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First, the "connection" between the radius r (in meters) in a satellites trajectory and orbital period time is given by the following equation:

$$\frac{r^3}{T^2}= \frac{k}{4\pi^2} $$

$K$ is a constant equal to $3.98 \cdot 10^{14}$.

I solved for $r$ and $T$ because that will be needed.
$$r=\sqrt[3]{\frac{T^2k}{4\pi^2}}$$ and $$T = \sqrt{\frac{r^3\pi^24}{k}} $$

The question: Satellite $X$ runs in a trajectory twice the radius of Satellite $Y$. Which satellite has the longest turnaround, and how much longer is it.

My thoughts: Now I know that this should be solved algebraically, instead of just testing with numbers and predicting that way. Any suggestions on how to proceed from here? Would the following be a good "strategy"?:

$$\sqrt{\frac{(2r)^3\pi^24}{k}} - \sqrt{\frac{r^3\pi^24}{k}} .$$

My primary motivation for posting is debating and discussing math because I'm re-doing old exams on my own. It's nice to get some input.

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To me the questions seems to be more about the quotient $T_2/T_1= \sqrt{\frac{(2r)^3\pi^24}{k}} / \sqrt{\frac{r^3\pi^24}{k}}$ than about the difference $T_2-T_1 = \sqrt{\frac{(2r)^3\pi^24}{k}} - \sqrt{\frac{r^3\pi^24}{k}}$. I guess you'll be able to simplify the quotient. –  Martin Sleziak Aug 7 '11 at 13:49
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There is no need to do any manipulation, in fact it is best not to. Just look at the given equation as it stands. The equation is

$$\frac{r^3}{T^2}=\frac{k}{4\pi^2}.$$

The right side is constant. So the left side must be constant.

If you double $r$, then the top, $r^3$, gets multiplied by $8$.

You must multiply $T$ by something so that the ratio $r^3/T^2$ remains unchanged. That something is $\sqrt{8}$. This is because any scaling factor $\alpha$ for $T$ scales the bottom, $T^2$, by $\alpha^2$, so we must have $\alpha^2=8$.

If this is not clear, let $\alpha$ be the number you must multiply $T$ by. Then $$\frac{r^3}{T^2}=\frac{(2r)^3}{(\alpha T)^2}=\frac{8r^3}{\alpha^2T^2}.$$ From this equation we obtain $$\alpha^2=8,$$ so $\alpha=\sqrt{8}=2\sqrt{2}\approx 2.828$.

Commment: It is impossible to give the difference between the turnaround times, in say, hours, unless we know an additional fact, say $r$, or $T$. So your proposed strategy of studying the difference $T_2-T_1$ is not the best approach. But as you can see, it is certainly possible to calculate the ratio of turnaround times. The given equation involves multiplication only: that is very common in physical laws. So the natural approach is multiplicative. You can do that multiplicative approach in the style suggested by @Martin Sleziak. That will also quickly tell you the ratio of the orbital times, and may feel more standard.

You might be interested to know that the equation that you started with is a specialization of Kepler's Third Law of planetary motion.

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Thanks. This approach didn't occur to me. Very elegant. –  Algific Aug 7 '11 at 14:45
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