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Gravitational potential energy problem

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data

    Question :-
    Two stars, each of a solar mass and radius ##10^7 m## are at a distance of ##10^{12} m## from each other. Find the speed of each star before collision if initial speed is negligible.

    2. Relevant equations

    $$V(r) = {-G Mm \over r}$$

    3. The attempt at a solution

    Initial total energy of a star would be $${-G m^2 \over 10^{26}}$$
    and final energy would be $${-G m^2 \over 10^{14}} + {1 \over 2 }mv^2$$

    use conservation of energy i got

    $${-G m^2 \over 10^{26}} = {-G m^2 \over 10^{14}} + {1 \over 2 }mv^2$$

    Solving for ##v## i got, ##3.6 \times 10^6 m/s##. But, the given answer is ##2.6 \times 10^6 m/s##.

    Now i can get the given answer by solving of ##v## in equation $${-G m^2 \over 10^{26}} = {-G m^2 \over 10^{14}} + mv^2$$

    I am not getting the concept that why i should drop the half from equation ? Please help me.
     
  2. jcsd
  3. Oct 30, 2016 #2

    BvU

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    How many stars are moving at impact time ?
     
  4. Oct 30, 2016 #3
    Two stars. But how does that help ?
     
  5. Oct 30, 2016 #4

    BvU

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    So how many times ##{1\over 2}mv^2 ## ?
     
  6. Oct 30, 2016 #5

    PeroK

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    Can you explain this fomula? What is ##V(r)## in this case?
     
  7. Oct 30, 2016 #6
    Gravitational potential energy.
     
  8. Oct 30, 2016 #7

    PeroK

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    ... of what?
     
  9. Oct 30, 2016 #8
    Of any one star.
     
  10. Oct 30, 2016 #9

    PeroK

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    Are you sure about that?
     
  11. Oct 30, 2016 #10
    But why do we add the Kinetic energy of other star, i am only considering Potential energy of one star.
     
  12. Oct 30, 2016 #11
    I think yes but i am getting a feel that i am wrong. But why ?
     
  13. Oct 30, 2016 #12

    PeroK

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    If ##m \ne M## do both stars have the same PE? Think about ##M >> m##.
     
  14. Oct 30, 2016 #13
    The bigger one will have less P.E, right ? But that does not answer why ##V(r)## is P.E of both stars combined.
     
  15. Oct 30, 2016 #14

    PeroK

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    You can derive it by looking at the force on each object and how that varies with the distance between them. That's a different exercise.

    For the time being:

    $$V(r) = {-G Mm \over r}$$

    Is the combined total PE of both stars.
     
  16. Oct 30, 2016 #15
    So if we have 3 stars of equal mass, then total PE would still be ##V(r) = {-G m^2 \over r}## or ##V(r) = {-3G m^2 \over r}## or ##V(r) = {-3G m^2 \over 2r}##. I think it is the third one.
     
  17. Oct 30, 2016 #16

    PeroK

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    What's ##r## if you have 3 stars?
     
  18. Oct 30, 2016 #17
    Distance between them. Like a vertices of a equilateral triangle. I just thought of this problem now.
     
  19. Oct 30, 2016 #18

    PeroK

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    Derivation of PE for two masses, ##m_1## and ##m_2##:

    ##F_1 = F_2 = \frac{Gm_1 m_2}{r^2}##

    ##dKE_1 + dKE_2 = \frac{Gm_1 m_2}{r^2}(-dr_1 - dr_2) = -\frac{Gm_1 m_2}{r^2}dr## (where ##r = r_1 + r_2##)

    ##dV = \frac{Gm_1 m_2}{r^2}dr##

    ##V(r) = -\frac{Gm_1 m_2}{r}##
     
  20. Oct 30, 2016 #19

    PeroK

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    By symmetry, the PE of each star due to each of the others is ##-\frac{Gm^2}{2r}##, so yes the total is ##-\frac{3Gm^2}{2r}##
     
  21. Oct 30, 2016 #20
    Ok i got this.

    Ok imagine three objects like $$\underbrace{\star}_{m_1} \overbrace{------}^{r} \underbrace{\star}_{m_2}\overbrace{------}^{r} \underbrace{\star}_{m_3}$$
    What should be the total PE of middle star ?

    it is not $$- {Gm_1m_2\over r} - {Gm_3m_2\over r}$$.
    We can't use symmetry argument here because masses are different.

    So how can i know the PE of mid star w.r.t to first star and w.r.t to third star separately ?
     
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