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?

Melody Nov 4, 2019

#1**+6 **

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

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

CalculatorUser Nov 4, 2019

#1**+6 **

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

CalculatorUser Nov 4, 2019

#3**+1 **

Great work CalculatorUser!

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.

Melody
Nov 4, 2019