In a mine 0.6 km deep, atmospheric pressure is 790 mm Hg. What is it equal to at this time on earth?

It has been experimentally established that with a change in the height (depth) h of the observer’s level relative to the earth’s surface with a zero level h₀ = 0 m (sea level) for every Δh = 12 meters, atmospheric pressure changes by Δp = 1 mm Hg. Art., while going up, it decreases, while going down, it increases. Then the ratio h: Δh = N – shows how many times (N) this value of h includes Δh. This means that the same number of times, when descending into the mine, the pressure increased by Δp and became by N · Δp more than it was at the zero level р₀, that is, р = р₀ + N · Δр. We get the formula for calculating the pressure at this time on the ground:

p₀ = p – N · Δp or p₀ = p – h · Δp: Δh.

It is known from the problem statement that the mine has a depth of h = 0.6 km = 600 m and that the atmospheric pressure in the mine is p = 790 mm Hg. Art., we get:

р₀ = 790 mm Hg. Art. – 600 m 1 mm Hg. Art. : 12 m;

р₀ = 740 mm Hg. Art.

Answer: at the moment the atmospheric pressure on earth is 740 mm Hg. Art.