SOLUTION: I have this question for a trig test and never seen anything like this in class before. Any help would be appreciated. A planet rotates on its axis through its poles and one rev

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Question 417610: I have this question for a trig test and never seen anything like this in class before. Any help would be appreciated.
A planet rotates on its axis through its poles and one revolution takes 1 day (1 day is 26 hours). The distance from the axis to a location on the planet 30 degrees north latitude is about 2447.5 miles. Therefore, a location on the planet at 30 degrees north latitude is spinning on a circle of radius 2447.5 miles. Compute the linear speed on the surface of the planet at 30 degrees north latitude. The linear speed on the surface of the planet is _______ miles/hr.
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
I have this question for a trig test and never seen anything like this in class before. Any help would be appreciated.
A planet rotates on its axis through its poles and one revolution takes 1 day (1 day is 26 hours). The distance from the axis to a location on the planet 30 degrees north latitude is about 2447.5 miles. Therefore, a location on the planet at 30 degrees north latitude is spinning on a circle of radius 2447.5 miles. Compute the linear speed on the surface of the planet at 30 degrees north latitude. The linear speed on the surface of the planet is _______ miles/hr.
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I have found the best way to work these kinds of problems is to use dimentional analysis: The circumference at 30 degrees latitude=2pi*radius=2pi*2447.5 mi
This is the distance traveled in one revolution which takes 26 hours.
Using dimentional analysis,
(revolution/26 hrs)*2pi*2447.5 mi/revolution*
revolution cancels out
2pi*2447.5 mi/26 hrs=591.47 mi/hr
ans:
The linear speed on the surface of the planet is 591.47 miles/hr.