Question

2. Atmospheric Mass Venus and Earth have about the same mass and radius, but the atmospheric pressure on the surface of Venus is measured to be 90 times larger than the pressure at the surface of the Earth. (a) Assuming Venus' atmosphere is pure CO2 and Earth's atmosphere is pure N2, estimate the atmospheric scale height for each planet. For simplicity you can assume that both atmospheres are isothermal with T = 550 K for Venus and T = 250 K for Earth. (b) Using the measured surface pressures, estimate the atmospheric density (in kg m-3) at the surface of each planet and at an altitude of 40 km above the surface. (c) Integrate the density profile to estimate the total mass of the atmosphere, from 0 to 40 km altitude, for each planet. Compare the ratio of atmospheric masses to the ratio of surface pressures, and discuss.

Answer 1

(a) Atmospheric scale height H = kT/(mg) ..................(1)

where k is Boltzmann constant, T is isothermal temperature of atmosphere , m is mass of gas molecule and g is acceleration due to gravity

Earh's atmospheric scale height H_e = 1.38 $\times$ 10^-23$\times$ 250 / ( 28 $\times$ 1.66 $\times$ 10^-27$\times$ 9.8) = 7574 m

Atmospheric scale height H_v of Venus = 1.38 $\times$ 10^-23$\times$ 550 / ( 44 $\times$ 1.66 $\times$ 10^-27$\times$ 9.8) = 10604 m

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Pressure P(z) at a height z is given by, P(z) = P_o e^-z/H

where P_o is pressure at surface of planet

Density $\rho (z)$ = P(z) / ( R_g T )

where R_g is gas constant , R_g = R / M , where R is universal gas constant and M is molar mass of gas molecules

Hence density $\rho (z)$ at a height z is given by

P(z) = Pen

gas density for Earth at a height of 40 km is calculated as follows

P_o = 1.013 $\times$ 10⁵ Pa , R_g = 8.314/ ( 28 $\times$ 10^-3 ) = 297

$\rho (z) = \frac{1.013 \times 10^{5}}{297 \times 250} e^{\frac{-40000}{7574}} = 6.94 \times 10^{-3} kg/m^{3}$

gas density for Venus at a height of 40 km is calculated as follows

P_o = 30 $\times$ 1.013 $\times$ 10⁵ Pa , R_g = 8.314/ ( 44 $\times$ 10^-3 ) = 189

$\rho (z) = \frac{30\times 1.013 \times 10^{5}}{189 \times 550} e^{\frac{-40000}{10604}} = 0.672.. kg/m^{3}$

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Planet

mass M of gas in atmosphere upto height 40 km is given by,

$M = \int_{0}^{40000} \rho (z) 4\pi z^{2}dz = \int_{0}^{40000}\frac{P_{o}}{R_{g}T}e^{\frac{-z}{H}} 4\pi z^{2}dz = \left ( \frac{4\pi P_{o}}{R_{g}T} \right )\int_{0}^{40000}e^{\frac{-z}{H}} z^{2}dz$

Above integration is performed using substitution y = z/H

we get ,

$M = \left ( \frac{4\pi P_{o}}{R_{g}T} \right ) \left \{ 2 - e^{\frac{-40000}{H}}\left [ \left ( \frac{40000}{H}+1 \right )^{2} +1\right ] \right \}$

Using the known values mass of atmosphere over a height of 40 km can be calculated using the above relation