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PHYS 175, FALL 2014
HW #5 S OLUTION
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11.2 In his 1964 science fiction story ”The Coldest Place,” author Larry Niven described the ”dark side” of Mercury as the coldest place in the solar system. What
assumption did he make about the rotation of Mercury? Did this assumption
turn out to be correct?
Mr. Niven assumed that Mercury’s spin-orbit resonance was the same 1:1 ratio as
Earth’s moon (one rotation per one revolution). This would mean that Mercury
had one side perpetually facing the Sun (very hot) and another the dark of space
(very cold). It turns out that Mercury has a more complicated spin-orbit resonance
(discovered a year after the story was published via radar observations) with a 3:2
ratio (three rotations per two revolutions). He wasn’t too far off in that Mercury
does exhibit the greatest temperature difference of all the planets in the solar system, ranging between about 600 K on the sunward face to 100 K on the dark side.
By comparison, the Moon’s dark side gets a little cooler, down to about 93 K.
11.4 Why was it so difficult to determine the rate and direction Venus’ rotation? How
was it finally determined? What is one proposed explanation for the slow, retrograde rotation of Venus?
Unlike Earth, with it’s patchy cloud cover, Venus is always completely shrouded
in clouds. This makes it impossible to image the surface in visible light. The same
radar technique used to determine Mercury’s exact spin-orbit ratio was employed to
determine Venus’ rotational character. Since Venus is in retrograde rotation (among
all the other planets, only Venus and Uranus rotate in an unusual fashion), it is likely
that a massive impact changed it’s rotation long ago, when Venus was still mostly
molten.
11.26 Why is it impossible for liquid water to exist on Mars today? If liquid water existed on Mars in the past, what must have been different then?
Mars is too cold and has too thin an atmosphere for water to exist in liquid form
on its surface for very long (the mean temperature is about -10◦ F). The equatorial
regions are well above freezing during the summer season (maximum high temperatures can reach 95◦ F!) ; however, even liquid water there would evaporate quickly
in the sparse atmosphere. In the past, Mars was likely warmer and had a much
denser atmosphere than is presently observed, since there is ample evidence for
water in the past (based upon land features observed there today). That said, this
was long ago (perhaps a billion or more years ago).
11.33 (a) Why is Mars red? (b) Why is the Martian sky the color of butterscotch?
(a) Mars’ surface is largely covered by rocks, sand and dust that are all rich in ironbearing minerals. The iron in them could combine with oxygen to form rust-like
compounds, which are reddish in color. It is likely that the loss of the martian atmosphere contributed to the ”rusting” of the surface, giving rise to its reddish hue in
PHYS 175, FALL 2014
HW #5 S OLUTION
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visible light. (b) The Martian atmosphere is so thin that very little scattering occurs
off the gases there, so it’s sky color is not determined by it. Rather, dust particles
containing magnetite are thought to make up 1% of the thin Martian atmosphere,
by volume, and scattering off these dust particles gives the sky its color.
11.41 Venus’ sidereal rotational period is 243.01 days and its orbital period is 224.70
days. Use these data to prove that a solar day on Venus lasts 116.8 days. (Hint:
Develop a formula relating Venus’ solar day to its sidereal rotational period and
orbital period similar to the first formula in Box 4-1.)
Let’s let the sidereal orbital and rotational periods be O and R. We’ll use S for the
length of the solar day (all of these will be measured in Earth days). At the end of
this exercise, we’ll see that the formula we need to find the solar day is given by
1
1
1
=
+ ,
S
O R
but that isn’t very informative as to how it was developed. We know that a planet
rotates 360◦ in its sidereal day, and revolves 360◦ in its sidereal year. From this we
◦
◦
write the two rates of rotation (in degrees per Earth day) as 360
and 360
. MultiplyR
O
ing these rates by a length of time gives us an angle that the planet passes through,
◦
S
in either rotation or revolution. So, in a solar day, the planet rotates through 360
R
◦
360
degrees on its axis and revolves through O S around the Sun. Since Venus is rotating opposite to the sense that it revolves, in one solar day it will go through an
◦
◦
S less than 360◦ as it revolves through the angle 360
S. These two angles
angle of 360
R
O
must be the same, so we have
360◦
360◦
S = 360◦ −
S
O
R
S
S
=1−
(divide each side by 360◦ )
O
R
1
1 1
= −
(divide each side by S)
O
S R
1
1
1
=
+
S
O R
We do a bit more math to solve for S as
S=
RO
243.01 · 224.70
=
≈ 116.8 days.
