infinite" distance, there is nothing to do for the other object. In that way, the gravitational potential energy of a single object is the same as the gravitational potential energy of the whole system.

    Due to symmetry, both masses have the same velocity, and 2*(1/2)=1. I don't get 2mv^2 anywhere.

    It is easier to consider the whole system, otherwise you have to split the gravitational energy in two parts without a real physical meaning.
  6. Mar 4, 2013 #5
    Thank you sir for the explanations they are clear. I appreciate all the feedback and thank you for brightening the start of my day.
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