Saturday night tricky math-err-matical problem

[Mark R]

A rope is tied around the equator of the earth and pulled tight.

One inch of slack is added to the rope. Then, at one point the rope is lifted vertically up, so that the maximum amount of rope is pulled taut to the ground.

Assume that the rope cannot stretch.

How high above the surface of the earth can the rope be lifted?

You may need the following information:
Radius of the earth: 3,958.8 miles

 

[myusername]

I read Saturday Night Live and thought it was going to be a funny vidcap post 🙁

 

[Zanix]

It's a radius problem, not a earth/equator problem.

Edit: I mean, I did this one in class last semister. It's about radius vs circumfrence.

 

[Eeezee]

> than 10 miles

 

[jchu14]

good problem, still trying to work it out as i type

Heres my approach and picture

So you the extended string held straight up off a sphere. You know that where the string and the sphere touches, it is tangent and makes 90 degree angle. Draw perpendicular lines from the point of tangency and the two lines meet at the center.

Let 'L'=lenght of each side of the protruded string and 'h' be the height off the ground, (Theta) be the angle made by the two perpendicular lines, and R is the radius of the Earth.

You know
eq1) 2*L-R*(theta)=1 inch ................. R(Theta)=arc length

from pythoreans,
eq2) L=sqrt( (R+h)^2-R^2)

look at the triangle formed by L, (R+h), and R
0.5(theta)=arctan(L/R)
eq3) theta=2*arctan(sqrt( (R+h)^2-R^2)/R)

plug 3) back to 1) and move around i got:

2sqrt(((R+h)^2-R^2))-R*arctan(sqrt( (R+h)^2-R^2)/R)-1=0

solve (i used matlab to solve numerically)

I got 413.19 inches O_O

ummm yea..... someone check me

 

[Mark R]

Winnar!

I did the geometry slightly differently, but the answer is the same.

I derived the 3 equations.
d = length of each side of the string where it is lifted off the ground
h = height off the ground at the point
theta = angle at the point where the string is lifted

1) h = sqrt (d^2 + R^2) - R
2) d = R*theta + 0.5
3) theta = asin (d / (R +h))

Then solved numerically with excel. Answer is, indeed, 413 inches.

 

[arcenite]

That would have to be a serious rope... a rope can only be so long before it can't even support itself.

 

[DrPizza]

Do I also have to assume that the earth's surface will not deform or otherwise allow the rope to be anywhere other than exactly on the surface?
I mean... if we actually could perform this experiment with a non-stretchable rope, the rope clearly will not ride on the surface of the ocean.

Come to think of it (I realized this after thinking about how the thickness of the rope affects the problem) (Here's some ideas to add if it's a homework problem) - IF the rope is unstretchable, then the rope can't be lifted, for if the rope is unstretchable, it cannot be pulled tight around the earth - it would have to be made in a circular shape. Pulling the rope above the earth would cause it to bend at some point... which causes stretching in the rope. Given rope of 1 inch diameter, I believe the top of the rope would have to be 2*Pi inches longer than the part of the rope touching the surface of the earth. Well... I suppose we could get around this by saying that the rope is compressible but not stretchable 🙂

And, are we to assume the radius of the earth to be exactly 3,958.8 miles at the equator? I think that's the mean radius of the earth (without looking for where that number came from)...

The IUGG value for the equatorial radius of the Earth is 6378.137 km (3963.19 statute miles Eric Weisstein's World of Physics), giving an equatorial circumference Eric Weisstein's World of Math of 40,075 km (24,901.5 miles)

The answer above appears to be correct, but if you looked up the radius of the Earth, you may want to change it to have the correct radius for the equator.

 

[coomar]

the earth is not a perfect sphere

 

[her209]

42 miles

 

[ijester]

If you simply pinch the rope between your fingers, and keep the entire rope tight, you could make a 1/2" high loop.

 

[TuxDave]

Oops, read the problem wrong..