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Asteroid deflected by Earth -- effect on Earth

  1. Aug 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose the asteroid of [other problem] has a mass of [itex]6 \times 10^{20} \textrm{kg} [/itex]. Find the proportional change in the kinetic energy of the Earth in this encounter. What is the change in the semi-major axis of the Earth's orbit? By how much is its orbital period lengthened?

    Other problem:
    Suppose the asteroid of [other problems] approaches the Earth with an impact parameter of [itex]5 R_E[/itex], where [itex]R_E[/itex] = Earth's radius, moving in the same plane and overtaking it. Find the distance of closest approach and the angle which the asteroid is scattered, in the frame of reference in which the Earth is at rest.

    2. Relevant equations

    3. The attempt at a solution

    The value of Earth's velocity is 28.9km/s and that of the comet is 38.6km/s. The angle of deflection was 18.4 degrees. I tried calculating
    [tex] \dot{\textrm{R}} - \frac{m_1}{M} \dot{\textrm{r}} [/tex] to work out the new velocity and from that the energy change. The book's answer was [itex]3\times10^{-6}[/itex] whereas mine was [itex]59\times 10^{-6} [/itex]. I'm not sure if my approach is correct.
  2. jcsd
  3. Aug 6, 2016 #2


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    The angle of deflection is not sufficient, you also have to know the initial or final direction or something else about the geometry of the problem.

    Where does your formula come from?
  4. Aug 6, 2016 #3


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    It is not clear what your reasoning or algebra is from that decsription. Please post the details of what you did and on what basis.
  5. Aug 7, 2016 #4
    OK I uploaded a couple of images. The first shows the set up and the second what has happened in the reference frame of the Earth.

    Here [itex]\dot{\textrm{R}}[/itex] is
    \frac{m_1 \dot{\textrm{r}}_1 + m_2 \dot{\textrm{r}}_2}{m_1+m_2}
    which is the velocity of the centre of mass and [itex]\dot{\textrm{r}} = \dot{\textrm{r}}_1-\dot{\textrm{r}}_2[/itex]. The formula I quoted is that of [itex]\dot{\textrm{r}}_2[/itex]. Here [itex]\dot{\textrm{r}}[/itex] is the velocity of the comet in Earth's reference frame. I find Earth's velocity components with this formula and then this speed to find the kinetic energy.

    Edit: for some reason the text did not show up, but here Earth is travelling at 29.8km/s, the comet at 38.6km/s, and the angle is 18.4 degrees.

    Attached Files:

  6. Aug 7, 2016 #5


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    The relative velocity between Earth and asteroid does not change, and conservation of momentum gives the final velocity of all components. Start there, then you can be sure to have the right formulas.
  7. Aug 7, 2016 #6
    What about the equation [itex] \mu \ddot{\textrm{r}} = \textrm{F}[/itex]? Wouldn't [itex]\dot{\textrm{r}}[/itex] change in this case? Also the deflection angle is the one you calculate from the previous question, which is confirmed in the solutions.
  8. Aug 7, 2016 #7
    Also the formula for [itex]\dot{\textrm{r}}_1[/itex] is [itex] \dot{\textrm{R}} + m_2/M \dot{\textrm{r}} [/itex] and we find that the total momentum is [itex] m_1 \dot{\textrm{r}}_1 + m_2 \dot{\textrm{r}}_2 = M \dot{\textrm{R}}[/itex] so that momentum is conserved according to them.
  9. Aug 7, 2016 #8


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    Do you know F as function of time? If not, it doesn't help.

    What are r1, r2? You are wildly combining various different formulas and symbols without introducing them or showing where they come from. That does not help.
  10. Aug 7, 2016 #9
    The previous question was from chapter 4 of the book and had a section on orbits from inverse square law forces. It had subsections on hyperbolic orbits and the deflection, from which I was able to obtain the previous result.
  11. Aug 7, 2016 #10
    Sorry I didn't notice your question. Here [itex]\textrm{r}_1[/itex] and [itex]\textrm{r}_2[/itex] are the positions of the comet and Earth, respectively.
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