R+O
243.01 + 224.70
PHYS 175, FALL 2014
HW #5 S OLUTION
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12.3 As seen from Earth, does Jupiter or Saturn undergo retrograde motion more frequently? Explain your answer.
Saturn is farther away, and according to Kepler’s laws travels slower than does
Jupiter (Saturn’s year is longer). This means that we will ”pass up” Saturn more
often in our orbit around the Sun, so we would observe more retrograde motion for
Saturn in a given period of time.
12.8 What would happen if you tried to land a spacecraft on the surface of Jupiter?
While there is a rocky core at the center of Jupiter, it is surrounded by a very massive
atmosphere and any spacecraft would be crushed (and likely melt) before it touched
down. (See Caution! on p. 328 of the text.)
12.13 Both Jupiter and Saturn emit more energy than they receive from the Sun in the
form of sunlight. Compare the energy sources of the two planets that produce
this emission.
Jupiter’s emissions are explained by the steady shrinking of the planet’s atmosphere
under its own gravitational attraction. As it shrinks, the atoms that make up its
atmosphere become hotter – this conversion of gravitational potential to thermal
energy is known as a Kelvin-Helmholtz contraction. The size of Jupiter and our understanding of thermal physics makes it relatively easy to explain this phenomenon
there. By the same understanding, we would have expected Saturn (much smaller)
to have shrunk more quickly and radiated away most of its thermal energy – it
shouldn’t be emitting much IR radiation today at all. An explanation was found
by way of helium raining out of the upper atmosphere of Saturn. Droplets of He
form in the cold upper layers, then begin to fall. As they collide with other atmospheric constituents they heat up and also give energy to what they collide with.
This mechanism, given the size and history of Saturn, is an adequate explanation of
the thermal (IR) emissions we observe today.
12.30 During the planning stages for the Pioneer 11 mission, when relatively little was
known about Saturn’s rings, it was proposed to have the spacecraft fly through
the Cassini division. Why would this have been a bad idea?
The Cassini division was initally thought to be empty, since it was observed as
a dark band in Earth-based images. When spacecraft (the Voyager missions and
Cassini) finally flew past, we could see light reflected off of the ”empty space” where
the Cassini division lay. This means that it is not empty and could damage a spacecraft during its passage. Even though the dust thought to be there is very sparse,
it would be enough to cripple a spacecraft and damage the sensitive instruments
carried aboard it.
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12.45 From the information given in Section 12-6, calculate the average density of Jupiter’s
rocky core. How does this compare with the average density of Earth? With the
density of Earth’s solid inner core? (See Table 9-1 and 9-3 for data about Earth.)
M⊕
MJc
dJc
ρ⊕
ρ⊕c
value
5.974 × 1024 kg
8M⊕
11000 km
5515 kg/m3
13000 kg/m3
ref.
Table 9-1
§12-6, p. 335
§12-6, p. 335
Table 9-1
Table 9-3
description
mass of Earth
mass of Jupiter’s solid core
diameter of Jupiter’s solid core
avg. density of Earth
avg. density of Earth’s solid core
Density is the ratio of mass to volume for an object. To calculate it, we use the
formula ρ = M
. We also need to convert the mass of Jupiter’s core into kg (it’s
V
expressed in terms of Earth’s mass), which gives us MJc = 4.78 × 1025 kg. Next
we divide the core’s diameter in half to find the radius (in m, so 1000x the number
of km) and use it in the volume for a sphere VJc = 43 πr3 = 6.97 × 1020 m3 . Finally,
Jc
ρJc = M
= 4.78×1025 /6.97×1020 ≈ 68580 kg/m3 . This is about 12 times the density
VJc
of Earth, and roughly 5 times the density of Earth’s own rocky core.