algebraic solution and other possible techniques?

[BSZDcZMX]

Hi there,

 

I've been doing some questions out of a book, and came across this question which I found interesting. I was wondering if there are other mathematical techiques that can be used to solve a problem like this?

 

Question:

 

The rotational period of Earth is 23.933 hours. A space shuttle revolves around Earth's equator every 2.231 hours. Both are rotating in the same direction. At the present time, the space shuttle is directly above the Galapagos Islands. How long will it take for the space shuttle to circle Earth and return to a position directly above the Galapagos Islands?

 

Solution:

 

The time taken to travel is [math]\frac{\theta}{\omega}[/math]. We know that the shuttle will complete one full orbit, and by the time it completes this orbit in 2.231 hours the Earth will have rotated a certain angle much less than that of the shuttle. We know that the shuttle will come back to the same point it was at in the beginning after completing a full orbit and travelling a further [math]\theta[/math] to catch up with the original point.

 

Taking [math]\omega _{e}, \omega _{s}[/math] to be the angular velocity of the Earth and shuttle respectively gives:

 

[math]\frac{\theta}{\omega _{e}} = \frac{2 \pi + \theta}{\omega _{s}}\rightarrow \frac{2 \pi + \theta}{\omega _{s}} - \frac{\theta}{\omega _{e}} = 0\rightarrow \theta = -\frac{2 \pi \omega _{e}}{\omega _{e} - \omega _{s}}[/math]

 

And then solving for time [math]t[/math], by dividing by [math]\omega _{e}[/math]:

 

[math]t = \frac{-\frac{2 \pi \omega _{e}}{\omega _{e} - \omega _{s}}}{\omega _{e}}\rightarrow t = -\frac{2 \pi}{\omega _{e} - \omega _{s}}[/math]

 

For this example my solution gave [math]t \approx 2.46[/math] hours.

 

 

Thanks