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Thread: Geometry in space

  1. #1
    Nov 2005

    Geometry in space

    I need help working out this problem
    While a planet P rotates in a circle about its sun, a moon M rotates in a circle about the planet, and both motions are in a plane. Let's call the distance between M and P one {lunar unit}. Suppose the distance of P from the sun is $\displaystyle 4.2X10^3$ lunar units; the planet makes one revolution about the sun every 3 years, and the moon makes one rotation about the planet every 0.11111111111 years. Choosing coordinates centered at the sun, so that, at time t=0 the planet is at $\displaystyle (4.2X10^3, 0)$ and the moon is at $\displaystyle (4.2X10^3, 1)$, then the location of the moon at time t, where t is measured in years, is $\displaystyle (x(t),y(t)) $ where
    $\displaystyle x(t)= $
    $\displaystyle y(t)= $
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  2. #2
    Moo is offline
    A Cute Angle Moo's Avatar
    Mar 2008
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Quote Originally Posted by viet View Post
    I need help working out this problem
    The planet describes a circle, within period of 3 years, and radius 4.2x10^3 years.

    So if we convert it to polar coordinates, we get :
    $\displaystyle r=4.2 \times 10^3$
    $\displaystyle T=3$ << period of the motion.
    And we know that $\displaystyle \cos \left(\frac{2 \pi}{T} \cdot t\right)$ has a period T, which is something that can be useful.

    Since it's a circle, $\displaystyle x(t)^2+y(t)^2=r^2$

    So if you let $\displaystyle x(t)=r \cos \left(\frac{2 \pi}{T}\cdot t\right)$ and $\displaystyle y(t)=r \sin \left(\frac{2 \pi}{T}\cdot t\right)$, you'll get the coordinates of the planet P.

    Now consider that the moon has a circular motion around P, but the coordinates of P are x(t) and y(t).

    Let $\displaystyle m(t)$ and $\displaystyle n(t)$ be the coordinates of the moon.
    It's a period $\displaystyle T'$ of 0.111...11 year.
    The coordinates satisfy this equation :
    $\displaystyle [m(t)-x(t)]^2+[n(t)-y(t)]^2=r'^2$, where r'=1 (lunar unit)
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