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Rectilinear motion of two attracting masses

  1. Dec 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider two masses of variable magnitude (M m) that are separated by a distance ( r ). Both masses are free to move. Calculate dr/dt.

    2. Relevant equations
    (See below)

    3. The attempt at a solution
    F = GMm / r2
    a = Gm / r2
    let k = Gm
    da/dr = -2kr-3
    dr/da = (-2k)-1 r3
    dr/da * da/dt = dr/dt

    Just need to solve for da/dt and I don't know how to do that.
     
  2. jcsd
  3. Dec 27, 2015 #2

    haruspex

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    Are you sure you have stated the question correctly? I would expect it to ask for d2r/dt2. There is no way to know dr/dt without further information.
    The acceleration you found, Gm/r2, is only for mass M.
     
  4. Dec 27, 2015 #3
    As far as stating the equation, I honestly can't say if I've done it correctly. The question is a matter of personal interest and didn't come from any textbook. I apologise for that. I just wanted to find a relationship between the time elapsed and the distance between two unbound masses. I find it really peculiar how you can solve for d2r/dt2 and not dr/dt. Why is that exactly? You said you needed more information. Would it help at all if we said that both masses were stationary at t=0?

    Thanks for taking the time to read :)
     
  5. Dec 27, 2015 #4

    haruspex

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    You can solve for dr/dt. The simplest is to find dr/dt as a function of r. You can use conservation of energy for that. Or you can derive it from the force=mass*acceleration equation using (d/dt)(v) = (dr/dt)(d/dr)(v) = v dv/dr. Finding dr/dt as a function of t (or r as a function of t) is harder.
    Yes and no. Your original formulation asked for dr/dt as a function of r. Knowing dr/dt at a given t does not help since we don't what what r is at that time. If you specify dr/dt at some initial r then you can find dr/dt as a function of r. If you want to know r or dr/dt as a function of t then you need an initial r and an initial t.
     
  6. Dec 27, 2015 #5
    Oh neat! I hadn't considered conservation of energy. So something like this?

    1/2 mv2 = -GMm / r
    I'm cancelling out the mass (m) out from both sides because I can, but I don't know if I should.
    1/2 v2 = -GM/r
    let k = -GM
    v = (2k/r)0.5
    Where v is presumably dr/dt? In your second method you seem to distinguish between dr/dt and v, leading me to think that I'm incorrect here.

    Also, a question regarding syntax: how is "dr/dt as a function of r," different from "dr/dt as a function of t?"
     
  7. Dec 27, 2015 #6

    haruspex

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    no, that's saying KE=PE. Conservation says the total is constant (need not be zero, or even positive).
    No, dr/dt is the same as v. Just used v to make the notation easier.
    dr/dt = sin(r) would be a differential equation with dr/dt as a function of r; dr/dt=kt would be one with dr/dt as a function of t. In the present problem, you can readily find out how dr/dt depends on r, but it is much harder to find how it depends on t.
     
  8. Dec 27, 2015 #7
    Oh, right. I don't know what I was thinking. So for the conservation of energy, that equation would look more like:

    1/2 mv12 + 1/2 Mv22 - GMm/ r = c (constant)
    (I'm including the kinetic energies of both masses because gravitational potential energy is written in terms of M and m )
    The problem now is that we have two velocities to describe the change in r.
    Would I be correct in saying that abs (v1) + abs (v2) = dr/dt ?

    In the future, I may try to find dr/dt as a function of t. What would you consider to be a good starting point for that?
     
  9. Dec 28, 2015 #8
    Yes. This equation can produced by time derivation of
     
  10. Dec 28, 2015 #9

    haruspex

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    I don't see how.
    Not quite. dr/dt can be negative, and the two velocities could have opposite signs. That equation covers neither of those cases.
     
  11. Dec 28, 2015 #10
    ##x_1##, ##x_2##, ##r## are functions of time.
     
  12. Dec 28, 2015 #11

    haruspex

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    Sure, but if you differentiate the energy equation m, M and G are not going to disappear. So how will you deduce an equation that contains none of those? The relationship between the two velocities and dr/dt is purely kinematic. It owes nothing to forces.
     
