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One extra metre: the rope around the Earth

A rope encircles the Earth. Add one metre: how far does it rise? A surprising result that does not depend on the radius.

Articles /rope-around-earth-one-extra-metre

6 min

Imagine a perfectly spherical Earth with a rope wrapped tightly around its equator. We add one metre of rope and arrange it in another circle with the same centre, at a uniform height above the surface.

How far does the rope rise? Almost nothing? A millimetre? Enough for a small ball to pass underneath?

Make a prediction before opening the solution. Do we need to know the Earth's radius?

Show the step-by-step solution

From circumference to radius

Let R be the original radius and h the uniform increase. Measure all lengths in metres. The initial circumference is C = 2πR. The new radius is R + h, so:

C + 1 = 2π(R + h)
2πR + 1 = 2πR + 2πh
1 = 2πh
h = 1/(2π) m ≈ 0.15915 m ≈ 15.9 cm

The rope rises by about 15.9 centimetres all the way around! R cancels out: we do not need to know the size of the Earth.

Why does intuition mislead us?

One metre seems negligible compared with the whole equator. Indeed, the percentage change in length is tiny. But we are looking for an absolute change in radius: each extra metre of circumference always produces the same increase, 1/(2π) metres.

The circumference formula is linear: every extra centimetre of radius requires 2π centimetres of rope, whatever the original circle.

The general formula and the inverse problem

If we add a length d, the uniform rise is:

h = d/(2π)
d = 2πh

To raise the rope uniformly by one metre, we therefore need 2π m ≈ 6.283 m of extra rope, not just one metre.

Earth, football or orange?

Repeat the ideal experiment around a football or an orange: adding one metre always increases the gap by about 15.9 cm. The percentage increase changes enormously, but the absolute increase does not.

Mind the assumptions

A spherical Earth and a circular rope form a mathematical model: we ignore mountains, elasticity, thickness and gravity, imagining supports that hold the rope in place. It is not supposed to float unsupported.

Lifting the rope at just one point is a different problem. Its shape is no longer a concentric circle, so this formula cannot be applied directly. Here “rope” means a length of string, not a geometrical chord joining two points on a circle.

Four variations to try

  1. Just one extra centimetre

    What is the uniform rise?

    Show the solution

    h = 1/(2π) cm ≈ 0.159 cm = 1.59 mm. Units matter!

  2. Letting a small ball through

    How much rope must we add to create a uniform 5 cm gap?

    Show the solution

    d = 2π · 5 cm = 10π cm ≈ 31.4 cm. A ball 5 cm in diameter fits exactly in the ideal model; extra clearance requires slightly more rope.

  3. Two metres instead of one

    Does the rise double or quadruple?

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    It doubles: h = 2/(2π) m = 1/π m ≈ 31.8 cm. The relationship is linear, not quadratic.

  4. A 1% increase

    If we increase the rope's original length by 1%, is the result still independent of the radius?

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    No: d = 0.01 · 2πR, so h = 0.01R. Adding a fixed length and adding a fixed percentage are different operations.