ASTR 1010 – Homework Assignment 12

Chapter 10 – Spring 2009

Question 1 - #5. If there was no greenhouse effect, a planet’s surface temperature would be determined by distance from Sun, reflectivity, and spin rate. The “no greenhouse” temperatures are lower than the actual temperatures because the greenhouse effect serves as a “blanket” to keep some of the surface heat in and thus raise the equilibrium temperature.

Question 2 - #9. Ozone is triatomic oxygen: O₃. Ozone absorbs UV radiation readily and thus heats the layer of the atmosphere it is mostly found in (the stratosphere for the Earth). Without ozone or a similarly UV-absorbing gas, a planet will not have a stratosphere.

Question 3 - #10. A magnetosphere is a region surrounding a planet in which charged particles are trapped by the planet’s magnetic field. The magnetosphere deflects the charged particles of the solar wind mostly around the planet, though some are made to spiral to the north and south magnetic poles, and some are trapped in the van Allen belts.

Question 4. - #16. The Moon and Mercury are basically too small to hold an atmosphere gravitationally. The idea that the Moon and Mercury might have some ice at the bottom of craters at their north and south poles is because sunlight never reaches there (the crater walls keep the bottoms in perpetual shadow) and the very cold temperatures on the crater floors mean that gases from impacting icy asteroids could be trapped there (because the escape velocity is less at colder temperatures). The idea that ice could be present there was popular in the early 90’s based on radar data, but it has somewhat fallen out of favor these days (I’m not sure why). Your book speaks of it as likely, but, remember, your book is about 5 years out of date.

Question 5. - #18. In the Goldilocks scenario, Mars primarily loses its atmosphere as the planet becomes geologically dead and there are no more mechanisms for putting CO₂ back in the atmosphere (recall rain “washes” the carbon dioxide out of the atmosphere). In a more sophisticated model, the geological death of the planet diminishes the magnetic field, and the weakening of the magnetosphere allows the full brunt of the solar wind to strip the atmosphere from the planet. Another factor is the lack of stratosphere which allows UV radiation to pound the atmosphere breaking up the gas molecules into their constituent atoms.

Question 6. - #19. In a runaway greenhouse effect a positive reinforcement loop is set up where, for instance, some greenhouse gas like water vapor produces a rise in temperature, which makes more water vapor, which raises the temperature, and so on, and so on. This happens on Venus because the planet is close enough to the Sun to prevent the formation of water oceans. This means the water is in the atmosphere (originally – it eventually goes away…but that is another story) and so we have a big greenhouse effect because the carbon dioxide is also stuck up there. On Earth, the formation of water oceans pulls out the greenhouse gas, water vapor, and the liquid water pulls out the other greenhouse gas, carbon dioxide, because the CO₂ dissolves in the water oceans. Thus, our greenhouse effect does not have that positive reinforcement loop and remains moderate.

Question 7. - #21. The carbon dioxide cycle is important on Earth because it acts like a thermostat. If the temperature goes up, there is more evaporation and thus more water vapor in the atmosphere. This leads to more rain which washes out some more carbon dioxide than usual and the greenhouse effect goes down a bit, thereby lowering the temperature. If the temperature goes down too much, then there is less water vapor in the atmosphere and the carbon dioxide is not washed out as readily, it builds up above normal levels, increases the greenhouse effect and the temperature goes up a bit. See Figure 10.34.

Question 8. - #58. A completely black planet has a reflectivity of 0. At 1 AU, the formula on page 302 gives 280 K. (Recall that in that formula, d is measured in AU, so you have the fourth root of 1/1, which is … 1). If the planet was completely reflecting, its reflectivity would be 1 which means its temperature would be 0. (The numerator in the quotient under the fourth root symbol would be 0/1 or just 0). This makes sense because if the planet reflected everything, no light would reach its surface and it couldn’t be heated from the outside (we are neglecting internal sources of heat here…). The freezing point of water is 273 K (approximately). So, for a planet at 1 AU) we just solve

273 = 280[(1 – a)/1²]^1/4

for a, where I have called the reflectivity “a” (sometimes, the reflectivity is called the “albedo”, hence the “a”). In this case, a = 0.10. Earth’s reflectivity is more like 0.29, so that would drive the temperature of the surface – in the absence of a greenhouse effect – to below freezing.

Question 9. - #60. At 1.38 AU, the no greenhouse temperature for a=0, d = 1.38 is

238 K

At 1.66 AU, the no greenhouse temperature for a=0, d = 1.38 is

217 K

So, that’s a difference of about 21 K.

This significant difference exacerbates Mars’ seasons compared to those of Earth.

Question 10. - #61. a) The radius of Venus is 6.051 x 10⁶ m. If we add 200 km, we get 6.251 x 10⁶ m. I don’t know why your book invokes Mathematical Insight 5.3. What we need is the formula for escape velocity (see page 139). We also need the mass of Venus: 4.87 x 10²⁴ kg. The escape velocity 200 km up from the surface is thus: 10.2 km/s .

b) For the thermal velocity, the formula is on p. 315.

V(thermal) for hydrogen: 2.4 km/s

V(thermal) for deuterium is the square root of 2 less than 1.7 km/s

c) Even though both thermal velocities are less than the escape velocity of Venus 200 km/s above the surface, we discussed in class how these gases still readily escape (recall the high-velocity tail argument – if you don’t recall it, see p. 315).

However, the fact that deuterium has a lower thermal velocity means there will be more of it than hydrogen in the Venusian atmosphere. Basically, if water vapor in the Venusian atmosphere was dissociated by ultraviolet light into its constituent atoms (hydrogen and oxygen for water, hydrogen, deuterium, and oxygen for “heavy” water), then the hydrogen would escape more readily, leading to a lower hydrogen/deuterium ratio than normal. This is what is seen in Venus’ atmosphere and thus lends credence to the idea that Venus lost its water vapor this way.