Fun challenge question.

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Fun challenge question.

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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

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#1

+2864

+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

#2

+130477

+3

Correct, CU !!!

Good job !!!

CPhill Nov 4, 2019

#3

+118703

+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

edited by Melody Nov 4, 2019

2 Online Users

CalculatorUser · Answer 1 · 2019-11-04T04:11:13

$2{\pi}r=C$

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

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Fun challenge question.

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