rope around the earth

crito2 · April 30th, 2012, 8:44 am

easy one:Let's say that you have a very long rope which goes all the way around the earth (assumed perfectly circular)Add one meter to that rope.what is the maximum height of a building you could build between that rope and the ground?(use R = 6200km)

rmax · April 30th, 2012, 8:47 am

QuoteOriginally posted by: crito2easy one:Let's say that you have a very long rope which goes all the way around the earth (assumed perfectly circular)Add one meter to that rope.what is the maximum height of a building you could build between that rope and the ground?(use R = 6200km)Makes no difference if it is the Earth Sun, or Universe

crito2 · April 30th, 2012, 8:55 am

This is not the famous problem for 2nd graders, once u added 1 meter to the rope, the rope is no longer circular:

samseh · April 30th, 2012, 2:21 pm

I would say about 120 m

AVt · May 1st, 2012, 4:38 pm

Hm ... I get h ~ 12.04

cm27874 · May 2nd, 2012, 4:44 am

0.5m; or didn't I get the joke?

DevonFangs · May 2nd, 2012, 8:47 am

QuoteOriginally posted by: cm278740.5m; or didn't I get the joke?Actually my first guess was 0.5m too but that's actually the minimum (if one doesn't want to put the rope on the ground)

frenchX · May 2nd, 2012, 8:47 am

So let's try.R=6200 kmThe perimeter of the Earth is P=2*Pi*R and the new length of the rope is P+0.001 (only 1 m)The figure is totally symmetric so I 'll focus on half the Earth. h is the height of the buildingL is the length of the rope from the top of the building to the tangent point.L²=(R+h)²-R²We need to calculate now the lengh of the remaning rope around the half earth.Let's call A the angle between the radius passing through the building and the one passing through the tangent point.cos(A)=R/(R+h)the complementary angle is then Pi-A and so the remaining rope is (Pi-A)*Rand then the equation is (Pi-arcos(R/(R+h))*R+sqrt((R+h)²-R²)=Pi*R+0.0005or simplified sqrt((R+h)²-R²)=0.0005+arcos(R/(R+h))*RI find something around 120,36 m as Outrun and Samseh said. EDIT : what about if the Earth is an ellipsoid with a flatness of 1/300 ? For simplicity the building would be build at a pole.

AVt · May 2nd, 2012, 5:50 pm

120 m, agreed - stupid be used km

frenchX · May 3rd, 2012, 7:56 am

What about for the ellipse ? I'll give it a try Rp is the small axis (polar)and Re=Rp*(1+1/300) is the big axis (equatorial)The perimeter is given by the elliptic integral P=4*int(sqrt(Re²*cos(t)^2+Rp²*sin(t)^2),t=0..Pi/2) which can be well approximated by the Ramanujan formula Pi*[3*(Re+Rp)-sqrt((3*Re+Rp)*(Re+3*Rp))]The perimeter is a bit bigger: 39020 km for ellipse compared to 38956 km for the perfect circle.That's where it gets harder and need to calculate a bit before posting.

AVt · May 3rd, 2012, 5:55 pm

I did it in a way different form yours (and the others I guess), find it appended,yours is the first solution given there (using Maple's notation and syntax).Roughly one needs a parametrisation in polar coordinates and the according lengthfor that curve (the latter is an [incomplete] elliptic integral, I guess), whilethe according 'radius' then is obvious. Which certainly is known (Mathworld? Classic geometry books?).The radius for the tangent point now is not a constant and the length not linear.Using that instead one will come up with 2 equations to be solved numericallyagain (since numerically elliptic functions are part of major systems that maybe fast). I would say. But have no desire to dig for the above :-)But the puzzling beauty of the 2 questions is already done using circles:Constant distance independent from radius for the 'classical' task and thehuge height for the 'peaked' task (which does not simply result from addingsouth pole and some waist from the equators)