## Around the earth

Posted by Zorglub on April 22, 2012 – 5:51 am

The earth is spherical with a radius of 6400 km.

A long chain is tighten around it. Then, an extra piece measuring 1 m is added to the chain and a point of the chain is pulled perpendicularly away from the earth until the chain is tight again. What is the distance between the earth and the point ?

April 22nd, 2012 at 7:06 am

about 15.9 centimeters

April 22nd, 2012 at 8:54 am

Just gonna make a quick guess and say .5 m. If the ends of the original chain were brought back together the 1 m addition would be folded in half making a point .5 m from the ground.

April 22nd, 2012 at 9:09 am

Try again, surprisingly it is much more than that.

April 22nd, 2012 at 10:02 am

I agree with asdf.

If c and r are the original circumference and radius, and C and R are the final values, then c = 2πr and C = 2πR, on subtracting we get C-c = 2π(R-r) => R-r = (C-c)/(2π).

But C-c = 1 m and R-r is the distance between the Earth and the chain, so R-r = 1/(2π) ≈ 15.9 cm.

OTOH, if the length of the chain was increased to get a distance of 1 m, then the extra length would be 2π ≈ 6.283 m.

April 22nd, 2012 at 12:23 pm

asdf and Chris, you have misread the question. It is not the entire chain which is pulled from the earth, but only a single point. The resulting chain will not form a circle.

April 22nd, 2012 at 1:59 pm

If I’m not mistaken, we gain approximately/roughly 212.5 m

April 22nd, 2012 at 4:25 pm

Hi Zorglub – guilty as charged. I’ll be back, no doubt to agree with Slavy.

April 22nd, 2012 at 6:45 pm

I get about 121.6447… m.

Regardless of whether Slavy or I am the closest, I like the surprisingly large answer. So thanks for an original (to this site) problem.

I suspect that Slavy used 1 m where he should have used 1/2 m. But then he should have got 193.0996… m. So I further suspect he used at least one approximation.

April 22nd, 2012 at 11:28 pm

Hi Chris, and Zorglub. Apparently, I also didn’t answer the posted problem , but I slightly different one (I blame the two Manchesters and the beer-wine concentration in my blood that they significantly increased last night). 212.5 m is approximately half of the distance where the chain doesn’t touch the ground. The height however is much smaller So my new answer is sqrt(213^2-212.5^2)=sqrt(851)/2 or around 14.5 m.

Chris, I don’t know for which functions you do Taylor expansion, but if you work with tan(x) and x^(1/3) (or if you use a scientific calculator for the arctan(x) and the cubic root, letting the computer approximate for you) you should see that the error for 212.5 is less than 0.1, meaning that 14.5 is pretty accurate

In conclusion, adding one meter allows us to squeeze a couple of elephants in between the earth and the chain Impressive, right? Thanks for the problem, Zorglub!

April 23rd, 2012 at 4:19 am

I like this problem because the answer is way larger than what I had expected. Chris has the answer – that’s a lot of elephants.

And yes Slavy, I worked with the cubic Taylor approximation of tan(x). I suspect that Chris has done the same.

April 23rd, 2012 at 4:55 am

Draw a circle centre of radius r and centre O. Draw a radial line from O, through a point, P, on the circle at 3 o’clock, extending it to a point H outside the circle. Draw a tangent to the circle from H. Let the tangent meet at T (at about 2 o’clock).

Length OT = length OP = r. Length PH = h. Length HT = t. Let A be the angle POT = angle HOT (in radians). We want to find h.

tanA = t/r => t = r tanA. The length of the arc PT is r A. We require that

r tanA – r A = 1/2, where r = 6.4*10

^{6}. Rearrange that as2 r (tanA -A) = 1 and bung 2 *6400000 * (tan(A) -A) = 1 into http://www.wolframalpha.com to get: A = 0.006165498935977687

Then t = 39459.6931902571965 and arc = 39459.1931902571968

=> T is about 40 km away from P (or H)

Pythagoras says, (r + h)² = r² + t² => h = √ (r² + t²) – r

=> h = 121.6447335…

April 23rd, 2012 at 5:04 am

Hi Zorglub, posts crossed. As you see, I didn’t use a Taylor series approximation for my final calculation. I did use tanA ≈ A + A

^{3}/3 while I was practising.I also like the off-the-wall 39.5 km to the point where the chain touches the Earth.

That’s so much better than the problem that I thought you were asking.

Slavy’s problem is that he doesn’t drink coffee

April 23rd, 2012 at 5:25 am

Yeah, I´m an idiot when arithmetics kicks in Having that arcPT=3*r*r/2, I have forgotten to square the radius in the actual computation, and my answer corresponds to arcPT=3r/2

That´s why I was so far away from the truth… The only remark to Chris´ solution – better use Taylor for tanA=A+A^3/3+higher-order-terms, instead of wolframalpha

April 23rd, 2012 at 5:27 am

… and beer tastes so much better than coffee

April 23rd, 2012 at 5:47 am

Hi Slavy, I know you’re not a geometry fan, and that you only make silly mistakes.

If I use the Taylor approximation tanA = A + A

^{3}/3, then I getA ≈ 0.006165530186, t = 39459.89319800 and arc = 39459.3931904

Then h = √ (r² + t²) – r => h = 121.646, so I agree, that’s pretty good.

I also just checked, and accept that including the fifth order term, makes for a hard time (without wolfram etc.)

April 23rd, 2012 at 5:12 pm

Yes, 121.6m is the correct solution. I think of this problem as a nice application of Taylor’s approximation.

April 26th, 2012 at 4:14 am

Oops. For anyone who didn’t follow the Taylor (srictly Maclaurin) series stuff: starting at 2 r (tanA – A) = 1, we have tanA ≈ A + A

^{3}/3=> 2rA

^{3}/3 ≈ 1 => A^{3}≈ 3/(2r) => A ≈ 0.006165530…