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Discovering mass of a star and orbital period of one of two planets

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be 41.7 km/s and 55.5 km/s. The slower planet's orbital period is 8.04 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?

    2. Relevant equations

    M= r2T/G
    T= 2[itex]\Pi[/itex]r3/2/[itex]GME\sqrt{}[/itex]

    3. The attempt at a solution

    I assume that the radius of the slower moving planet must be found first. Also, as for finding the period, does that same r need to be found, or is this going to be a very elaborate "plugging in the variables" type of equation?
     
    Last edited: Sep 26, 2011
  2. jcsd
  3. Sep 26, 2011 #2

    gneill

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    Staff: Mentor

    You can arrive at the mass of the star without first finding an orbital radius with suitable manipulation of "standard" formulas so that only the given data is required. You're given a circular orbit velocity and a period, so consider formulas involving those.

    The formulas that you've stated look rather dubious to me.
     
  4. Sep 26, 2011 #3

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Hello TmrK,


    Your LaTeX code didn't work, because you cannot use the [noparse] and [/noparse] tags within TeX tags. They are meant to be used on plain text. Instead, to achieve superscripts, subscripts, and fractions in LaTeX, try the following (right click the equation below and select the option from the menu to see the source code for it):

    [tex] T = \frac{2\pi}{\sqrt{GM}}r^{3/2}~~~~~~~~~~[1] [/tex]

    Since the orbits are circular, the orbital speed is constant. This makes things easy, since you know both the period and speed of the slower planet, and speed = distance/time. In this case, the time is the orbital period, and the distance is the distance around the circle (i.e. the circumference of the orbit):

    [tex] T = \frac{2\pi r}{v}~~~~~~~~~~[2] [/tex]

    Also, you know that in this situation, gravity is what provides the centripetal force that keeps the object in a bound orbit. Hence you can equate the expressions for the centripetal force and the gravitational force:

    [tex] \frac{GMm}{r^2} = \frac{mv^2}{r} [/tex]

    which yields:

    [tex] \frac{GM}{r} = v^2~~~~~~~~~~[3] [/tex]

    The interesting thing is that you can combine [2] and [3] together to derive [1]. This shows how Kepler's 3rd Law [1] actually follows from Newton's laws (although we have only derived it for the special case of circular orbits here). However, for the purposes of part (a) of this problem, I think it's easier just to use [2] and [3] separately. Use [2] to solve for r, and then use that result in [3] to solve for M.

    EDIT: or, as gneill pointed out, you can use [2] and [3] together to eliminate r as a variable, resulting in an expression for M in terms of only v and T. This method is equivalent.
     
  5. Sep 26, 2011 #4
    Sorry, I'm really not used to using the "Show/Hide Latex Reference" button. I'll post the equations again.

    M=v2r/G
    T=2[itex]\Pi[/itex]r3/2/Square Root of GM

    Where G is the Universal gravitational constant, M is the mass of the sun, and T is the period (of the second planet).
     
  6. Sep 27, 2011 #5
    So if I were to combine [2] and [3], I would get something like:

    M=v2[itex]\frac{vT/2\Pi}{G}[/itex]
    or
    M=v3[itex]\frac{T}{2G\Pi}[/itex]
     
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