Pressure at Depth — Mariana Trench (Worked Example)

Given

• h = 10,900 m

• ρ = 1025 kg/m³

• g = 9.8 m/s²

• P₀ = 101,325 Pa

Find

The absolute pressure P.

Principle

P = P₀ + ρgh — at trench depths the ρgh term utterly dominates the atmosphere.

Formula

Solution

1. Substitute: $P = 101325 + (1025)(9.8)(10900)$

2. Gauge part: $109{,}490{,}500\ \mathrm{Pa}$

3. Add: $P = 109{,}591{,}825\ \mathrm{Pa}$

Unit check

ρgh gives N/m² = Pa and P₀ is Pa. Both terms are pressures — the sum is valid.

What this means physically

≈110 MPa — over a thousand atmospheres, like holding 50 jumbo jets on your head. Only specially built submersibles survive; fish there have no air spaces to crush.