Suppose the Earth is perfectly spherical. A string is wrapped tightly around the equator of the Earth. Another string, 1 metre longer, is placed around the Earth so that it forms a larger circle, staying the same distance above the ground everywhere.
How high above the ground is the second string?
Repeat this for: (i) the Moon (ii) Jupiter (iii) a volleyball. Are you surprised by the three answers? Why or why not?
Answer: 2π(R + x) − 2πR = 1 ⇒ x = 12π m ≈ 0.159 m ≈ 16 cm. The same 16 cm for (i) the Moon, (ii) Jupiter and (iii) a volleyball, because the radius R cancels.
Step-by-step solution
Idea: Circumference = 2π × radius, so adding 1 m to the circumference always adds 12π m to the radius, whatever the original radius.
- Let the Earth’s radius be R m and the second string be x m above the ground. Its radius is R + x.½ mark
- Length difference: 2π(R + x) − 2πR = 1 ⇒ 2πx = 1.1 mark
- x = 12π ≈ 16.283 ≈ 0.159 m, about 16 cm.1 mark
- (i) Moon, (ii) Jupiter, (iii) volleyball: the working is exactly the same with a different R, and R cancels each time, so the string is again 12π m ≈ 16 cm high in all three cases.1½ marks
- Most people are surprised: 1 m seems tiny next to the Earth’s 40,000 km equator, yet it lifts the string 16 cm, the same as for a volleyball. It is not really surprising once we see that circumference grows by 2π for every unit of radius, whatever the size.1 mark
Height = 1/(2π) m ≈ 0.159 m ≈ 16 cm, the same for the Earth, the Moon, Jupiter and a volleyball, because the radius cancels out.
Check: Volleyball radius about 0.105 m: circumference 0.66 m, longer string 1.66 m, radius 1.66 ÷ 2π ≈ 0.264 m, gap 0.264 − 0.105 ≈ 0.159 m ✓.
Answer to write in the exam
2π(R + x) − 2πR = 1
2πx = 1 ⇒ x = 12π m ≈ 0.159 m
R does not appear in x
∴ ≈ 16 cm for Earth, Moon, Jupiter and volleyball alike
Common mistakes that cost marks
- Working out the huge Earth circumference and expecting a tiny gap, without simplifying the algebra first.
- Writing 2πx = 1 as x = 2π.
- Answering in km after using the Earth’s radius in km while the extra length is in metres.
How this can come in the exam
MCQ (1 mark)
A circular running track’s lane is moved 1 m outward all round. By how much does the lane length increase?
- 1 m
- 2 m
- π m
- 2π m
Show answer
(D) 2π m
2π(r + 1) − 2πr = 2π m, whatever the radius.
Try one yourself
How much longer must the string be to stand 1 m above the ground all round the equator?
Show answer
2π × 1 ≈ 6.28 m longer.
More questions like this
- We have a cylinder with a base radius of r cm and height h cm. A square pyramid is fitted inside it. The square base of the pyramid lies on the base of the cylinder, its corners on the boundary of the cylinder. The apex of the pyramid lies on the top of the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.
- What is the change in volume when:
- Given a cube with volume V, express its total surface area S in terms of V.
- Given a cube with total surface area S, express its volume V in terms of S.
- A cylindrical glass of height 25 cm and radius 4 cm has water up to a height of 16 cm. A crow wants to drink water from this glass. The water must be at a height of 20 cm for the crow to reach it. There are some marbles lying around. How many marbles of radius 1 cm should the crow drop into the glass to make the water reach the required height?