Course:MATH102/Question Challenge/1997 December Q5

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Question

As a moon orbits a planet, its angular position is given by Kepler's equation

$kt=\theta -{\frac {1}{2}}\sin(\theta )$

where $k$ is some constant.

(a) If it takes 50 days for the moon to complete an orbit, what it the value of $k$?

(b) Find the rate of change ${\frac {d\theta }{dt}}$ when $\theta ={\frac {\pi }{3}}$.

Moon orbiting a planet in an elliptical orbit.

Hints

Hint 1
Remember that 1 revolution is $2\pi$ radians.

Hint 2
At $t=50$, the moon has made 1 full revolution so $\theta =2\pi$ radians and $\sin(\theta )=0$. You can use this to find $k$.

Hint 3
To find ${\frac {d\theta }{dt}}$, consider differentiating both sides of the equation with respect to time, and use the chain rule on the right hand side. Then isolate (solve for) ${\frac {d\theta }{dt}}$. Only plug in the value $\theta ={\frac {\pi }{3}}$ at the very end.

Solutions

Solution

Part A) When $t=50$ the moon has completed one orbit. This means that $\theta =2\pi$: $k(50)=2\pi -1/2sin(2\pi )$ $50k=2\pi -0$ $k={\frac {2\pi }{50}}$ $k={\frac {\pi }{25}}$ Part B) To solve for ${\frac {d\theta }{dt}}$ when $\theta =\pi /3$ we will derive with respect to time and isolate for ${\frac {d\theta }{dt}}$: ${\frac {\pi }{25}}={\frac {d\theta }{dt}}-{\frac {1}{2}}cos(\theta ){\frac {d\theta }{dt}}$ ${\frac {\pi }{25}}={\frac {d\theta }{dt}}(1-{\frac {1}{2}}cos(\pi /3))$ ${\frac {\pi }{25}}={\frac {d\theta }{dt}}(1-{\frac {1}{2}}*{\frac {1}{2}})$ ${\frac {\pi }{25}}={\frac {d\theta }{dt}}(1-{\frac {1}{4}})$ ${\frac {\pi }{25}}={\frac {d\theta }{dt}}(3/4)$ ${\frac {d\theta }{dt}}=4\pi /75$