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Homework 3
Author: Student, Amazing A.
October 17, 2016
Chapter 5
Problem 54
Suppose the surface of the Sun were about 12000 K, rather than 6000 K.
a. How much more thermal radiation would the sun emit? From the StefanBoltzmann Law 1, we know that the Energy/Power emitted from thermal radiation is ∝ σT 4 . That means that a small change in temperature T will cause a
large change in the energy emitted.
Problems like this are most easily solved by using ratios. How much more
energy will be given off by the ‘Hot’ Sun relative to the ‘Cool’ Sun is determined
by the temperature :
4
Energy(‘HotSun0 )
σThot
(12000K)4
=
= 16
→
4
Energy(‘CoolSun0 )
σTcool
(6000K)4
(1)
So although the temperature only increased by a fact of 2, the energy output
of the Hotter Sun is 16 times greater.
b. What would happen to the sun’s peak wavelength emission? From Wien’s
Law (Law 2 in the book) for Thermal radiation, we know that the energy output
peaks at different wavelengths as the temperature changes. At higher energies
(higher temperatures) the visual color shifts to more blue colors. Bluer colors
have shorter wavelengths:
λmax (nm) =
2.9 × 106
T
(2)
This can be rewritten in a ratio like part (a), but it does not have to be in
this simpler case. At a temperature of 6000 K, the energy output of the Sun
peaks at 480 nm. This is a blue-ish green in Figure 5.15 in the book. The
Hot sun peaks at a wavelength of half of that (λ = 240 nm) which is in the
Ultra-Violet part of the spectrum not visible to the human eye.
c. Do you think it would still be possible to have life on earth? Explain.
As scientists explore what and where life can exist on Earth, it becomes
increasingly clear that some species (like Tardigrades), can exist in the most
extreme conditions imaginable. However, for the diversity of life on the surface
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of the earth, a hot sun would certainly spell doom. Not only would the energy
output of the sun be 16 times what it currently is, but the type of radiation
(UV) is deadly to extant species. Humans could certainly no longer survive on
such an Earth. This would be a case for Extreme Global Warming.
Chapter 6
Problem 23
How much greater is the light-collecting area of a 6-meter telescope than that of
a 3-meter telescope? (a) two times, (b) four times (c) six times.
We must consider that a telescope is essentially a light bucket. The greater
the surface area of the bucket, the more light it can collect. The area of the
collecting region is the square of its Diameter. The relative light gathering
power of two telescopes (A and B) can be found by:
∝
2
(6 m)2
DA
→
=4
2
DB
(3 m)2
(3)
So the answer is (b). By doubling the size of the mirror, we collect 4 times
the amount of light.
Problem 26
The Hubble Space Telescope obtains higher-resolution images than most groundbased telescopes because it is (a) larger (b) closer to the stars (c) above Earth’s
atmosphere.
The most correct answer is (c). Option (a) is not correct, as there are
much larger ground based telescopes than the Hubble. Although option (b) is
technically correct, the relative distance change of the Hubble Space Telescope
from the Earth’s surface is entirely negligible when compared to the vast distance
to our nearest star. The atmosphere is highly volatile and distorts the images
of the best instruments on Earth. Space based telescopes have an enormous
advantage in this respect.
Chapter 7
Problem 26
Which of the following three kinds of objects resides closest to the sun on average? (a) comets (b) asteroids (c) Jovian planets?
Asteroids (b) are the best answer. Jovian planets begin with Jupiter, which
orbits at R > 5 AU away, while comets exist in the Kuiper belt primarily
(R > 40 AU). Asteroids prefer orbits nearer to the sun than Jupiter, orbiting
in a belt between the gas giant and Mars (R < 5 AU).
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Problem 28
Planetary orbits are (a) very eccentric (stretched out) ellipses and in the same
plane (b) fairly circular and in the same plane (c) fairly circular but oriented
in every direction.
Most planets are in the same plane as one another, and have fairly low
eccentricities (apart from Mercury with e = 0.2). Astronomers believe this is a
fundamental clue to the formation mechanism of planets, which are thought to
arise from a disk of material. Therefore (b) is the most correct answer. Comets
are better suited to (a).
Extra Credit: Problem 39
How many Earths could fit inside Jupiter (assuming you could fill up all the
volume)? How many Jupiters could fit inside the Sun? The equation for the
volume of a sphere is V = (4/3)πR3 .
The easiest way to do this problem is to compute the volumes based on the
relative radii of each object with respect to each other. In this case, we can
adopt the books values of the size of the Sun and Jupiter in terms of Earth
radii. The radius of the sun is 108 earth radii, while Jupiter is 11.2 times larger
than the Earth. Then we can find the number of Earths that fit inside Jupiter
by using a ratio of their volumes:
NEarths =
3
(4/3)πRJup
VJup
=
3
VEarth
(4/3)πREarth
(4)
Doing the problem this way simplifies things, as the constants cancel out so
that:
3
RJup
(11.2 REarth )3
NEarths = 3
≈ 1405
(5)
=
REarth
(1 REarth )3
So more than 1400 Earth volumes could fit inside Jupiter’s volume. We can
do the same exercise for the Sun and Jupiter:
NJupiters =
3
RSun
(108 REarth )3
=
≈ 897
3
RJupiter
(11.2 REarth )3
(6)
This ratio could have been rewritten in units of Jupiter radii (approximately
10 Jupiter radii = 1 solar radii) and the results would be the same. The volume
in km3 could also be computed for the Earth, Jupiter, and the Sun (1.0×1012
,1.5×1015 , 1.4×1018 , km3 respectively), but the results look messier in the intermediate steps.
Chapter 8
Problem 25
How many of the planets orbit the Sun in the same direction as Earth does? (a)
a few (b) most (c) all
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All of the planets orbit the sun in the same direction and in the same plane
(c). This is a direct consequence of the formation of the sun, which was thought
to accrete material from a disk. The planetesimals grew from this original disk.
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