[superprismatic]

Suppose the earth were a perfectly smooth

sphere with radius 6,371,000 meters. Now,

there's an old problem which supposes that

a rope girdles the earth along a great

circle. An extra meter of rope is then

spliced into it. You then assume that the

rope is raised the same amount above the

earth's surface all the way around. You

are asked to determine what the amount of

the rise is. It's pretty easy to see that

the rise is independent of the radius of

the earth and is 1/(2π) meters. But we

consider a different problem here:

Suppose the earth were a perfectly smooth

sphere with radius 6,371,000 meters. Now,

suppose a rope girdles the earth along a

great circle. An extra meter of rope is

then spliced into it. Now, grab the rope

in one spot and pull it taut directly away

from the center of the earth. How far

above the surface of the earth will it be

able to be pulled? Give the answer to the

nearest meter.

[Guest]

???

I would think that the number is so insignificant, it to the nearest meter would be zero.

[Peekay]

A quick calculation results in 1,346,596 meters. Definitely need to check this once more. But, I have to leave now...

Edited June 2, 2011 by Peekay

[EventHorizon]

Where the rope starts away from the earth it will be tangent to the surface. This means that there's a right angle formed be the spot you're pulling, the spot it is tangent, and the center of the earth.

If you let x be the distance of the rope from the spot it is tangent and the spot you are pulling, you get the following equation (where theta/2 is the angle of the right triangle at the center of the earth, and r is the earth's radius):

tan(theta/2) = x/r

The length of rope had the rope been laying on the earth would then be theta*r (circle segment length equation).

We know 2x = theta*r+1 simply because they are the differences in the length of the rope.

Now we have 2 equations and 2 unknowns.

6,371,000 * theta + 1 = 2x

x = 6,371,000 * tan(theta/2)

6,371,000 * theta + 1 = 2 * 6,371,000 * tan(theta/2)

theta + 1/6,371,000 = 2 tan(theta/2)

Plugging it into my calculator gives theta = .0123496791.

What we want is the hypotenuse of the right triangle minus the radius of the earth.

So 6,371,000/cos(theta/2) - 6,371,000 = 121.4607233

So unless I've made some mathematical errors, the answer is 121 meters when rounded to the nearest meter.

[Guest]

Assuming the rope is perfectly taut, it would be 1/2 meter right? An extra meter is in the rope, so that is what is pulled, and it needs to reach a point above the surface and return, so 1/2 meter above the surface.

[EventHorizon]

Assuming the rope is perfectly taut, it would be 1/2 meter right? An extra meter is in the rope, so that is what is pulled, and it needs to reach a point above the surface and return, so 1/2 meter above the surface.

Lets adapt the problem to a smaller scale. You have an 8 meter rope tied to two posts 8 meters apart. You add in 2 meters more of rope. You go to the center and pull the rope up. How high could you pull it?

Your solution would be that it is 1 meter up, right? This means that the rope would need to still be along the ground along the whole length of the rope, except for the 2 meters of rope you are holding.

My understanding is that the problem is that when you lift up the rope, the rope will only touch the ground at the ends where it is tied down. The rope will then form two straight lines (post to you to other post). This allows you to lift it up a lot further, and the rope will end up taut at 3 meters up (can you see why?).

Try finding analogous solution for the rope around the earth.

Also, it's interesting there are 4 responses and 4 different answers

[Guest]

Also, it's interesting there are 4 responses and 4 different answers

Let me offer my 5th solution for which I didn't do any math:

Slightly more than 1/2 a meter.

I didn't do the math to confirm it, but if you pull the rope tightly around the earth and then add a meter in one spot, that rope goes up half a meter and then goes back down. But now you get to pull it taught which would bring the point slightly more than half a meter above the Earth's surface. I didn't do the math to get the actual number but I think it's a very small amount more than half a meter.

[Peekay]

Made a stupid mistake in my hurry. No wonder the result was ridiculous!

Eventhorizon's solution checks out all right! Good work.

[Guest]

I was standing on the rope while pulling it in my answer.

To solve based on your explanation, I would say pulling the rope creates 2 equal right triangles at that point, with the other points being the center of the earth and the "poles" that are perpendicular to you. The length of one side is r, the other side is r+x where x is distance above the surface. The hypotenuese is based on the circumference. Half the circumference worth of rope is used on the other side of the world, so we have half left plus the extra meter. Half of that is used by each triangle for its hypotenuese, for 1/2 meter plus 1/4*2*pi*r, or 1/2*pr*r + 1/2 meter. Using Pythagorean Theorem gives r² + (r+x)² = 1/4(pi*r)² + 1/2(pi*r) + 1/4

Leaving the (r+x)² on one side and moving r² to the other, and then plugging in for r and solving, then taking the square root of both sides and subtract r from both sides leaves x = 1,346,596 meters.

