The Earth’s surface has an estimated volume of 1.38 billion km3 of water. Suppose the Earth is a perfect sphere and all this water forms a uniform layer completely covering the Earth’s surface, like a thin water bubble. Estimate the thickness of this water layer. The Earth’s radius is ~6371 km.
- (i) Write an expression that gives the thickness of this water layer.
- (ii) Simplify the expression in (i) using a calculator.
Answer: (i) t = ∛(R3 + 3V4π) − R, or, since the layer is thin, t ≈ V4πR2. (ii) t ≈ 1.38 × 1094π × 63712 ≈ 2.7 km.
Step-by-step solution
Idea: The water is a hollow spherical shell: outer sphere (radius R + t) minus the Earth (radius R). A thin layer has volume ≈ surface area × thickness.
(i) Write an expression that gives the thickness of this water layer.
- Let t be the thickness. Water volume V = 43π(R + t)3 − 43πR3, so t = ∛(R3 + 3V4π) − R.1 mark
- Because t is tiny compared with R, the layer is like a thin skin: V ≈ 4πR2 × t, so t ≈ V4πR2.1 mark
t = ∛(R3 + 3V4π) − R ≈ V4πR2
(ii) Simplify the expression in (i) using a calculator.
- 4πR2 = 4 × 3.1416 × 63712 ≈ 5.10 × 108 km2 (the Earth’s surface area).1 mark
- t ≈ 1.38 × 1095.10 × 108 ≈ 2.7 km. (The exact cube-root formula gives 2.704 km, practically the same.)1 mark
About 2.7 km
(i) t = ∛(R³ + 3V/(4π)) − R ≈ V/(4πR²). (ii) t ≈ 2.7 km.
Check: Shell volume with t = 2.704: 43π(6373.7043 − 63713) ≈ 1.38 × 109 km3 ✓.
Answer to write in the exam
(i)
V = 43π(R + t)3 − 43πR3
∴ t = ∛(R3 + 3V4π) − R ≈ V4πR2
(ii)
4πR2 = 4π(6371)2 ≈ 5.10 × 108 km2
t ≈ 1.38 × 1095.10 × 108
∴ t ≈ 2.7 km
Common mistakes that cost marks
- Dividing the water volume by the Earth’s volume instead of its surface area.
- Using the diameter 12 742 km as R.
- Writing 1.38 billion as 1.38 × 106.
How this can come in the exam
MCQ (1 mark)
A ball of radius 10 cm is given a coat of paint 0.1 cm thick. The volume of paint is about
- 40π cm3
- 4π cm3
- 400π cm3
- 0.4π cm3
Show answer
(A) 40π cm3
Surface area × thickness ≈ 4π × 100 × 0.1 = 40π cm3.
Try one yourself
A spherical ball of radius 7 cm is covered with a layer of chocolate 0.5 cm thick. Find the volume of chocolate (exact shell). (π = 227)
Show answer
43 × 227 × (7.53 − 73) = 8821 × 78.875 ≈ 330.5 cm3 (thin-layer estimate 4π × 49 × 0.5 = 308 cm3).
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