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Gravitation: Looks easy at first

  1. Nov 12, 2005 #1
    Hi, all. I have a problem I wish to solve. It's not from a text book, I thought of it myself.

    If two masses of 1 kilogram each are 1 metre apart in space, how long will it take for them to reach each other from their attractive gravitational forces?

    Remember that the force (and hence the acceleration) acting on each mass increases as the distance between them decreases, (by F=Gm1m2/r^2).

    Assume that no friction or other gravitational forces act on the masses.

    Thanks.
     
  2. jcsd
  3. Nov 12, 2005 #2

    Astronuc

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    Well, assume each sphere must travel half the distance, and the acceleration is given by g = Gm/d2, where d is the separation of CM's.

    So if x is the separation distance, each sphere must travel x/2 and the separation d = x + 2r, where r is the radius of the spheres, assuming they are the same size.
     
  4. Nov 12, 2005 #3

    daniel_i_l

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    Gold Member

    Since the acceleration isn't constant, then you could find the average acceleration (by using calculus - integrate and divide) and use that to find the time.
    Or I'm not sure if you can do this (I'd use cal) but you can say that:
    [tex]
    a = \frac{Gm}{d^2} [/tex]
    [tex]
    d = \frac{Gmt^2}{2d^2} [/tex]
    [tex]
    d = \sqrt[3]{\frac{Gmt^2}{2}}
    [/tex]
    for d you want 2r. Put that in and solve for t.
     
    Last edited: Nov 12, 2005
  5. Nov 13, 2005 #4
    Ok, let me show you what I did before posting here.
    [tex]
    F = \frac{{Gm_1 m_2 }}{{r^2 }}
    [/tex]
    But [tex]
    m_1 = 1, m_2 = 1
    [/tex]
    and [tex]
    F = ma
    [/tex]
    So:
    [tex]
    a = \frac{G}{{r^2 }}
    [/tex]
    Now, r is the distance between the two masses, so taking the initial position of the first mass as the origin, r can be expressed as:
    [tex]
    r = d_2 - d_1
    [/tex]
    But, [tex]
    d_2 = 1 - d_1
    [/tex]
    So:[tex]
    r = 1 - 2d_1
    [/tex]
    And therefore: [tex]
    a = \frac{G}{{(1 - 2d)^2 }}
    [/tex]
    Integrate twice to get:
    [tex]
    d = \frac{G}{{2(1 - 2d)^2 }}t^2
    [/tex]
    Now, total distance travelled (D) can be expressed as the integral of distance (s) from 0 to a certain time (t) with respect to time.
    [tex]
    \int_0^t s \,dt = D
    [/tex]
    We already know that D = 1/2. It's the distance one of the masses travels before colliding with the other mass. So we get the integral:
    [tex]
    \int_0^t s \,dt = \frac{1}{2}
    [/tex]
    However, the problem arises when I try and rearrange this to have d in terms of t:
    [tex]
    d(1 - 2d)^2 = \frac{G}{2}t^2
    [/tex]
    It's obviously a cubic, which makes it difficult to rearrange.
    Do I need to rearrange it to integrate it? Is there a simpler method of doing this?
    Thanks.
     
    Last edited: Nov 13, 2005
  6. Nov 13, 2005 #5

    Kurdt

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    The problem with this is that the acceleration equation is an instantaneous equation. We know the acceleration necesarily depends on time because as the two objects get closer the force between them gets larger and therefore the acceleration increases too. The time dependance comes merely from the distance between the two particles. You can model this by saying da/dt (that is the rate of change of acceleration) is a constant and integrating thrice to obtain how r(t) changes with time. The only problem then comes in solving what da/dt is. I suspect you could put r(t) into the instantaneous acceleration equation then you have two unknowns (da/dt and t) and two equations. I'll have a bash at it later on and see what i get because I'm busy right now but your question caught my attention.

    Good luck solving it!
     
  7. Nov 13, 2005 #6
    Hmm, on second thoughts what I've done is completely wrong.
    When I integrated this:
    [tex]
    a = \frac{G}{{(1 - 2d)^2 }}
    [/tex]

    I accidently assumed that d was constant, which it isn't. It's a function of t, so it has to be integrated itself.
    So the big problem is finding d in terms of t.
    Any ideas?
     
  8. Nov 13, 2005 #7

    Kurdt

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    Like i say if you assume the rate of change of acceleration is constant then r(t)= (1/6)(da/dt)t^3 plus some arbitrary constants. Its a very difficult problem to solve but very intriguing.
     
  9. Nov 22, 2005 #8
    So has anybody actually solved this yet?
     
  10. Nov 22, 2005 #9

    dx

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    No, its either
    [tex]\int{ds}[/tex]
    or
    [tex]\int{v} {dt}[/tex].
     
  11. Nov 22, 2005 #10

    mezarashi

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    To spiffy:

    Due to the symmetry, it is easiest to analyze the problem from the perspective of one of the masses. The force acting on it is from the equation governing gravitation. And the acceleration is simply that divided by mass. Now as for the kinematics analysis:

    Create an axis in the middle between the two masses. Create two vectors r1 and r2 from this center to the masses. Also, create a starting point where mass 1 originally is such that [tex] s = 1-r_1, ds = -dr_1[/tex]. You can then apply the integral formula:

    [tex] \frac{d^2s}{dt^2} = a [/tex]

    and solve for s accordingly where [tex] a = \frac{Gm_2}{(r_1+r_2)^2}[/tex], and also due to symmetry [tex]r_1 = r_2[/tex].
     
  12. Nov 23, 2005 #11

    Kurdt

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    mezarashi:

    How do you propose to solve the problem that the acceleration changes with time?
     
  13. Nov 23, 2005 #12

    mezarashi

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    The general equation

    [tex] \frac{d^2s}{dt^2} = a(r_1) [/tex]

    is applicable to non-constant acceleration. By using the change of variables [tex] r_1 \longrightarrow s [/tex], we can then solve the differential equation.
     
    Last edited: Nov 23, 2005
  14. Nov 23, 2005 #13

    Astronuc

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    Be careful there.

    Remember F = G (m1m2)/r2 and using F = ma, then a = G mi/r2, where r = distance between CM's.

    Also look at what mezarashi and dx have written.

    In an equation, pay attention to units, e.g. (1 - 2d) would only be valid if 1 has dimensions of length, assuming d is some distance (length).

    Also in these types of problems, be aware of the diffence between position and displacement (distance). Displacement (Distance) = difference detween two positions.
     
  15. Nov 23, 2005 #14
    Well, I gave the problem to my physics teacher.
    He solved it by considering the kinetic energy at the start and end of the system.
    It gave a value of about 37 hours.

    I tried it by assuming the rate of change of acceleration was a constant, which gave an alright approximation (24.05 hrs), but that seems to be a bad assumption to make in light of the other solution.
     
  16. Nov 23, 2005 #15

    mezarashi

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    Problems involving changing forces and changing acceleration can be solved! It's nothing mysterious, but the mathematics is more 'involved'. Take for example the case of the spring. You attach a mass m to a spring which experiences a constantly changing force F = -kx which is a function of position, the same way your gravitational scenario was. Using Newton's second law, can you find how long it will take for the mass to go up and come back down to its same position again (asking for a time, similar to your situation)?

    Solving this, you will find out how the period formula [tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex] came about.
     
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