1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Archived Transformation of Gravitational to Kinetic Energy

  1. Nov 28, 2013 #1
    I am not following the template for the reason that this is a generic question.

    Consider that the change in kinetic energy is 1J.
    Suppose further you have two particles, both of equal mass that are gravitationally attracted to each other (and the change in energy comes from the fact that they moved closer to each other -- these are planetary objects).

    What is their final velocity? Assume they start from rest.

    First thing to notice: their velocities will be exactly the same throughout. so (vf)^2 = (-vf)^2 = vf

    So two approaches:

    dk = (1/2)(m)(v^2)

    Rearrange for v and divide by two because the velocity "splits" into two for both the masses.
    You will find a factor of root 2 over 2.

    Approach two:

    dk = (1/2)(m)(v^2) + (1/2)(m)(v^2) = m v^2.
    Rearrange for v. Here we have no need for dividing by two, we straight up get the v. This v has no factor infront of it. It is just root of (dk/m).

    So two contradicting pieces. The first approach is in fact correct. I have confirmed it.
    Can someone point out the flaw in the logic of the other piece?

    [EDIT]
    Okay I found one mistake in my reasoning:
    The change in energy is for the system. I should mention that.
    So the way I got it was 2 *(Gm^2/r), r is the distance between their centres. Multiply by two for both.

    So the change in kinetic energy of the SYSTEM is actually: (1/2)(2m)(2*vf)^2.

    Wait... no okay, I got it! vf of the system has to remain the same. Makes no sense if it changed by conservation of energy.

    Thanks for letting me write. I was stuck on this problem for a while!
     
    Last edited: Nov 28, 2013
  2. jcsd
  3. Jun 27, 2016 #2

    BiGyElLoWhAt

    User Avatar
    Gold Member

    Assuming both masses are the same, then they will both experience the same change in velocity.

    The important thing to note with method 1 is that the mass (denoted as 'm') is the mass of the system, whereas in method 2, the masses (also denoted 'm') are the masses of the 2 individual "particles". 'm' from method 1 = 2x 'm' from method 2.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted