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Pathways to Astronomy
Solutions to 1st End of the Unit Problem
Student Version
Unit 1
1. 0.017 mm
Unit 2
1. a. 60 cm, about the size of a beach ball.
1. b. 1,800 cm or 18 m.
1. c. 540,000 cm or 5,400 m
Unit 3
1. 7  105 / 6.4  103 = 7/6.4  105−3 ≈ 1  102 or about 100 times larger.
Unit 4
1. It might be hypothesized that lightning in a rainstorm is caused by raindrops carrying
an electrical charge down to Earth—making an electrical difference between the ground
and air that causes an electrical discharge. This is suggested by the association of
lightning with strong rain storms. It is also suggested by an analogy: the frictional effects
between the raindrops and the air might be like a balloon rubbed against a sweater, with
the raindrops collecting electrons as they rub against molecules as they fall to the ground.
We could disprove this hypothesis by showing that rain drops have no electric charge.
Unit 5
1. The point along the celestial equator that comes closest to your zenith is where it
crosses the meridian. The distance of this point from your zenith is the same as your
latitude—42 from the zenith at 42N, and 34 from the zenith at 34S.
Unit 6
1. The seasonal changes associated with the tilt would disappear. Every part of Earth
would have 12-hour days and nights, but the equatorial regions would still have the most
direct sunlight and be hottest. There would be very little temperature change during the
year, although it would generally be slightly hotter everywhere on Earth at the beginning
of January when the Earth is closest to the Sun.
Unit 7
1. The 8766 hours in a year would be divided into 180 days of 48.7 hours, or 48 hours 42
minutes. A point on the Earth would rotate by about 2 (= 360 / 180) less to complete a
rotation with respect to the stars, so the sidereal day would be 358/360  48.7 = 48.4
hours long.
1
Unit 8
1. The first-quarter moon rises at about noon and sets about midnight.
Unit 9
1. The middle of 2050 CE will be the year 5810 in the Jewish calendar, 1472 in the
Islamic calendar, and 4747 in the Chinese calendar.
Unit 10
1. One example that its possible to observe for yourself is what happens to the
appearance of ships near the horizon on the ocean (or any other large body of water). If
you are on shore close to water level the waterline of the ship will not be visible (or
possibly a large fraction or even the whole ship). Climbing up to a greater height, the
lower parts of the ship become visible again. If the ocean were flat, this would not
happen.
Unit 11
1. From its average distance of 1.52 AU from the Sun, the greatest elongation is
approximately 41. Note, however, that the greatest elongation will vary depending on
where Mars is in its orbit because its orbit is quite elliptical.
Unit 12
1. You can use Kepler’s 3rd Law, a3 = P2. For P= 64 years, we find a3 = 642, so a = 16
AU
Unit 13
1. The main part of Orion, for example, is about 20 tall and 10 wide.
Unit 14
1. Inertia is a property of matter than opposes changes in velocity. The more matter or
mass an object has, the greater is this effect of inertia. Therefore a bowling ball has more
inertia than a small inflated balloon. Moving toward you at one meter per second, the
balloon would be easier to catch, because it would require less force to stop.
Unit 15
1. 8.3 meters/second2 [90 kph = 90000meters per hour = 90000/3600 meters per second
= 25 meters per second. Acceleration = 90000 meters per second / 3 seconds = 8.3
meters per second squared.
Unit 16
1. The gravitational force would be one quarter the strength you feel at the surface of
Earth. On average, when you stand on the surface of the Earth, you are 6370 km from the
center of the Earth, so 6370 km farther away is twice the distance. Since Gravity follows
an inverse square law, if you double the distance, you reduce the force by four times,
making it one quarter as strong.
