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 $\frac{d\theta }{dt}$: $\frac{\pi }{25}=\frac{d\theta }{dt}-\frac{1}{2}cos(\theta )\frac{d\theta }{dt}$ $\frac{\pi }{25}=\frac{d\theta }{dt}(1-\frac{1}{2}cos(\pi /3))$ $\frac{\pi }{25}=\frac{d\theta }{dt}(1-\frac{1}{2}*\frac{1}{2})$ $\frac{\pi }{25}=\frac{d\theta }{dt}(1-\frac{1}{4})$ $\frac{\pi }{25}=\frac{d\theta }{dt}(3/4)$ $\frac{d\theta }{dt}=4\pi /75$