# Geometry in space

• January 27th 2009, 03:08 PM
viet
Geometry in space
I need help working out this problem
Quote:

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 $4.2X10^3$ 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 $(4.2X10^3, 0)$ and the moon is at $(4.2X10^3, 1)$, then the location of the moon at time t, where t is measured in years, is $(x(t),y(t))$ where
$x(t)=$
$y(t)=$
• January 28th 2009, 01:37 AM
Moo
Hello,
Quote:

Originally Posted by viet
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 :
$r=4.2 \times 10^3$
$T=3$ << period of the motion.
And we know that $\cos \left(\frac{2 \pi}{T} \cdot t\right)$ has a period T, which is something that can be useful.

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

So if you let $x(t)=r \cos \left(\frac{2 \pi}{T}\cdot t\right)$ and $y(t)=r \sin \left(\frac{2 \pi}{T}\cdot t\right)$, 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 $m(t)$ and $n(t)$ be the coordinates of the moon.
It's a period $T'$ of 0.111...11 year.
The coordinates satisfy this equation :
$[m(t)-x(t)]^2+[n(t)-y(t)]^2=r'^2$, where r'=1 (lunar unit)