2
Unit 17
1. about 14 km/s and 11 years. (Figures are rounded.) [V=2π x d/P
V=6.28 x (7.5 1011m)/(3.47 x 107 sec.), where 3.47 x 107 sec. is Jupiter’s period in
seconds. This results in V = 1.36 x 104 m/sec or about 14 km/sec. For the Period,
Kepler’s Third Law is 1 = a3/P2, or a3 = P2, and knowing that Jupiter’s average distance
from the Sun is approximately 5 times that of Earth (a=5), we have 53 = P2 , which is P2 =
125 or P=125 ≈ 11. So the period is roughly 11 years.
Unit 18
1. 7350 m/s. [Using the calculation example at the bottom of page 138, change R to 6.37
X 106 m plus 106 m = 7.37 X 106 m and follow through with the calculation.
Unit 19
1. Roughly midnight and noon for high tides; 6 a.m. and 6 p.m. for low tides.
Unit 20
1. 2 m/s. [KE=mV2/2, or V = √(2KE/m), substitute 1J for KE, and .5kg for m and solve
for m/s.
Unit 21
1. 500 sec. [150 X 106 km/300 X 103 km/sec.
Unit 22
1. 6 X 1014 Hz. [λ x ν = c: ν = 3 x 108 m / 5 x 10-7m = 6 X 1014 Hz.
Unit 23
1. 970K. [T=2.9 X 106Knm/300nm = 970K.
Unit 24
1. Although the spectrum of light from Venus is produced by reflected sunlight, and
hence is very similar to the Sun’s spectrum, elements and compounds in the planet’s
atmosphere can absorb certain characteristic wavelengths, resulting in dark, non-solar,
absorption lines. Such lines either are not in the Sun’s spectrum, or intensify the same
lines already there, thus showing the presence of such elements and compounds in the
atmosphere of the planet.
Unit 25
1. 229 km/sec toward us. [V=((655.5-656)/656) X 3 X 105 km/sec = -229 km/sec.
3
Unit 26
1. 2 X 106 m; 0.2 mm [The ratio of the Keck diameter to resolution is 10/500 X 10-9 or
20,000,000. Thus the diameter is 20 million times greater than the resolution wavelength.
To find the required diameter for the a radio telescope, multiply the required resolution
by 20 million to get 2 X 106 meters 2000 km. For the gamma ray telescope, multiply the
0.01nm resolution by 20 million to get 2 X 10-4 or 0.2mm.
Unit 27
1. 100 times. [How faint an object you can see depends on collecting area, which
depends on the square of the diameter of the aperture. Here the diameters are 8 cm for the
binoculars and 8 mm for your eyes. The binocular aperture is 10 times greater than that of
your eyes, so the increase is 102 or 100.
Unit 28
1.
Unit 29
1. 0.02 arcseconds; 0.1 arcseconds. [Resolution angle = (0.02 X wavelength in
nm)/Diameter in cm, so for the 240nm light, R = 0.02 X 240/240 or 0.02 arc seconds; for
the 1.2 micron light, R=0.02 X 1200/240 or 0.1 arcseconds.
4
Unit 30
1.
Geometry shows that there is less atmosphere between the observer and a star when the
star is high overhead than when it is near the horizon. The increased volume of air near
the horizon both absorbs some of the star light making the star appear dimmer, and
causes the image to be unsteady due to turbulent motions in the air. The latter effect is
known as “scintillation” or “twinkling.”
Unit 31
1. The iris is a special set of muscles that controls the size of the eye opening or pupil
(aperture). The pupil corresponds to the opening of a telescope tube (in telescopes with
tubes). Not all telescopes have irises, but some do. The cornea and lens together
correspond to the objective lens or mirror of a telescope, essentially the light collecting
and focusing apparatus. The retina corresponds to the focal plane of a telescope, the
region where images are formed. Rods and cones are similar in purpose to a CCD device,
which detects the incoming light. The brain is the "central processing unit," like the
computer that records and to some extent interprets data.
