How High the Air Goes

solved PressureAtDepth

Q. The density of the atmosphere at sea level is 1.29 kg/m³ $ρ = 1.29$ kg/m³. Assume that it does not change with altitude. Taking g as 9.8 m/s² $g = 9.8$ m/s² and the pressure at sea level as 1.01 × 10⁵ Pa $P = 1.01 \times 10^{5}$ Pa, how high would the atmosphere extend $P_{a} = 0$ Pa, in metres?

Inputs

density

ρ

1.29 kg/m³

gravity

g

9.8 m/s²

sea level

P

1.01e+05 Pa

Outputs

height

h

How it's solved

Solved with PressureAtDepth

Given

$ρ = 1.29$ kg/m³
$g = 9.8$ m/s²
$P = 1.01 \times 10^{5}$ Pa
$P_{a} = 0$ Pa from “how high would the atmosphere extend”

Find $h$ .

Working

pressure adds the surface s(sea level) $P = P_{a} + P_{g}$
$101000 = 0 + P_{g}$
$P_{g} = 101000$ Pa
pressure of the fluid above(sea level) $P_{g} = ρ \cdot g \cdot h$
$101000 = 1.29 \cdot 9.8 \cdot h$
$h = 7989.24 m$
$h \approx 7.99 \times 10^{3}$ mto 3 significant figures

Conclusion

Sea level sits at the bottom of an ocean of air; if the air were as dense all the way up, its pressure would be used up within a few kilometres.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :how_high_the_air_goes do4  kind :numerical5  describes "The density of the atmosphere at sea level is %{density}. Assume that it does not change with altitude. " \6            "Taking g as %{gravity} and the pressure at sea level as %{sea_level}, %{height}, in metres?"7  given density: "1.29 kg/m3"8  given gravity: "9.8 m/s2"9  given sea_level: "1.01e5 Pa"10  asks height: "how high would the atmosphere extend"11end

solution.rb

1require_relative "../questions/how_high_the_air_goes"2require_relative "../science/mechanical_properties_of_fluids/pressure_at_depth"34solve :how_high_the_air_goes, using: PressureAtDepth do5  objects :sea_level67  take density: rho8  take gravity: g9  take sea_level: sea_level.pressure1011  deduce_from "how high would the atmosphere extend", P_a => "0 Pa"1213  height = sea_level.depth14  answer height:, in: "m", figures: 31516  conclusion "Sea level sits at the bottom of an ocean of air; if the air were as dense all the way up, " \17             "its pressure would be used up within a few kilometres."18end

playground.rb

1require_relative "../solutions/how_high_the_air_goes"23play_with :how_high_the_air_goes do4  title "How High the Air Goes"5end

The laws it drew on

quantity	written	units	within	from
mass	$m$	kg	0 … ∞	Pressure
gravity	$g$	m/s²	—	Pressure
thrust	$F$	N	—	Pressure
area	$A$	m²	—	Pressure
pressure	$P$	Pa	0 … ∞	Pressure
pressure at the surface	$P_{a}$	Pa	—	PressureAtDepth
density	$ρ$	kg/m³	0 … ∞	PressureAtDepth
depth	$h$	m	—	PressureAtDepth
gauge pressure	$P_{g}$	Pa	—	PressureAtDepth
pressure of ground	$P_{g}$	Pa	0 … ∞
thrust on ground	$F_{g}$	N	—
area of ground	$A_{g}$	m²	—
pressure of sea level	$P$	Pa	0 … ∞
thrust on sea level	$F$	N	—
area of sea level	$A$	m²	—
gauge pressure on ground	$P_{g,g}$	Pa	—
depth of ground	$h_{g}$	m	—
gauge pressure on sea level	$P_{g}$	Pa	—
depth of sea level	$h$	m	—
mass of sea level	$m$	kg	0 … ∞

P_{s} = \frac{F_{s}}{A_{s}}

pressure is force over area

P_{g,p} = ρ \cdot g \cdot h_{p}

pressure of the fluid above

P_{p} = P_{a} + P_{g,p}

pressure adds the surface s

P_{g} = \frac{F_{g}}{A_{g}}

pressure is force over area(ground)

P = \frac{F}{A}

pressure is force over area(sea level)

P_{g,g} = ρ \cdot g \cdot h_{g}

pressure of the fluid above(ground)

P_{g} = ρ \cdot g \cdot h

pressure of the fluid above(sea level)

P_{g} = P_{a} + P_{g,g}

pressure adds the surface s(ground)

P = P_{a} + P_{g}

pressure adds the surface s(sea level)