  13. Dec 28, 2015 #12
    Disapear because ##F_1+F_2=0## by 3o Newton's Law.
     
  14. Dec 28, 2015 #13

    haruspex

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    The relationship between the two velocities and dr/dt has nothing to do with Newton's Laws. It is a simple fact of geometry and kinematics. You could change the gravitational law (and, consistently, energy conservation etc.) and the relationship would still hold. If you have a way of obtaining that relationship by differentiating the energy equation please post it, preferably in a private mail, not on this thread. I can imagine a way, but it goes something like this:
    - differentiate the energy equation with respect to distance, so a force equation
    - Apply N3 so that what you should end up with is 0=0, but along the way use kinematics to feed in the relationship between the velocities and the distance, and
    - voila, all you have left is that kinematic relationship
     
  15. Dec 28, 2015 #14
    This is fascinating stuff. It's pretty clear to me that I've got a lot more to learn in way of calculus and mechanics.
    Thank you all for the responses!
     
  16. Dec 28, 2015 #15

    rcgldr

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    The two objects accelerate towards a common center of mass. Let m1 be on the left accelerating to the right, and m2 on the right accelerating to the left.

    a1 = F/m1 = (G m1 m2 / r) / m1 = G m2 / r2
    a2 = F/m1 = -( G m1 m2 / r) / m2 = - G m1 / r2
    a2 = a1 (a2 / a1) = (-m1/m2)a1
    v2 = v1 (v2 / v1) = (-m1/m2)v1
    let v = rate of closure = v2 - v1 = (-m1/m2)v1 - v1 = - (1 + m1/m2)v1 = -((m1 + m2)/m2)v1
    v1 = -m2/(m1+m2) v
    v2 = -(m1/m2) v1 = -(m1/m2)(-m2/(m1+m2)) v = m1/(m1+m2) v
    Let r0 = initial distance:
    1/2 m1 v12 + 1/2 m2 v22 = G m1 m2 / r - G m1 m2 / r0 = G m1 m2 (r0 - r)/(r0 r) =
    1/2 m1 (v m2/(m1+m2))2 + 1/2 m2 (v m1/(m1+m2))2 = G m1 m2 (r0 - r)/(r0 r)
    1/2 (v/(m1 + m2))2 ((m1 m22) + (m12 m2)) = G m1 m2 (r0 - r)/(r0 r)
    1/2 (v/(m1 + m2))2 (m1 + m2) = G (r0 - r)/(r0 r)
    1/2 v2 / (m1 + m2) = G (r0 - r)/(r0 r)
    1/2 v2 = G (m1 + m2) (r0 - r)/(r0 r)
    v = - sqrt(2 G (m1 + m2) / r0) sqrt((r0 - r) / r)

    Alternate method:
    let a = d(dr/dt)/dt = a2 - a1
    a = -G (m1 + m2 ) / r2
    use chain rule to get to v as a function of r based on v(r(t))
    a = dv/dt = (dv/dr)(dr/dt) = v dv/dr
    v dv/dr = -G (m1 + m2 ) / r2
    v dv = -G (m1 + m2 ) dr / r2
    integrate and use - G (m1 + m2 ) / r0 as constant
    1/2 v^2 = G (m1 + m2 ) / r - G (m1 + m2 ) / r0 = G (m1 + m2 ) (r0 - r)/(r0 r)
    v = - sqrt(2 G (m1 + m2) / r0) sqrt((r0 - r) / r)

    Contuning from here leads to a tricky integral:
    v = dr/dt = -sqrt(2 G (m1 + m2) / r0) sqrt((r0 - r) / r)
    ##- \ \sqrt{\frac{r_0}{2 G (m1 + m1)}} \sqrt{\frac{r}{r_0-r}} dr = dt##
    one method is to define θ as
    ##r = r_0 \ sin^2(θ)##
    ##dr = 2\ r_0 \ sin(θ)\ cos(θ) dθ##
    ##θ = {sin^{-1}\left ( \sqrt{\frac{r}{r_0}} \right )}##
    ##\sqrt{\frac{r}{r_0-r}} = \sqrt{\frac{r_0 \ sin^2(θ)}{r_0-r_0 \ sin^2(θ)}} = \sqrt{\frac{r_0 \ sin^2(θ)}{r_0(1\ -\ sin^2(θ))}} = \sqrt{\frac{r_0 \ sin^2(θ)}{r_0\ cos^2(θ)}} = tan(θ)##
    resulting in
    ##-\sqrt{\frac{r_0}{2 \ G \ (m1 + m1)}} \ (2 \ r_0 \ sin^2(θ)) dθ = dt##
    ##t = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ (m1 + m2)}} \int_{θ_0}^{θ_1}2 \ sin^2(θ) dθ = -\frac{{r_0}^{3/2}}{\sqrt{2 \ G \ (m1 +m2)}} \left [ θ \ -\ sin(θ)cos(θ) \right ]_{θ_0}^{θ_1}##
     
    Last edited: Dec 29, 2015
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