In consultation with the other members of your group, solve the problem outlined below. If you finish your assigned problem quickly, try one of the other problems as well.
1. Several of the highest peaks on the planet Earth are located in the Himalayas. Hikers regularly attempt to reach the summit of Mount Everest at 8850 m above sea level; many of those who succeed bring a supply of oxygen with them to supplement that available in the thin atmosphere. To calculate how much oxygen is available in the air at the summit of Everest, first calculate the scale height of O2 in the Earth's atmosphere, assuming that the typical temperature is 253 K. Next, calculate the percentage of oxygen in the atmosphere below 8850 m, assuming that the content decreases uniformly. For comparison, calculate the percentage of water vapor (H2O) at that height.
h = 8850 m
m = 16 + 16 = 32 x 1.67 x 10-27
g = 9.8 m/s2
T = 253 K
k = 1.38 x 10-23
H = kT/gm = 6.67 km
Since n is proportional to P for an ideal gas, we can calculate the
ratio of the pressures to determine the percentage of oxygen:
P(h) = P(ho) exp(-h/H)
P(h)/P(ho) = exp(-h/H) = 0.26
Therefore, oxygen is only 26% of sea-level!!!
For H2O: m = 18 x 1.67 x 10-27 kg
H = k T/ g m = 11.8 km
P(h)/P(ho) = exp(-h/H) = 0.47
2. As expected for a small moon, Phobos has very little atmosphere. To calculate the expected lifetime for CO2 on the surface of this moon, first determine the surface temperature at aphelion. Mars's orbital parameters are: semi-major axis=1.524 AU, eccentricity = 0.093. Phobos has an albedo of 0.018. Next, calculate the mean molecular speed of CO2 at that temperature. Finally, calculate the ratio of the escape speed from Phobos to the mean molecular speed, and comment on the expected lifetime of CO2. RPh=12 km, MPh = 9.6 x 1015 kg.
Tss = 394(1-A)0.25(rp)-0.5
rp = 1.524(1+0.093) = 1.666 AU
Tss = 394(1-0.018)0.25(1.666)-0.5 = 304 K
Vrms = (3 K T/m)0.5 = 414 m/s
Vesc = (2 G M/R)0.5 = 10.3 m/s
Vesc/Vrms = 0.02 so the lifetime is very short.
3. The Very Large Telescope (VLT) is a set of four large optical telescopes located on Cerro Paranal, at a height of 2635 m above sea level. The Paranal mountain was chosen for this new facility because of its atmospheric conditions and its remoteness. While not the highest observing site in the world, the Paranal site is still sufficiently high that water vapor and carbon dioxide levels are reduced. To understand this, first calculate the scale height for H2O and CO2 in the Earth's atmosphere, assuming that the typical temperature at the telescope site is 270 K. Next, calculate the percentage of each of these gases in the atmosphere below 2635 m, assuming that the content decreases uniformly for each gas.
h = 2635 m
m = 1 + 1 + 16 = 18 x 1.67 x 10-27
g = 9.8 m/s2
T = 270 K
k = 1.38 x 10-23
H = kT/gm = 12.6 km for water vapor.
Since n is proportional to P for an ideal gas, we can calculate the
ratio of the pressures to determine the percentage of water:
P(h) = P(ho) exp(-h/H)
P(h)/P(ho) = exp(-h/H) = 0.81
For CO2: m = 44 x 1.67 x 10-27 kg
H = k T/ g m = 5.17 km
P(h)/P(ho) = exp(-h/H) = 0.60
4. Suppose a new moon were discovered in orbit around Venus with an albedo of 0.07. (i) Calculate the expected noontime temperature on the surface of this moon at aphelion. Venus's orbital parameters are: semi-major axis = 0.723 AU, eccentricity = 0.007. (ii) Next, determine the molecular speed of N2 at that temperature. (iii) Assuming that the moon has an escape speed of 2.0 km/s, comment on the expected lifetime of N2 on its surface.
Tss = 394(1-A)0.25(rp)-0.5
rp = 0.723(1+0.007) = 0.728 AU
Tss = 394(1-0.07)0.25(0.728)-0.5 = 453 K
Vrms = (3 K T/m)0.5 = 633 m/s
Vesc/Vrms = 3.2 so the lifetime is a few thousand years.