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Need help - body in central force-field

  1. Dec 17, 2007 #1
    [SOLVED] need help - body in central force-field

    Hi,

    I need help for the following issue:

    A body B is falling freely on a straight line towards a body A, which emits a central force-field. The formula for acceleration is:

    a(r)=Gm/r^2

    I want to derive a formula both for:

    v(t)=[?]

    and

    s(t)=[?]

    where at t_0=0 the velocity of body B is v(t_0)=0 and distance s(t_0)=0 (the initial moment of body B falling on body A).

    Thanks for any help in advance,

    chemical master.
     
  2. jcsd
  3. Dec 17, 2007 #2

    Shooting Star

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    Write a = dv/dt = (dv/dr)(dr/dt) = vdv/dr = Gm/r^2. Now you can integrate. (Are you sure there wasn't a minus sign on the RHS?)
     
  4. Dec 17, 2007 #3


    Really thanks, shooting star.

    Yeah, I think it was a(r)=-Gm/r^2

    I'm really ashamed of myself, but I now have a problem with doing that.
    If I integrate: Integrate[-Gm/r^2, {r, -Infinity, Infinity}] i get a(r)=Gm/r and vdr/dt=Gm/r

    First of all I doubt that it's correct, for second I don't know what to do next. How can I integrate -Gm/r^2 with respect to t if there's no t. Could you help?
     
  5. Dec 17, 2007 #4

    Shooting Star

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    Why the plus and minus infinities? You have been given the initial conditions. If you integrate, you'll have the constants of integration, whcih you can find by puting in the given conditions.

    I just noticed that you have written S=0 at t=0 . Look at the problem again. There was probably a given distance S at t=0 and v=0. The body is just falling from rest from a certain distance.
     
  6. Dec 17, 2007 #5
    First of all I would like to underline that this is not a homework nor a problem. I'm just very curious about the evolution of velocity from r=a (for instance) in time and the relation between the distance from a to b (on r) and time intervals. I knew from the first time it has something to do with derivatives, but still can't figure out how to integrate just do obtain the v(t) and s(t) of the falling body.

    Any ideas?
     
  7. Dec 18, 2007 #6

    Shooting Star

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    Formulate the problem a bit more precisely. Suppose S is the initial dist of the falling body at t=0, where you have just dropped it from rest, which means v = 0. You want to know how r and v changes with time. When the falling body reaches the pulling body, the speed becomes infinite. So, we will avoid putting r=0.

    We got previously v.dv = -(GM/r^2)dr. Integrating, we get,

    v^2/2 = -GM/r + A, where A is a constant.

    Evaluate A by putting in the initial conditions. Then integrate again .
     
  8. Dec 18, 2007 #7
    Ok, I get

    v^2/2 = -GM/r + v_0^2

    v(r)=Sqrt(-2Gm/r + 2v_0^2)

    Is that correct?

    Suppose the initial r is 5 (r=5, t_0=0, v_0=0)

    I want to get v(t) and r(t), not v(r). I want the formula to have a time variable.

    Any thoughts?

    Thanks.
     
  9. Dec 18, 2007 #8

    Shooting Star

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    v = dr/dt, so you can integrate again. Your v(0) = 0 will make things simpler.
     
  10. Dec 18, 2007 #9
    How can I integrate: Integrate[-Gm/r, dt] if in "-Gm/r" there's no "t"?

    :confused:
     
  11. Dec 18, 2007 #10

    Shooting Star

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    Bring all the 'r's to one side and take t to the other side.

    ( I'd posted this line more than one hour back, but somehow it didn't take it.)
     
  12. Dec 18, 2007 #11
    Can you give me a direct result, I'm getting more and more confused! Pls.

    :confused:
     
  13. Dec 18, 2007 #12
    Anybody?
     
  14. Dec 18, 2007 #13
    v(r)=Sqrt(-2Gm/r + 2v_0^2)

    now put v=dr/dt

    dr/dt=sqrt(-2Gm/r + 2v_0^2)

    or, dt=dr/sqrt(-2Gm/r + 2v_0^2) now solve...
     
    Last edited: Dec 18, 2007
  15. Dec 18, 2007 #14
    How? You cannot solve such an eqation! If so, show me how... :smile:
     
  16. Dec 18, 2007 #15

    rl.bhat

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    Since the acceleration is the function of r only, you cannot exprexss v in terms of t only, but you can express v and s in terms of r only.
    The first integration gives v^2/2 = -GM/r + A, where A is a constant. If B is falling from a distance R from A, v = o when r = R. So Constant A = GM/R and v^2 = 2(GM/R - GM/r) Similarly v = ds/dt = ds/dr*dr/dt = ds/dr*v of ds = dr. When you integrate you get s = r + C. When r = R, s = 0. So C = -R and s = r - R
     
  17. Dec 18, 2007 #16
    thanks for this rl.bhat
     
  18. Dec 18, 2007 #17
    Ok, thanks, but suppose I wish to measure the time from when s_0=0 (v=0) to s_1. Just how could I calculate how long would it take?

    I cannot calculate it neither from v nor from a since it's changing in terms of r.

    Any hints?
     
  19. Dec 18, 2007 #18

    Shooting Star

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    What is 's' and 'r' that you've used?
     
  20. Dec 18, 2007 #19
    I would like to indicate that i need a solution to velocity and distance from r=R to the center IN TERMS OF TIME, not r. Any suggestions would be gratly appreciated.

    Thanks,

    chemical master
     
  21. Dec 18, 2007 #20

    Shooting Star

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    All right, let's start from where we had left.

    v^2/2 = GM/r + A. If v=0 at r=R, then A=-GM/R. This means that at a dist of R from the centre, you are simply allowing the body to fall with initial speed zero, ina st line towards the centre of force. I presume that's what you had wanted.

    v = sqrt(2GM(1/r -1/R) => sqrt(2GM/R)dt = dr*sqrt(r/(R-r)), (where v= dr/dt)

    Can you do this integral by substituting r=Rsin^2y? You’ll get ‘t’ in terms of r. Unfortunately, that eqn can’t be inverted analytically. :grumpy:

    I might as well give you the answer, since I’ve spent quite some time doing it.

    Sqrt(2GM/R)*t = arcsin(sqrt(r/R)) –sqrt[r(R^2-r^2)/R^(3/2)] - pi/2.

    You may recheck my calculation to find any careless algebraic mistakes.
     
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