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If you put two small stationary spherical objects (say 1 meter wide weighting 1kg) 1 light year across and let them collide after some time.

Will the energy of collision be the same as if you put them 10 light years across or 100 light years across?

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  • No, but the difference is unmeasurably small and doesn't matter. Not sure why it matters that they are 1m wide, though... 1kg packed into the size of 1m is about the density of air. So what you are really saying is that you make some really thin balloons really far away. For what? There is nothing to be learned from such a Gedankenexperiment. – CuriousOne Jan 27 '16 at 15:17
  • I'm working on my own answer to this, but, until then, http://physics.stackexchange.com/questions/14700 might be helpful, as may http://tinyurl.com/gq67m53 which shows a simpler (but still not really solvable) version of your question. –  Jan 27 '16 at 18:15

1 Answers1

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The energy of the collision is equal to the kinetic energy $K$ of the two bodys just before they collide.

The mechanical energy is conserved during the whole process, so $$ V_1+K_1=V_2+K_2 $$ where $V$ is the gravitational potential energy of the bodies and $K$ their kinetic energy. Subscripts $1$ and $2$ refer to the initial time and the time of the collision respectively.

Since both bodies are initialy at rest, $K_1=0$. Hence, $$ K_2=V_1-V_2 $$ The gravitational potential energy of two bodies of mass $M_a$ and $M_b$, separated by a distance $x$ is $$ V=-G \frac{M_a M_b}{x_i} $$ Hence, the energy of the collision is $$ K_2=-G M_a M_b \left( \frac{1}{x_1}-\frac{1}{x_2} \right) \\ ~~=G M_a M_b \left( \frac{1}{x_2}-\frac{1}{x_1} \right) $$ As you can se from the equation above, the energy of the collision increases as the initial distance between the two bodies increases. However, putting your numbers, the difference between 10 l.y. and 100 l.y. is negligible.

Art
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