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Calculating two masses colliding

  1. Mar 14, 2008 #1
    I have been wondering for a long time now on how to calculate the time taken for two masses in space to collide by force of gravity.

    Now what ive basically done was integrate the formula a = GM / R^2 between the two points corresponding to the distances appart from the masses (assuming the masses are equal.) And with the acceleration, T = sqrt ( 2S / a ) (S being the distance between masses).

    And ive gotten values of, say two 1 tonne masses at a distance of 1 Km would take approx 70 years to collide with eachother.

    Now ive got no idea whether ive gotten it right, or im way off. It seems like a fair answer (then again they only weigh one tonne).

    Can anyone confirm my working is correct and or what im doing wrong. Thanks

  2. jcsd
  3. Mar 14, 2008 #2


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    HOW did you integrate that? That's the crucial point and you don't say. You can't just integrate the two sides independently- the right is a function of R and the left side is a derivative with respect to t.

    Again, I can't say what you are doing wrong because you don't say what you did! Simply saying you "integrated" an equation doesn't help since I don't know how you tried to itegrate it.

    Here's how I would do that problem: First I'm going to put it in the "center of mass" system. Since the two objects have the same mass that will be the point exactly half way between them. If r is the distance from each mass, the distance between the two masses is 2r so we have, for each mass, [itex]dr^2/dt^2= -GM/4r^2[/itex]. Since the independent variable "t" does not appear in that explicitely, we can reduce the order of the equation by letting v= dr/dt. By the chain rule, [itex]dr^2/dt^2= dv/dt= (dv/dr)(dr/dt)= v(dv/dr)[/itex] so the equation becomes [itex]v(dv/dr)= -GM/r^2[/itex] for v as a function of r. We can write that as [itex]vdv= -(GM/4r^2)dr[/itex] and NOW integrate the left side with respect to v and the right side with respect to r. We get [itex](1/2)v^2= GM/4r+ C[/itex]. (If you move the "GM/4r" to the left side and multiply by M, you might recognize that as "conservation of energy".) If, initially, v= 0 and r= R, C= -GM/4R and we have [itex]v= dr/dt= \sqrt{GM/4r- GM/4R}= (\sqrt{GM}/2)\sqrt{1/r- 1/R}= (\sqrt{GM}/2R)\sqrt{(R-r)/r}[/itex]. That can be separted as [itex]\sqrt{r/(R-r)}dr= \sqrt{GM/2R}dt[/itex]. I don't have time now to complete that but if what you did is not like that, it's probable that your result is wrong.
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