That is if the rope leaves the surface at the poles. But that is nearly the same answer I get if instead of 1 extra meter added to rope, I add zero meters, and I do not think that is right. So my answer is probably based on the incorrect assumption that the rope leaves the surface at the poles.

Lets adapt the problem to a smaller scale. You have an 8 meter rope tied to two posts 8 meters apart. You add in 2 meters more of rope. You go to the center and pull the rope up. How high could you pull it?

Your solution would be that it is 1 meter up, right? This means that the rope would need to still be along the ground along the whole length of the rope, except for the 2 meters of rope you are holding.

My understanding is that the problem is that when you lift up the rope, the rope will only touch the ground at the ends where it is tied down. The rope will then form two straight lines (post to you to other post). This allows you to lift it up a lot further, and the rope will end up taut at 3 meters up (can you see why?).

Try finding analogous solution for the rope around the earth.

Also, it's interesting there are 4 responses and 4 different answers

[EventHorizon]

A funny application of this new found phenomena: When people pull the waist-line of their pants out to show how much weight they've lost, you can show them the math and convince them they didn't really lose much circumference.

edit: Just don't start with, "assume the radius of your waist is that of the earth..." as it might make them feel bad.

Edited June 3, 2011 by EventHorizon

[Guest]

A funny application of this new found phenomena: When people pull the waist-line of their pants out to show how much weight they've lost, you can show them the math and convince them they didn't really lose much circumference.

edit: Just don't start with, "assume the radius of your waist is that of the earth..." as it might make them feel bad.

So wait, are you sure your answer is correct EventHorizon? I don't see how it could be that large of an answer. It's certainly less intutive to me then the 1/2pi solution.

[Guest]

x

Edited June 3, 2011 by Dej Mar

[Guest]

Eventhorizon, I think you messed up at the last step. I'll bold and edit your aswer here:

Where the rope starts away from the earth it will be tangent to the surface. This means that there's a right angle formed be the spot you're pulling, the spot it is tangent, and the center of the earth.

If you let x be the distance of the rope from the spot it is tangent and the spot you are pulling, you get the following equation (where theta/2 is the angle of the right triangle at the center of the earth, and r is the earth's radius):

tan(theta/2) = x/r

The length of rope had the rope been laying on the earth would then be theta*r (circle segment length equation).

We know 2x = theta*r+1 simply because they are the differences in the length of the rope.

Now we have 2 equations and 2 unknowns.

6,371,000 * theta + 1 = 2x

x = 6,371,000 * tan(theta/2)

6,371,000 * theta + 1 = 2 * 6,371,000 * tan(theta/2)

theta + 1/6,371,000 = 2 tan(theta/2)

Plugging it into my calculator gives theta = .0123496791.

What we want is the hypotenuse of the right triangle minus the radius of the earth.

So 6,371,000/cos(theta/2) - 6,371,000 = 121.4607233 I think you messed up here by using the circumference of the Earth accidentally instead of the radius. I think the correct is:

6,371,000/(2*pi*cos(theta/2)) - 6,371,000/(2*pi) = 19.3310745 meters

So I think 19 meters is the correct answer, although that's still a lot larger than I thought the answer would be just guessing.

So unless I've made some mathematical errors, the answer is 121 meters when rounded to the nearest meter.

[Guest]

Lets adapt the problem to a smaller scale. You have an 8 meter rope tied to two posts 8 meters apart. You add in 2 meters more of rope. You go to the center and pull the rope up. How high could you pull it?

Your solution would be that it is 1 meter up, right? This means that the rope would need to still be along the ground along the whole length of the rope, except for the 2 meters of rope you are holding.

My understanding is that the problem is that when you lift up the rope, the rope will only touch the ground at the ends where it is tied down. The rope will then form two straight lines (post to you to other post). This allows you to lift it up a lot further, and the rope will end up taut at 3 meters up (can you see why?).

Try finding analogous solution for the rope around the earth.

Also, it's interesting there are 4 responses and 4 different answers

The circumference of the given ‘spherical’ Earth is 12742000 π meters. The length of the rope is 12742000π + 1 meters. The rope will hug the Earth 9556500 π meters before it pulls away from the surface. The remaining length of the rope will form an isosceles triangle with like side lengths of approximately 5003772.199 meters and a base length, the length of the chord of the great circle where the rope is tangent, of 6371000∙√(2) meters.