Unit 32
1. 5.25 kg/l for Venus; 1.24 kg/l for Jupiter. Venus is overall more dense than the
average rock, where as Jupiter is considerably less dense. In fact Jupiter, being composed
mostly of gases, is only slightly more dense than water overall. The published density
figure for Jupiter is 1.33 kg/l. The discrepancy between the established figure and the
calculated value here is due to the fact that Jupiter is not a sphere, but rather an “oblate
spheroid,” with a greater equatorial radius than its polar radius. Since the calculated value
used the equatorial radius, the volume thus calculated is slightly inflated, making the
density calculation too low.
5
Unit 33
1. The approximate U238/Th234 ratios after 1, 2, 3 and 4 billion years are 6/7, ¾, 5/8 and
27/50. In percentages of U238 remaining after the stated time periods, this would be 86%,
73%, 63% and 54%.
Unit 34
1. The orbital period of these planets would be about 0.03 years or about 11 days; 1 year
(same as Earth); and about 32 years. Example: According to Kepler’s Third law, 1=A3/P2,
or P = A3/2. So at 0.1 AU from a one solar mass star, the period in years would be (0.1)3/2
which is about 0.03. The distance these planets travel in their orbits is 1/10th, the same as,
and 10 times farther than the Earth in its orbit. The resulting speeds are therefore about
100 km/sec, 30 km/sec and 10 km/sec.
Unit 35
1. 25.5 hours. [The equatorial diameter of Earth is 12756 km. Travel time is distance
divided by speed, to T = 12756 km/(500 km/hr) or 25.5 hours.
Unit 36
1. The mass of hydrogen in the oceans exceeds the mass of the atmosphere by about 31
times. [Since there are 2 hydrogen atomic masses for every 16 oxygen atomic masses, the
ratio is 1/8, meaning that one out of every 9 atomic masses in water is due to hydrogen.
Thus 1/9 or about 11.1 percent of the ocean’s mass is hydrogen. The ocean’s total mass is
1.4 X 1021 kg, so the mass of hydrogen is one ninth that, or about 1.6 X 1020 kg. Now,
the total mass of the atmosphere is about 5.1 X 1018 kg, so the ratio of oceanic hydrogen
to the entire atmosphere is 1.6 X 1020 / 5.1 X 1018, or about 31. .
Unit 37
1. It may be easier to understand the math for part b, where we can solve for the fraction
of the volume by calculating the contributions to the Moon’s total mass = 3.0 VSi + 7.9
VFe = 3.3 V(Moon). Since VSi + VFe = V(Moon), we can solve for either volume fraction
VSi /V(Moon) or VFe/V(Moon). This gives 6.1% of the Moon’s volume being iron, and
93.9% silicate. The fraction of the masses in (a) is solved by finding the total volume =
MSi /3.0 + MFe /7.9 = M(Moon)/3.3. This gives mass fractions of 14.7% iron and 85.3%
silicate.
Unit 38
1. 1 1/3. one day = ¾ one year, or D = .75y. Thus y=D/.75. If it is 58.65 earth days, then
the year would be 58.65/.75 or about 78 earth days, which is 1 and 1/3 times the planet’s
day.
Unit 39
1. About 3.7 X 1020 kg [The molecular mass of CO2 is 44 (C is 12, two Os are 32) Of the
total mass of 4.6 X 1020 kg, 32/44ths are Oxygen: (32/44) X 4.6 X 1020 kg = 3.3 X 1020
kg kg H2O--> H is about 11 percent; O is about 89 percent Total mass = 3.3X1020 kg/.89
= 3.7 X 1020 kg.
6
Unit 40
1. About 0.69, 0.35, 2.69, and 2.35 AU for the largest and smallest opposition
separations, plus the largest and smallest superior conjunction separations. [Example:
Earth’s total variation is about 3 percent, or 1.5 percent plus or minus from the average of
1 AU. Thus, the minimum is 0.985 AU. For Mars the change is about 10 percent either
way, so its minimum is 1.52 AU *1.1 = 1.67 AU. At minimum separation at opposition,
the Earth must be at its maximum distance from the Sun (aphelion), while Mars must be
at its least (perihelion): 1.67-0.985 = 0.69 .
