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!

Energy in Meteor Impact

  1. Dec 4, 2007 #1
    1. The problem statement, all variables and given/known data
    A meteor of mass about 3.4 x 10^12 kg is heading straight for Jupiter. When it hits there will be a huge release of energy, visible her on earth. Assuming it has fallen from far away, how much energy will be released when it hits Jupiter? The radius of Jupiter is 7 x 10^7 m and its mass is 1.9 x 10^27 kg.
    A) 3.4 x 10&19 J
    B) 5.8 x 10^18 J
    C) 5.4 x 10^22 J
    D) 6.8 x 10^19 J
    E) 6.1 x 10^21 J

    3. The attempt at a solution
    I'm somewhat lost as to how to approach this problem. When I've worked with energies, it has always involved velocity. In the chapter on gravitation in my text I only found mention of gravitational potential energy, [tex]U = -\frac{Gm_{e}m}{r}[/tex], but I don't have the distance between the masses. I'm guessing the collision would be inelastic, but I'm not sure how that helps. I can only think of ways of solving this if I'm given the velocity of the asteroid. Any help on this would be greatly appreciated.
  2. jcsd
  3. Dec 4, 2007 #2
    After looking it over again, I think maybe I could use an integral.
    [tex]W_{grav} = -Gm_{e}m\int_{r_{1}}^{r_{2}}\frac{dr}{r^2}[/tex]
    That is listed in our text.
    So maybe...
    [tex]W_{grav} = -Gm_{e}m\int_{\infty}^{0}\frac{dr}{r^2}[/tex]
    I don't think that's at all right. Just thinking aloud

    If anyone has any ideas, any at all, please share. I'm turning this in completed or not in a few minutes. =/
    Last edited: Dec 4, 2007
  4. Dec 4, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper

    Compare the gravitational potential energy of the meteor on Jupiter's surface with it's energy at r=infinity (where U=0). You don't need to integrate anything.
    Last edited: Dec 4, 2007
  5. Dec 4, 2007 #4
    I asked my physics prof before class and he said to use the escape velocity in the kinetic energy equation:

    [tex]V = \sqrt{\frac{2GM_{J}}{R_{J}}}[/tex]
    [tex]K = \frac{1}{2}MV^2 = \frac{1}{2}M\frac{2GM_{J}}{R_{J}} = \frac{GMM_{J}}{R_{J}}[/tex]

    Your way works better, and is easier to understand. Thanks :)
  6. Dec 4, 2007 #5


    User Avatar
    Science Advisor
    Homework Helper

    Sure. But it's the same thing. The calculation of escape velocity comes from potential energy. Which in turn is also the same thing as the integral you proposed earlier. Integrating the force over distance gives you the potential.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?