Half the length of the chord and the height of the triangle forms the two sides of a right triangle where the hypotenuse is the length of the side of the isosceles triangle. Using the Pythagorean theorem, the altitude of the right triangle is approximately 2177829.6199 meters.

The distance from the surface of the Earth is this distance less the length of the sagitta (versine) of the chord. The sagitta is the difference between the radius of the circle and the apothem. Given that the central angle is π/2 radians (a right angle), the apothem is equal to half the length of the chord, which is the radius/√(2). The length of the sagitta, then, is approximately 1866022.697 meters. The distance above the surface of the Earth then calculates to be approximately 311801 meters.

[Guest]

Ignore my prior post. I see where I made an error.

[Guest]

I reworked it and agree with EventHorizon's solution. I wonder if Use The Force mistook 6371 km for the circumference when it is given as the radius?

I did this after reading EventHorizon's solution, so it is my working through after seeing the solution.

My prior answer assumed an incorrect point where the rope leaves the planet. It should leave at the point where the tangent line is pointing directly to the point x above the surface that the rope is pulled to. Any point before this and the line will pass through the planet. There is a right angle at this point, with side A being the radius, side B being the line from the tangent point to where the rope is held, and line C being the hypotenuse which is r+x with x being the distance above the surface that the rope reaches. Side B will be 1/2 meter plus the length of the rope had it lain on the surface from this point to you. If the angle between A and C is theta in radians, then B = 1/2 + 2pi*r * (theta/2pi) = 1/2 + r*theta

We also know that tan(theta) = B/A, so B = r*tan(theta)

1/2 + r*theta = r*tan(theta)

tan(theta) - theta = 1/2r

theta = 0.00617484

To get C, note that cos(theta) = A/C so C=r/cos(theta) = 6371121.461

and subtracting r to get x, is 121 meters.

B, for the curious, is 39340 meters.

These values for A, B, and C check with Pythagorean theorem.

[Guest]

Let

A and B be the points at which the rope loses contact with the earth.

C is the point of pulling.

O is the radius of the earth.

x = distance between the point of pulling of the rope and A (or B for that matter)

r = Earth’s radius

theta = angle subtended at the earth's centre by AB (in radians)

Remember, theta << 1

Therefore, tan (theta/2) = theta/2 = x/r

Theta = 2x/r

Now,

L1 = Length of the rope in contact with the earth (the longer portion between A and B = 2*pi*r* ((2*pi – theta)/(2*pi))

L1 = r*(pi+theta) = r*(Pi + 2x/r)

Total length of the extended rope = 2*pi*r + 1 metres = L1 + 2x

Therefore, 2*pi*r + 1 = r*(Pi + 2x/r) + x

Assuming x << r, and solving this,

x = ½ metres (irrespective of r).

The length in question, i.e. how far above the surface of the earth, the rope will be able to be pulled,

Let it be y.

Therefore, from the right angled triangle COB, (y+r)^2 = r^2 + x^2

As r>> x as well as y, this gives y = 0

(I have tried to evaluate to 10 decimal places, still 0)

One useful, but evident conclusion, the rope will leave the earth for 2x metres = 1 metre (the extension provided)

EventHorizon, started well, but perhaps swayed somewhere in between. See the Spoiler.

[Guest]

ideally it should form a equilateral triangle, where 1 extra meter is used by the two sides.

Since the distance of vertex from opposite side for such a triangle is approx. 0.433 meter, the same should be the height in the question.

Edited June 3, 2011 by KlueMaster

[EventHorizon]

Let

A and B be the points at which the rope loses contact with the earth.

C is the point of pulling.

O is the radius of the earth.

x = distance between the point of pulling of the rope and A (or B for that matter)

r = Earth’s radius

theta = angle subtended at the earth's centre by AB (in radians)

Remember, theta << 1

Therefore, tan (theta/2) = theta/2 = x/r

Theta = 2x/r

Now,

L1 = Length of the rope in contact with the earth (the longer portion between A and B = 2*pi*r* ((2*pi – theta)/(2*pi))

L1 = r*(pi+theta) = r*(Pi + 2x/r)

Total length of the extended rope = 2*pi*r + 1 metres = L1 + 2x

Therefore, 2*pi*r + 1 = r*(Pi + 2x/r) + x

Assuming x << r, and solving this,

x = ½ metres (irrespective of r).

The length in question, i.e. how far above the surface of the earth, the rope will be able to be pulled,

Let it be y.

Therefore, from the right angled triangle COB, (y+r)^2 = r^2 + x^2

As r>> x as well as y, this gives y = 0

(I have tried to evaluate to 10 decimal places, still 0)

One useful, but evident conclusion, the rope will leave the earth for 2x metres = 1 metre (the extension provided)

EventHorizon, started well, but perhaps swayed somewhere in between. See the Spoiler.

tan(theta/2) = .0061749181

theta/2 = .0061748396

When r is so big, those small errors make a big difference. And you are essentially assuming y=0 by saying x<<r, when it is really about 1/162 of the length (x != 1/2).

