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Two stars rotating about a centre of mass -- Find speed

  1. Sep 21, 2016 #1
    1. The problem statement, all variables and given/known data
    The diagram shows a binary star system consisting of two stars each of mass 4 * 1030 kg separated by 2 * 1011 m. The stars rotate about the centre of mass of the system.

    c62fe2da4ee5.jpg

    (a) (i) Copy the diagram and, on your diagram, label with a letter L a point where the gravitational field strength is zero. Explain why you have chosen this point.
    (a) (ii) Determine the gravitational potential at L. (G = 6.7 * 10-11 N m2 kg-2.)
    (b) (i) Calculate the force on each star due to the other.
    (b) (ii) Calculate the linear speed of each star in the system.
    (b) (iii) Determine the period of rotation.

    Answers: (a) (ii) -5.36 * 109 J kg-1, (b) (i) 2.68 * 1028 N, (ii) 2.59 * 104 m s-1, (iii) 2.43 * 107 s

    2. The attempt at a solution
    (a) (i) L should be the middle point between these two stars since the point where gravitational field strength is zero is the point where two gravitational fields cancel out.

    (a) (ii) Using the formula: U = - Gm / r = - 6.7 * 10-11 * (4 * 1030 + 4 * 1030) / [(2 * 1011) / 2] = - 5 360 000 000 J kg-1.

    (b) (i) F = Gm1m2 / r2 = (6.7 * 10-11 * 4 * 1030 * 4 * 1030) / (2 * 1011)2 = 2.68 * 1028 N.

    (b) (ii) v = √Gm1 / 2 / r = √(6.7 * 10-11 * 4 * 1030) / (2 * 1011) / 2) = 51 768.7 m s-1. Doesn't fit the answer, though if we divide this by 2 we get 25 884 m s-1, same as the answer.

    (b) (iii) T = 2πr / v = (2 π ((2 * 1011) / 2) / (2.59 * 104) = 24 259 403 s.

    Is my logic in (a) (i) correct? What's wrong with (b) (ii)? Other calculations should be right, but if not, what did I miss?
     
  2. jcsd
  3. Sep 21, 2016 #2

    Orodruin

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    (a) is correct. Please explain your logic in (b), otherwise it is difficult to see exactly why you get a factor of 2 wrong.
     
  4. Sep 21, 2016 #3

    gneill

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    Your (a) (i) is okay.

    For (b) (ii), the gravitational parameter (##\mu##) for the system depends upon both masses. You need to use right value for ##\mu## if you want to apply the equation ##v = \sqrt{\frac{\mu}{r}}## . You can check this yourself by equating the gravitational force to the centripetal force for one of the masses, solving for v.
     
  5. Sep 21, 2016 #4

    Orodruin

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    This only accounts for a factor sqrt(2) (the reduced mass of the system is half of the individual masses). He has also used the distance from the center of mass instead of the separation distance.

    The reason I wanted him to state his reasoning was to make him think about theprerequisites of the formula rather than just inserting numbers without thinking first.
     
  6. Sep 21, 2016 #5
    if you place the origin on one of the star and take rotation of the other one around it ,what will be the reduced mass of the system? each one is rotating about the other one so to reduce this two body system to a single particle moving under the gravitational field one has to take recourse to the transformation from two-body to a single body system .
     
  7. Sep 21, 2016 #6
    v is velocity, G is the gravitational constant, m is the mass of one star and r is the radius of the star from the star to the point L.

    If I understood you correctly, it should be not m1 / 2 but (m1 + m2) or (m1 * m2)? In both cases the answer is wrong (73 212 m s-1 and 1.04 * 1020 m s-1).

    Hm, I re-calculated (m1 * m2) and (m1 + m2) and also used both the length to L and also the full length but in the four cases I didn't get the correct answer.

    ---

    Update:
    KE = Gm1m2 / 2r
    Since KE = 0.5 m1 / 2 v2:
    2m1 / 2v2r = 2Gm1m2
    v = √Gm1m2 / m1 / 2r = √(6.7*10-11 * 4*1030 * 4*1030) / (4*1030 * 2*1011) = 36 606 m s-1, but again wrong.
     
    Last edited: Sep 21, 2016
  8. Sep 21, 2016 #7

    Orodruin

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    Again, do not just use a formula and plug in numbers. Go back to basics and think of how the formula is derived.
     
  9. Sep 21, 2016 #8
    That particular formula is based on a satellite of mass m orbiting the Earth with speed v along a circular path of radius r. F = mv2 / r = GmmEarth / r2. But looks like this formula is of no use to this situation.

    And the KE is also not much of use.
    KE = Gm1m2 / 2r
    Since KE = 0.5 m1 / 2 v2:
    2m1 / 2v2r = 2Gm1m2
    v = √Gm1m2 / m1 / 2r = √(6.7*10-11 * 4*1030 * 4*1030) / (4*1030 * 2*1011) = 36 606 m s-1, but again wrong.

    Update: F = ma → a = F / m = 2.68*1028 / 4*1030 = 0.0067 m s-2 → a = v2 r → v = √ar = √0.0067 * 1011 = 25 884 m s-1. Looks correct.
     
    Last edited: Sep 21, 2016
  10. Sep 21, 2016 #9

    gneill

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    [
    Don't just try to apply a formula blindly. Go back to basics. You have a diagram of your situation; Identify the distance that must be used to find the gravitational force. Identify the radius of rotation that must be used to find the centripetal force on one of the stars. Note that the two distances may not be the same!
     
  11. Sep 21, 2016 #10
    try to analyze the motion of the two bodies around each other....

    quote;
    physical forces are generally between two bodies; and by Newton's third law, if the first body applies a force on the second, the second body applies an equal and opposite force on the first. Therefore, both bodies are accelerated if a force is present between them; there is no perfectly immovable center of force. However, if one body is overwhelmingly more massive than the other, its acceleration relative to the other may be neglected; the center of the more massive body may be treated as approximately fixed. For example, the Sun is overwhelmingly more massive than the planet Mercury; hence, the Sun may be approximated as an immovable center of force, reducing the problem to the motion of Mercury in response to the force applied by the Sun. In reality, however, the Sun also moves (albeit only slightly) in response to the force applied by the planet Mercury.
    so when you try to apply satellite equation here you do not get correct result.....
    if you reduce to equivalent one body problem then 'reduced mass will come in and it will be 1/2 of mass.
    see
    https://en.wikipedia.org/wiki/Classical_central-force_problem#Newton.27s_theorem_of_revolving_orbits
     
  12. Sep 21, 2016 #11
    But in our case we have two stars of the same mass. v = √G(m/2) / r = √(6.7 * 10-11 * (4*1030)/2)) / (2 * 1011) = 25 884 m s-1. But the situation is not "one body is overwhelmingly more massive than the other".

    I think it's easier to derive the answer with F = ma → a = F / m = 2.68*1028 / 4*1030 = 0.0067 m s-2 → a = v2 r → v = √ar = √0.0067 * 1011 = 25 884 m s-1.
     
  13. Sep 21, 2016 #12

    Orodruin

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    The point is that you can reduce the two body problem to a mathematically equivalent situation - but you need to take care about what the effective mass and potential are.

    Of course, both approaches give the same result and I would use the force approach if teaching at basic level.
     
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