Variation on a theme.?

Question

This is the alternative question to Irish Dubliner's original.

A company wishes to lay a cable all the way round th Earth's equator which we will assume is exactly 40,000km. However, the cable is 34.54m too short. A simple solution to this problem is to move the cable a certain amount north or south of the equator so that the cicumference is a little shorter. But how much (either way) should the cable be moved so that it can join up around the circumference of the Earth? (Assume the Earth to be a perfect sphere).

Answer 1

The circle would have to be smaller in circumference by 34.54 meters. If the earth is 40,000,000 meters in circumference, the circumference of the cable circle will have to be 39,999,965.46 meters. Therefore, the ratio of the radius of the allowed cable circle to the circumference of the earth is 0.9999991365. Then we need to take the arccosine of that. The arccosine of this is 0.0013141538146442906265277165645457. We then multiply this by 40000000, and divide by 2 pi, giving 8366.1630233483698395659816511899,
or 8.366 kilometers.

Answer 2

Let R be radius of sphere earth. Consider a sectional cut (parallel to diameter) in lower half of earth at a distance of (x) from centre and having section radius (r).

Given 2*Pie*R = 40000000 meter = N1 and
2*Pie*r = (40000000 - 34.54) meter = N2 (say)

We have to fine x in Km

we have R^2 = x^2 + r^2, we get

x^2 = (N1^2 - N2^2) / 4*(pie)^2= 70063664.02 meter^2

x = 8370.40 meter = 8.370 km answer

so we will move along poles (either sides) by 8.370 km to fix the short cable

Edit : there was minor calculation mistake now corrected.

Answer 3

The radius and thus the circumference is proportional to the cosine of the latitude. The distance north along the surface is the product of the radius and the latitude. Call the circumference C and the cable deficit S then, relating these to latitude (L):

C*cos(L) = C-S

L = acos((C-S)/C)

And the distance from the equator is: D = L*C/(2π) = acos((C-S)/C)*C/(2π) = 8.366 km.

Going one step farther, you can eliminate the arccosine since the angle is so small. Use a Taylor series approximation:

acos(1-x) = sqrt(2x)

D = sqrt(2S/C) * C/2π = sqrt(SC/2)/π

Answer 4

Imagine the inverted cone formed by the cable wrapped around the earth at some latitude theta, with the cone apex at the earth centre. Let radius of Earth=R, radius of cable =r
Then r/R=sin(90-theta)= costheta
2*pi*r/2*pi*R=0.999991365
So theta= 0.2381deg= 4.15572x1o^-3rad
So cable should be moved through 0.2381deg or R*theta(rad)=2645.55555m

Answer 5

34.54=2pi r
r = 5.5

The radius would have to be 5.5 km shorter