Earth Science: The Physical Setting
1st Edition
ISBN: 9780133200409
Textbook solutions
All Solutions
Page 53: Practice Questions
Exercise 1
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2
Exercise 2
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4
Exercise 3
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4
Exercise 4
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3
Exercise 5
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4
Exercise 6
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2
Exercise 7
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1
Exercise 8
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3
Exercise 9
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1
Exercise 10
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3
Exercise 11
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2
Exercise 12
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2
Exercise 13
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Exercise 14
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4
Exercise 15
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4
Exercise 16
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Exercise 17
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Exercise 18
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Exercise 19
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3
Exercise 20
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1) a galaxy is a collection of billions of stars whereas a constellation is a collection of only a very few stars.
2) galaxy does not form any imaginary shape whereas a constellation can form pattern shapes i.e animals.
3) there are billions of galaxies in this universe but only about 88 constellations are known.
Exercise 21
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1) Planets are celestial objects orbiting the sun (a star) whereas moons are celestial objects orbiting the planets.
2) Planets are larger than moons.
Exercise 22
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1
Exercise 23
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4
Exercise 24
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3
Exercise 25
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The following are the given values:
$$begin{array}{c c c} hline
text{Planet} & text{Ave Distance from the Sun (Au)} & text{Ave Orbital Speed (km/s)} \ hline
text{Mercury} & 0.4 & 48.0 \
text{Venus} & 0.7 & 35.0 \
text{Earth} & 1.0 & 30.0\
text{Mars} & 1.5& 24.0 \
text{Jupiter} &5.2& 13.0 \
text{Saturn} & 9.6 & 10.0 \
text{Uranus} & 19.0 & 7.0\
text{Neptune} & 30.0 & 5.1 \
text{Pluto} & 39.0 & 4.7 \
hline
end{array} $$
$$begin{array}{c c c} hline
text{Planet} & text{Ave Distance from the Sun (Au)} & text{Ave Orbital Speed (km/s)} \ hline
text{Mercury} & 0.4 & 48.0 \
text{Venus} & 0.7 & 35.0 \
text{Earth} & 1.0 & 30.0\
text{Mars} & 1.5& 24.0 \
text{Jupiter} &5.2& 13.0 \
text{Saturn} & 9.6 & 10.0 \
text{Uranus} & 19.0 & 7.0\
text{Neptune} & 30.0 & 5.1 \
text{Pluto} & 39.0 & 4.7 \
hline
end{array} $$
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The graph below shows the plot of the average distance from the sun and the average orbital speed for each of the eight planets from the data above.
Exercise 26
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there is an inverse relationship between a planet’s average distance from the sun and the planet’s average orbital speed. when a planet is closer to the sun, its orbital speed velocity is greatest, and when a planet is farthest from the sun, its orbital speed velocity is slowest.
Exercise 27
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the two planets are Mars and Jupiter.
Exercise 28
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Vesta Asteroid. (because it is the closest to the sun)
Exercise 29
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$a) :eccentricity :of:an:ellipse:(E)=dfrac{distance:between:foci}{length:of:major:axis}=dfrac{d}{L}$
$b):E=dfrac{3.45}{6.8}$
$c):E=0.507$
$d);eccentricity;of;spacecraft;orbit;is;greater;than;the;earth,; the; Earth’s; orbit; is; nearly; circular.$
Exercise 30
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There is a red shift in the light from stars in distant galaxies.
Exercise 31
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±200 million years.
Exercise 32
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a) Betelgeuse.
b) Polaris.
c) Aldebaran.
d) Sirius.
e) the sun.
b) Polaris.
c) Aldebaran.
d) Sirius.
e) the sun.
Exercise 33
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Because most galaxies are redshifted, lights wavelength is stretched/frequency decreases. This occurs when the galaxies are currently moving away from each other. Working backwards, this implies that there was one point in time when all mass was at a single point. This is the basic essence of the big bang theory.
Exercise 34
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nearly 190 million Km
Exercise 35
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526 days.
Exercise 36
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oblate spheroid.
Exercise 37
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4 billion years.
Exercise 38
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the rotation rate of thelma planet.
Exercise 39
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due to its small mass and diameter, and the near distance to the sun and it has a solid surface.
Exercise 40
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the magnitude of gravitation between thelma and the sun is greater than the gravitation between the sun and mars. due to the difference in mass and the distance from the sun.
Exercise 41
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it would cause a large crater.
Exercise 42
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luminosity increases, then decreases.
Exercise 43
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Procyon B.
Exercise 44
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The red-shift of light from distant galaxies indicates that they are moving away from us, i.e., the universe is expanding from its original size. and it was a strong evidence to big bang theory.
Exercise 45
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The estimated age of Earth and our solar system is 4.6 billion years and these distant galaxies are 12 billion years old. so, the age of the distant galaxies are older than the age of our solar system.
Exercise 46
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The given diagram below shows the position of Halley’s Comet and Asteroid 134340 at various times in their orbits.
Step 2
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The eccentricity of the asteroid’s orbit is 0.250. From the diagram, we can identify the length of the major axis which is 10.6 units. We can also identify the center. To plot the position of the second focus of the asteroid’s orbit, we need to calculate the distance between the foci.
Step 3
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The formula to calculate the eccentricity of an ellipse is given by:
$$ begin{aligned}
text{e} = frac{d}{L} tag{1}
end{aligned} $$
where,
$$ begin{aligned}
e &= text{eccentricity of an ellipse} \
d &= text{distance between foci} \
L &= text{length of major axis} \
end{aligned} $$
We can derive from this equation the distance between foci:
$$ begin{aligned}
d= eLtag{2}
end{aligned} $$
$$ begin{aligned}
text{e} = frac{d}{L} tag{1}
end{aligned} $$
where,
$$ begin{aligned}
e &= text{eccentricity of an ellipse} \
d &= text{distance between foci} \
L &= text{length of major axis} \
end{aligned} $$
We can derive from this equation the distance between foci:
$$ begin{aligned}
d= eLtag{2}
end{aligned} $$
Step 4
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Plug the known values in equation (2) to calculate the distance between foci.
$$ begin{aligned}
d&= 0.250 times 10.6 \
&=2.65 text{ units}
end{aligned} $$
Dividing this by 2 will give its distance from the center.
$$ begin{aligned}
text{distance from the center}&= frac{2.65}{2} \
&=1.325 text{ units}
end{aligned} $$
$$ begin{aligned}
d&= 0.250 times 10.6 \
&=2.65 text{ units}
end{aligned} $$
Dividing this by 2 will give its distance from the center.
$$ begin{aligned}
text{distance from the center}&= frac{2.65}{2} \
&=1.325 text{ units}
end{aligned} $$
Step 5
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Plotting these in the diagram will give us the position of the second focus of the asteroid’s orbit which is marked by ‘X’.
See the explanation.
Exercise 47
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Halley’s comet was traveling faster, because it was the closest to the sun.
Exercise 48
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due to Halley’s Comet Orbits the Sun. and inside kuiper belt.
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