Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    [itex]\vec{F}[/itex]=m[itex]\vec{a}[/itex]

    x(t)=x0+v0t+(1/2)at2

    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:

    t=((2d3/(G(m1+m2)))(1/2)

    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

    phyzguy

    User Avatar
    Science Advisor

    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

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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=(.5)m1v2
    K2=(.5)m2v2

    U1(x)=G(m1m2)/(x)
    U2(x)=G(m1m2)/(x)

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Time for Gravity to force two objects in a vacuum together
Loading...