In a planetary atmosphere in hydrostatic equilibrium, the number density ’n’ of
ID: 1442047 • Letter: I
Question
In a planetary atmosphere in hydrostatic equilibrium, the number density ’n’ of a gas at a height h, n(h) is given by n(h) = n(0) exp [- (h-h0)/H] where H is called the ‘scale height’ (=kT/Mg) k is the Boltzmann’s constant T the temperature(2000 K), M the atomic mass; M is 4 atomic mass units (2 protons & 2 neutrons) for helium g is the gravity(= 0.8 x earth’s gravity, for Venus). n (0) is the number density at a reference height of h0. By what fraction does the density of helium changes between the altitudes of 500 km and 2000 km in the atmosphere of Venus?
(Remember an AMU = # of proton plus # of neutrons. We can assume mass of a neutron is about the same as proton (p+). So you can convert 4 AMU using mass of P+
- You can keep the keep the n(o) as particles/cm3 since the exp{-h-ho)/H} will be a decimal with no units.
- Can calculate the number density for He at 2 or 3 heights between 500 m & 2000 km using the n(h) equation. (Remember to show your work)
- Plot the values you calculated on the graph. The density at 500km is 108 per cm?3 (Look on the Internet for a linear-log graph paper - also called semi log paper)
In a planetary atmosphere in hydrostatic equilbium, the number density 'n' of a gas at a height h, n(h) is given by n(h) = n(0) exp [-(h-ho)/H] where H is called the 'scale height (kT/Mg) k is the Boltzmann's constant T the tempea(2000 K), M the atomic mass; Mis 4 atomic mass units (2 protons & 2 neutrons) for helium g is the gravity(0.8 x earth's gravity, for Venus) n (0) is the number density at a reference height of ho. By what fraction does the density of helium changes between the altitudes of 500 km and 2000 km in the atmosphere of Venus?Explanation / Answer
Here, H = kT/mg
= (1.38 * 10-23 * 2000)/(6.64 * 10-27 * 9.8 * 0.8)
= 53.012 * 104 m
=> n(h) = n(0) exp(-1500000/53.012 * 104)
= n(0) exp(-2.8295)
=> n(h)/n(0) = 0.05904
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