SCCC 115
Homework 6 - Due 3 Nov. 2005
Prof. Christina Lacey
R&D: 10-19, 10-20, 11-8, 12-1, 12-7     P: 11-5, 12-9

R&D: 10-19:

Since Mars has an atmosphere and it is composed mostly of a a greenhouse gas, why isn't there a significant greenhouse effect to warm its surface?

While Mars' atmosphere is mostly CO$_2$, a greenhouse gas, there isn't enough atmosphere to create an effective insulating layer to hold in the trapped IR radiation.

R&D: 10-20:
Compare and contrast the evolution of the atmosphere of Mars, Venus, and Earth.

All three planets have secondary atmospheres produced by volcanoes. Venus, which is closest to the Sun, was too warm for liquid water to form. Without liquid water, the CO$_2$ could not settle out of the atmosphere and dissolve into the oceans. So Venus became a victim of the runaway greenhouse effect and became warm with a very dense, highly insulating atmosphere. The H$_2$O in the atmosphere was destroyed by UV sunlight.

Earth was far enough from the Sun for liquid water to form and precipitate out into the oceans. The CO$_2$ in the atmosphere dissolved into the Oceans and eventually was also removed from the atmosphere by life processes. The small amounts of H$_2$O and CO$_2$ in the atmosphere create an insulating layer that is just enough to keep Earth warm and mostly liquid.

Mars, farthest, from the Sun, likely had liquid water early on in its evolution. So CO$_2$ was able to dissolve into water, but Mrs is less massive than the Earth or Venus, and the effect of losing the CO$_2$ was to thin the atmosphere to the point that most of the IR radiation escaped. Mars, due to its lesser mass, has less tectonic activity, volcanism stopped early in its formation and the atmosphere was not renewed with from CO$_2$ and H$_2$O volcanoes.

R&D: 11-8:

Why has Jupiter retained most of its original atmosphere?

Jupiter has a large enough mass to retain its original atmosphere. Additionally, Jupiter is located far from the Sun, so it is so cold, that the atoms and molecules in the original atmosphere were not able to attain large enough average molecular velocities to escape Jupiter.

R&D: 12-1:

Seen from Earth, Saturn's rings sometimes appear broad and brilliant but at other times seem to disappear. Why?

Saturn has a tilt of 27$^\circ$ and as it orbits the Sun, the position of the tilt relative to the orbit and the position of the Sun. The rings are in the equatorial plane of Saturn, so the situation is similar to Earth's tilt and the change of seasons. We sometimes see the rings tilted almost 27$^\circ$, so they appear broad, and at other times we see the rings almost edge-on so that the rings appear to disappear. The orbit of Saturn takes over 30 years, so the orientation of the rings changes every 7 - 8 years.

R&D: 12-7:

What mechanism is responsible for the relative lack of helium in Saturn's atmosphere, compared to Jupiter?

Helium has precipitated (rain) out of Saturn's atmosphere, after Saturn initially cooled off. This precipitation mechanism accounts for the lack of helium in Saturn's atmosphere and the extra energy coming from Saturn.

P: 11-5:

Given Jupiter's age and current atmospheric temperature, what is the smallest possible mass the planet could have and retain its hydrogen atmosphere?

To retain hydrogen (from pg. 208):

$$6\times v_{average \; hydrogen} \le v_{escape}$$ (1)

$$v_{escape} = \sqrt{\frac{2GM}{R}}=11.2 km/s \sqrt{\frac{ M\;(M_\oplus)}{R\; (R_\oplus)}}$$ (2)

On Jupiter at the surface: T= 124 K, $R_{Jupiter}= 11.2 R _\oplus$.

$$v_{ave \; hydrogen} = 0.157  km/s \sqrt{\frac{T(K)}{M_{H_2} (M_H)}} =0.157  km/s \sqrt{\frac{124 K}{2 (M_H)}}=1.23  km/s$$ (3)

So using the first relationship, we can write and solve for $M\;(M_\oplus)$:

$$6\times v_{average \; hydrogen} \le v_{escape}$$

$$6\times 1.23  km/s \le 11.2 km/s \sqrt{\frac{ M\;(M_\oplus)}{R\; (R_\oplus)}}$$

$$\mbox{Square both sides of the equation}:$$

$$(6\times 1.23)^2 \le (11.2)^2 \frac {M (M_\oplus)}{11.2 \; (R_\oplus)}$$

$$M (M_\oplus) \ge \frac {(6\times 1.23)^2}{(11.2)} = \frac{55.0}{11.2}= 4.9\; M_\oplus$$ (4)

So if the mass of Jupiter is greater than or equal to $4.9\; M_\oplus$, Jupiter will retain its hydrogen atmosphere.

P 12-9

Show that Titan's surface gravity is about 1/7 that of Earth's. What is Titan's escape speed?

Surface gravity= $a$ = acceleration; (a=g= 9.8 m/s$^2$ at Earth's surface)

From pg. 56:

$$a = \frac{v^2}{r} \quad v=\frac{GM}{r}$$

$$a = \frac{GM}{r^2}$$

$$a_{Titan} = \frac{GM_{Titan}}{r^2_{Titan}}; \quad a_{Earth} = \frac{GM_{Earth}}{r^2_{Earth}}$$

$$\frac{a_{Titan}}{a_{Earth}} = \frac{\frac{GM_{Titan}}{r^2_{Titan}}} {\frac{GM_{Earth}}{r^2_{Earth}}}= \frac{M_{Titan}r^2_{Earth}}{M_{Earth}r^2_{Titan}}$$ (5)

From Table 12.2 (pg. 316):

$$M_{Titan} = 1.8 \times M_{moon}= 1.8 \times 7.4 \times 10^{22} kg = 1.3 \times 10^{23} kg$$

$$r_{Titan} = 5,000 km/2= 2,500 km$$

$$M_{Earth} = 6 \times 10^{24} kg$$

$$r_{Earth} = 6,500 km$$ (6)

$$\frac{a_{Titan}}{a_{Earth}} = \frac{M_{Titan}r^2_{Earth}}{M_{Earth}r^2_{Titan}} = \frac{1.3 \times 10^{23} kg (6,500 km)^2}{6 \times 10^{24} kg (2,500 km)^2} = 0.147$$

$$a_{Earth} = g = 9.8  m/s^2$$

$$a_{Titan} = .147 \times 9.8  m/s^2= 1.44 m/s^2$$ (7)

The escape velocity at the surface of Titan is:

$$v_{escape} = \sqrt{\frac{2GM_{Titan}}{R_{Titan}}}= \sqrt{\frac{2  6.67 \times 10^{-11} Nm^2/kg^2 1.3 \times 10^{23} kg}{2500 km \frac{1000 m}{1 km}}}$$

$$= \sqrt{\frac{1.7 \times 10^{13} m^2}{2.5 \times 10^6  s^2}}=\sqrt{ 6.9 \times 10^6  m/s^2}= 2.6 \times 10^3 m/s = 2.6 km/s$$ (8)