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Time for Gravity to force two objects in a vacuum together

  1. Mar 7, 2012 #1
    If I have two, 10 kg objects, 10 meters apart, in a vacuum, How long will it take for the gravitational force between them to pull them together so that there is no room between?

    Please help I have been trying to figure how to do this math forever!!! I know the physics formula for the gravitational force is G(6.67x10^-11)m1m2/d^2... So i think you need to find the second integral? of 6.67*10^-10/(10-2p)^2 to find the time.
  2. jcsd
  3. Mar 8, 2012 #2
    It's not as hard as it seems. From Newton's law of gravitational force you have:

    [itex]\vec{F}[/itex]1→2=-G (m1m2)/(d)2 [itex]\widehat{d}[/itex] = -[itex]\vec{F}[/itex]2→1

    Now, from Newton's equations (of dynamics and cinematics) you can also get the following:



    I'm assuming you're considering 1D movement (along the x axis) and that v0=0 m.s-1. You will need to solve for a (with both objects):

    [itex]\vec{a}[/itex]1=[itex]\vec{F}[/itex]2→1/m1 [itex]\Leftrightarrow[/itex] [itex]\vec{a}[/itex]1= G (m2)/(d)2 [itex]\widehat{d}[/itex]
    [itex]\vec{a}[/itex]2=[itex]\vec{F}[/itex]1→2/m2 [itex]\Leftrightarrow[/itex] [itex]\vec{a}[/itex]2=-G (m1)/(d)2 [itex]\widehat{d}[/itex]

    Considering object 1 at the origin and object 2 at distance d from the origin:

    x1(t)=(1/2)a1t2 [itex]\Leftrightarrow[/itex] x1(t)=(1/2)G(m2)/(d)2 t2
    x2(t)=d+(1/2)a2t2 [itex]\Leftrightarrow[/itex] x2(t)=d-(1/2)G(m1)/(d)2 t2

    Finally, you equate both expressions (x1(t)=x2(t)) and you get:


    From that result you realize that the time it'll take for the two objects to come together (collide) is aproximately 108 seconds (about 3 years!), which is comprehensible acounting for the mass of both objects (20kg).
  4. Mar 8, 2012 #3


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    Mathoholic's answer is incorrect. When you write, x(t)=x0+v0t+(1/2)at^2, this assumes that a is constant, which it isn't in this problem. Since a is also varying with time, you have to set up a differential equation and integrate, as the OP said. If you write x as the particle separation, then since particle m1 experiences an acceleration:
    [tex]m_1 \ddot x = \frac{G m_1 m_2}{x^2}[/tex]
    and similarly for m2, if you add these together, you get:
    [tex]\ddot x = \frac{G (m_1+m_2)}{x^2}[/tex]
    Since you can write:
    [tex]\ddot x = \dot x \frac{d \dot x}{dx}[/tex]
    You can integrate once to get xdot as a function of x, then integrate a second time to get x as a function of t. I don't want to do the problem for you - can you try it yourself? By the way, I think Mathoholic also did the numbers wrong. I get that the final answer should be:
    [tex] t = \frac{2}{3 \sqrt{G(m_1+m_2)}}(x_0)^{3/2}[/tex]
    which, when I put in the numbers, comes out about 6E6 seconds.
  5. Mar 8, 2012 #4
    Yes, I understand my mistake (how could've I forgotten?!), nevertheless, our answers only differed by a factor of (≈(2/(3(2)(.5))) , which is interesting given we took different paths. Also, I think it's 6E5. When I first computed my expression I didn't use any calculator so (1E8) drifted a bit from the right answer (scumbag brain).
    Last edited: Mar 8, 2012
  6. Mar 8, 2012 #5
    Thank you both! I will try to solve this but Im still shaky on my calculus. I haven't really learned integrals, but have a basic idea of how to do them and am trying to teach myself. I calculated how long it would take IF accelertion was constant and i got a little over a week... I will keep working!
  7. Mar 8, 2012 #6


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    It's easier if one notes that the total energy E is a constant of the motion (thereby providing one of the integrals right off the bat).
  8. Mar 8, 2012 #7
    Like so:



    K1=-U1(x) → (.5)m1(dx/dt)2=-Gm1m2/x
    K2=-U2(x) → (.5)m2(dx/dt)2=-Gm1m2/x

    Sum on both sides and reverse the equation:

    (dt/dx)2=x/(G(m1+m2)) → dt=(x/(G(m1+m2)))(.5)dx
    t0t[itex]\int[/itex]dt=x0x[itex]\int[/itex](x/(G(m1+m2)))(.5)dx → t=(2/3)(x0)(3/2)/(G(m1+m2))(.5)

    When you integrate, t0 and x (on the limit) are considered to be equal to zero.
    Last edited: Mar 8, 2012
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