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avatar+108608 

Here is a fun question for some people.

If it is very easy for you then please do not answer. 

It is intended to be a interesting challenge question.

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Assume the Earth is a perfect sphere.

A peice of string is wrapped around the equator to make a snug fit.

Now another string exactly one metre longer than the first one is wrapped around the earth so that the 2 peices of string form 2 concentric circles.

How far off the surface of Eath is the second peices of string?

 Nov 4, 2019

Best Answer 

 #1
avatar+2824 
+5

\(2{\pi}r=C\)

 

That is the formula for circumference.

 

Now to find how far off, we find the difference between the two radii when the concentric circles are formed.

 

We make a system of equations

 

\(2{\pi}x=C\)

\(2{\pi}y=(C+1)\)

 

We have to evaluate \(|y-x|\)

 

Substituting: \(2{\pi}y-1=2{\pi}x\)

 

Simplifying: \(1=2{\pi}y-2{\pi}x\)

 

Solving: \(\frac{1}{2{\pi}}=y-x\)

 

Solving for \(|y-x|\),

 

\(\boxed{\frac{1}{2{\pi}}}\)

 

 

That was my attempt laugh

 Nov 4, 2019
 #1
avatar+2824 
+5
Best Answer

\(2{\pi}r=C\)

 

That is the formula for circumference.

 

Now to find how far off, we find the difference between the two radii when the concentric circles are formed.

 

We make a system of equations

 

\(2{\pi}x=C\)

\(2{\pi}y=(C+1)\)

 

We have to evaluate \(|y-x|\)

 

Substituting: \(2{\pi}y-1=2{\pi}x\)

 

Simplifying: \(1=2{\pi}y-2{\pi}x\)

 

Solving: \(\frac{1}{2{\pi}}=y-x\)

 

Solving for \(|y-x|\),

 

\(\boxed{\frac{1}{2{\pi}}}\)

 

 

That was my attempt laugh

CalculatorUser Nov 4, 2019
 #2
avatar+109061 
+3

Correct, CU  !!!

 

Good job  !!!

 

 

cool cool cool   

CPhill  Nov 4, 2019
 #3
avatar+108608 
+1

Great work CalculatorUser!     cool cool

 

You maths is excellent,

 

Just for a slightly different presentation I could say:

 

\(C=2\pi r\\ \frac{C}{2\pi}=r\\ \frac{C+1}{2\pi}=\frac{C}{2\pi}+\frac{1}{2\pi}=r+\frac{1}{2\pi} \)

 

So if the circumference is increased by 1 unit, the radius is increased by  \(\frac{1}{2\pi}\;units\)

 

So if the circumference is increased by 1 metre, the radius is increased by \(\frac{1}{2\pi}\;metre\)

 

And this equals an approximate radius increas of  0.159 metres or approx  16cm

 

This distance between the 2 concentric circles will always be the same.

 It does not matter how big or small the original circle is.     cool

Melody  Nov 4, 2019
edited by Melody  Nov 4, 2019

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