String Girdling Earth Calculator

Solve the famous rope around Earth puzzle - calculate additional string length or resulting gap

Calculate String Girdling Effects

Length of string to be spliced into the original rope

Gap Results

Famous Examples

Classic Problem: Lift 1 meter

Question: If you lift a rope around Earth 1 meter off the ground, how much extra string do you need?

Answer: ΔL = 2π × 1m = 6.283 meters ≈ 6.3 meters

Surprising fact: Earth's radius (6,371 km) doesn't matter!

Reverse Problem: Add 1 meter of string

Question: If you add 1 meter to the rope around Earth, what's the gap?

Answer: d = 1m ÷ (2π) = 0.159 meters ≈ 16 cm

Result: A cat could fit through the gap! 🐱

Earth Reference Data

Radius6,371 km
Circumference40030 km
Circumference24874 miles

The formula works for any spherical object, not just Earth!

Key Formulas

Gap from Length
d = ΔL ÷ (2π)
d = gap height, ΔL = added length
Length from Gap
ΔL = 2π × d
ΔL = required length, d = desired gap
Circumference
C = 2π × r
C = circumference, r = radius

Key Insights

🤯

The object's size doesn't matter - same result for Earth, a coin, or the sun!

📏

Adding 1 meter creates only a 16cm gap around Earth

🧮

The relationship is always linear: gap = length ÷ (2π)

🏃

This principle applies to athletics track starting lines

Understanding the String Girdling Earth Problem

The Classic Puzzle

Imagine a string tightly wrapped around Earth's equator. If you lift this string uniformly 1 meter off the ground all around the equator, how much longer would the string need to be? The answer is surprisingly small - only about 6.3 meters!

Why It's Counterintuitive

  • Earth's circumference is ~40,000 km (25,000 miles)
  • Yet adding just 6.3m lifts the entire rope 1 meter
  • The Earth's size doesn't affect the calculation!

Mathematical Explanation

Original circumference: C₁ = 2πR

New circumference: C₂ = 2π(R + d)

Additional length: ΔL = C₂ - C₁

Simplified: ΔL = 2π(R + d) - 2πR = 2πd

Key Insight

The radius R cancels out completely! The additional length depends only on the height increase (d), not the original size of the object.

Real-World Application

Athletics tracks use this principle - outer lanes have staggered starting lines offset by 2π × lane width to ensure equal distances.

Any Sphere Works

This formula applies to any spherical object - a marble, basketball, planet, or star. The size doesn't matter!

Linear Relationship

Double the height, double the required length. The relationship between gap and additional string is perfectly linear.