How High the Air Goes

Q. The density of the atmosphere at sea level is 1.29 kg/m³. Assume that it does not change with altitude. Taking g as 9.8 m/s² and the pressure at sea level as 1.01 × 10⁵ Pa, how high would the atmosphere extend, in metres? expects 7.99 × 10³ m

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

density

ρ

1.29 kg/m³

gravity

g

9.8 m/s²

sea level

P

1.01e+05 Pa

Outputs

height

h

The solution

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 Ruby

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

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

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

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)