Expert: Paul Klarreich - 10/19/2007

Question
Imagine a steel Band wrapped around the Equator of Earth like a belt.
Imagine that 1 meter of steel is added to the band in a manner so that
the entire belt is raised off the ground the same distance all the way
around.

will the band be high enough so that you can roll a baseball under it?

If the Eiffel Tower were on the equator, then how much length would
need to be added to the belt so that the Eiffel Tower would fit under it


Answer
Questioner:   erica
Category:  Advanced Math
Private:  No
 
Subject:  word problem
Question:  Imagine a steel Band wrapped around the Equator of Earth like a belt.
Imagine that 1 meter of steel is added to the band in a manner so that the entire belt is raised off the ground the same distance all the way around.

will the band be high enough so that you can roll a baseball under it?

If the Eiffel Tower were on the equator, then how much length would need to be added to the belt so that the Eiffel Tower would fit under it
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Hi, Eri,

You will need the formula:  C = 2pi r

Your first band has a length equal to  2pi ER, where ER is the Earth Radius.

You are adding 1 meter to that length, so the new circumference is 2pi ER + 1.  What is the New Radius, which I will call NR?

2pi NR = 2pi ER + 1   

NR = ER + 1/(2pi)   << divided by 2pi

Now 1/(2pi) is the increase in radius, and the height of the (floating?) band of metal.  1/2pi is about 1/6.28, or about  39.37/6.28, or about 6 inches.  

Baseball?  Si.  Basketball?  No.
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Your Eiffel Tower problem is similar, but you have the additional radius given, and find the additional circumference.

C = 2pi(NR + ET)   << ET is the height of the Eiffel Tower.

C = 2pi NR + 2pi ET

2pi is the old circumference, so the increase is 2pi ET.  I leave it to you to determine the height of the Eiffel Tower.  [DON'T GO TO PARIS!  THERE ARE OTHER WAYS.]