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Phun Question, or Using an Orbit to Find the Mass of a Sun

Golden LegGolden Leg Registered User
edited February 2009 in Help / Advice Forum
So here's the deal.

You are an alien on an alien planet orbitting the planet's sun in a circular orbit. You want to find the mass of your sun. You "triangulate" the center-to-center distance between your planet and sun to be 3.52E+10 meters. The period of motion of your planet (the length of your year) is 1.81E+7 seconds. You know G = 6.67E−11 Nm^2/kg^2. What is the mass of your sun?

Given: R = 3.52E+10 m, T = 1.81E+7 s, G = 6.67E−11 Nm^2/kg^2
Want: M (as opposed to m)
Equation: F = GmM/r^2, ?, ?

We have three unknowns, M, m, and F. With T, one could potentially solve for angular velocity (omega = 2PI(1/T)) and go from there.

Anyone up for some phun?

Golden Leg on

Posts

  • thisisntwallythisisntwally Registered User regular
    edited February 2009
    Golden Leg wrote: »
    So here's the deal.

    You are an alien on an alien planet orbitting the planet's sun in a circular orbit. You want to find the mass of your sun. You "triangulate" the center-to-center distance between your planet and sun to be 3.52E+10 meters. The period of motion of your planet (the length of your year) is 1.81E+7 seconds. You know G = 6.67E−11 Nm^2/kg^2. What is the mass of your sun?

    Given: R = 3.52E+10 m, T = 1.81E+7 s, G = 6.67E−11 Nm^2/kg^2
    Want: M (as opposed to m)
    Equation: F = GmM/r^2, ?, ?

    We have three unknowns, M, m, and F. With T, one could potentially solve for angular velocity (omega = 2PI(1/T)) and go from there.

    Anyone up for some phun?

    this sounds more like someone's homework than phun

    thisisntwally on
    #someshit
  • Golden LegGolden Leg Registered User
    edited February 2009
    You're crazy!

    Golden Leg on
  • thisisntwallythisisntwally Registered User regular
    edited February 2009
    actually I'm an economist. If I live a good life, perhaps I can come back as a physicist in my next life, and help you with your homework.

    thisisntwally on
    #someshit
  • FuzzywhaleFuzzywhale Registered User
    edited February 2009
    I'll try to help you out

    you are given a circular orbit which is pretty good; that makes things easy. I don't think Newton's law of universal gravitation is the right way to go on this one. this was just a quick google search so ill just hook you up here. Apparently:

    the orbital period of something in a circular orbit is: T=(2pi)(r^3/(G*M))^(1/2). where M's the mass of the thing you are orbiting. You should be able to algebrate that and solve for what you need.

    Fuzzywhale on
  • Sir Headless VIISir Headless VII Registered User regular
    edited February 2009
    That formula is derived using newtons law of gravitation. This Site and This Site should give you everything that you need that's not in the OP. Just be mindful of the difference between angular velocity and tangential velocity.

    Sir Headless VII on
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  • SentrySentry Registered User regular
    edited February 2009
    all these threads do is make me realize that I know nothing.

    Absolutely nothing.

    Sentry on
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    'Fuck yeah, me too. What little kid ever pretended to be part of the lynch-mob?'
  • shadydentistshadydentist Registered User regular
    edited February 2009
    Use centripetal force and set it equal to gravitational force.

    shadydentist on
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  • Golden LegGolden Leg Registered User
    edited February 2009
    You guys are terrific.

    The equation F = GmM/r^2 can be set to solve for M, rendering M = Fr^2/Gm. F is equal to the centripetal force, and Fcentripetal = (OMEGA)^2(r). Angular velocity, or OMEGA, equals 2PI/T. Solving for OMEGA and plugging into the equation for M gives us M = [(OMEGA)^2*(r)^3]/G.

    Turns out the mass of that sun equals 6.55x10^29 kg.

    Golden Leg on
  • AridholAridhol Daddliest Catch Registered User regular
    edited February 2009
    Sentry wrote: »
    all these threads do is make me realize that I know nothing.

    Absolutely nothing.

    Aridhol on
  • locomotivemanlocomotiveman Registered User
    edited February 2009
    I did a similar lab last night with Jupiter using the motion of its moons. Unfortunantly for this I turned said lab in last night as well. The problem is none of your formulas look anything like what we used last night. Your final answer looks good though. There is nothing sadder than watching people try to do calculations like this on their celphones.

    locomotiveman on
    aquabat wrote:
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    Can you like, permanently break the forums?
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