Does it even make intuitive sense that it you would only be able to raise it 1 meter? Lets say you have it held down all around the earth except for a 4 meter length where you have 5 meters of rope. Using simple 3-4-5 triangles, you can easily see that you can lift the center 1.5 meters (eg, two 1.5-2-2.5 triangles)

If you had 10 meters between where it was held down it would be sqrt((11/2)^2 - (10/2)^2) = 2.2 meters up if help up at the center. And it only gets higher with more length to work with.

ideally it should form a equilateral triangle, where 1 extra meter is used by the two sides.

Since the distance of vertex from opposite side for such a triangle is approx. 0.433 meter, the same should be the height in the question.

an isosceles triangle. An equilateral needs all 3 the same length.

[Guest]

I worked and reworked this problem (as I made a big error the first time), and I came to the same conclusion as EventHorizon. With most fractions given rounded to the sixth decimal place, the precision actually used was much greater.

θ ≈ 0.01234967890883 radians

rope in contact with the Earth = (2π – θ)∙6371000

length of arc segment between the two tangent points of the rope and the Earth = θ∙6371000 ≈ 78679.804328

length of rope segment not in contact with the Earth = θ∙6371000 + 1 ≈ 78680.804328

An isosceles triangle is formed with two halves of the extended rope and the chord that is formed at the two tangent points. Let s = half the length of the extended rope

s ≈ 39340.402164

A right triangle is formed with s and R, which gives the distance from the center of the Earth to the point of the rope that would be pulled. Using the Pythagorean theorem, this gives the approximate length of 6371121.460720. Subtracting the radius of the Earth gives the distance the rope is pulled away from the surface:

6371121.460720 – 6371000 ≈ 121.460720 meters.

[Guest]

It is amazing to imagine: a rope is tightly wrapped around the earth. You cut it at a point, tie another rope to increase its length by just a meter, and then you can pull it up all the way to 121 metres high!!!

Also, the rope will lose contact with the earth for as long as 78,680 metres.

tan(theta/2) = .0061749181

theta/2 = .0061748396

When r is so big, those small errors make a big difference. And you are essentially assuming y=0 by saying x<<r, when it is really about 1/162 of the length (x != 1/2).

Does it even make intuitive sense that it you would only be able to raise it 1 meter? Lets say you have it held down all around the earth except for a 4 meter length where you have 5 meters of rope. Using simple 3-4-5 triangles, you can easily see that you can lift the center 1.5 meters (eg, two 1.5-2-2.5 triangles)

If you had 10 meters between where it was held down it would be sqrt((11/2)^2 - (10/2)^2) = 2.2 meters up if help up at the center. And it only gets higher with more length to work with.

an isosceles triangle. An equilateral needs all 3 the same length.

You are right EventHorizon. See the Spoiler

[dark_magician_92]

i think

under ideal stretching, it shud be 0.5 metres, and in real case it wud be slightly greater than 0.5 m if the thickness of rope is <<1 meters, dunno something feels me there is more to it .

[Guest]

Someone needs to recheck some assumptions I think. Adding 1 m of rope, logically, I don't beleive can create 128+ meters of rope. Now if you mean that adding 1 meter to the RADIUS then I may agree. As proven in a previous post, adding 10 meters of rope to a 40,000 rop provides just 1.6 ft of clearance.

[dark_magician_92]

i must get rid of posting answers as soon as i see the question, without dee thinking. once again am wrong

[Peekay]

Going by all the posts so far, I think EventHorizon deserves a BIG round of applause for keeping a cool head and going through the calculations properly. I agree with him that even a small rounding error will result in a big difference in the solution.

[Guest]

Let

r be the radious of tha sphere.

y be the distance from the ground to the top of where you pull the rope

a be the angle from where you stand to where the rope is tangent to the sphere

x be the distance from where the rope is tangent to the earth to where you are pulling the rope

tan (a)=x/r

hence

x = r*tan(a)

2*pi*r - 2*a*r + 2x = 2*pi*(r+1)

substitute x

2*pi*r - 2*a*r + 2*r*tan(a) = 2*pi*(r+1)

2*r*tan(a)-2*a*r=2*pi

a~~0.6528°

We also know:

r^2 + x^2 = (r+y)^2

substitute x = r*tan(a)

r^2 + (r*tan(a))^2 = (r+y)^2

solve for y

y ~ 413 m