Unit 41
1. 3 to 6 years. [Using Kepler’s Third law, 1=A3/P2, or P = A3/2. 2.13/2 = 3; 3.33/2 = 6.
Unit 42
1. The primary differences between the planets that support the nebular theory are the
differences in composition, number of satellites, and existence of rings between the
terrestrial and Jovian planets. The stark differences in composition would be expected at
various distances from the Sun if the planets formed from the same mix of material but
subjected to different temperatures. The lower temperatures far from the Sun allowed
easier accretion of materials especially ices. This lead to more massive planets, higher
gravities and a higher likelihood of capturing and retaining icy moons and rings.
Unit 43
1. About 45,000 km/hr or about 12.6 km/sec. [Given a rotational period of 9.9 hours and
the equatorial diameter of 71,492 km, the circumference is 2 times pi times the radius, or
about 449,000 km. Dividing the latter by the period gives rotational speed: 45353 km/hr.
Divide that by the number of seconds in an hour to get : 12.6 km/sec.
Unit 44
1. 92% and 118 percent that of Earth, respectively. [surface gravity =GM/R2 . To find
values relative to Earth, substitute the relative values for Earth into the equation. For
example, the mass of Uranus is 14.54 times that of Earth, and its radius is 3.98 times
Earth’s. Substituting in gives 14.54/3.982 = 0.917 or about 92 percent that of Earth. Thus,
the surface gravity of Uranus is about 92 percent that of Earth. A 60 kg person would
“weigh” roughly 55 kg on Uranus .
Unit 45
1. Density of 1.66 kg/l. This gives Enceladus considerably higher density than Tethys,
indicating that in addition to icy materials, it must have some silicates and possibly even
some iron. Enceladus’ higher density and a composition other than nearly pure ice (as
with Tethys), may indicate that it had some internal heat in the form of radioactive decay.
This could provide the energy for some of the surface features including smooth terrain
that apparently has been (relatively) recently renewed. [Assume Enceladus is a sphere
(although this is not quite true). Volume is 4/3 X Pi X Radius squared. This amounts to
about 6.5 X 1016 cubic meters or 6.5 X 1019 liters. Density is Mass/Volume, so we have
1.08 X 1020 kg / 6.5 X 1019 l, or about 1.66 kg/l.
7
Unit 46
1. 2 and 4 times longer, respectively, for Europa and Ganymede [A generalized view of
Kepler’s Third Law yields P = A3/2 . Plug in the sizes of Europa’s and Ganymede’s
orbits, relative to Io’s, to get their periods relative to Io’s. Example, Europa’s orbital
radius is 671100/421800 or about 1.59 greater than Io’s, and 1.593/2 = 2.
Unit 47
1. about 17.9 AU. [As shown before, Kepler’s Third Law yields A = P2/3. If the Period is
76 years, then the semimajor axis (A) in AU is 762/3 or 17.9.
Unit 48
1. about 43 meters for density of 3 kg/l; about 62 meters for a density of 1 kg/l. Chances
are that the rock meteoroid would hit the earth and form a crater, perhaps one-half to a
kilometer across, causing serious destruction in the immediate area. Because of its lower
density, the ice meteoroid likely would heat up and explode in the air, causing damage
through the airblast, but forming no crater. [Volume = 4/3 X π X R3, and Density = M/V.
Thus we have D = M/(4/3πR3). Rearranging this we get R = (3M/4D)1/3 . Substituting in
the given values for D and M, the R is about 43 meters for a density of 3 kg/l.
Unit 49
1. 1.31 X 106 km [Using the formula from Unit 10 L = d X (A/57.3), where d is the
known distance and A is the angle subtended, the diameter is L. With d = 1.5 X 108 km
and A equal to 0.5 degrees, this is 1308901.
Unit 50
1. 1.4 g/cc. Jupiter’s density is 1.33 g/cc, which means that Jupiter is slightly less dense
than the Sun and it likely made of the same materials (mostly hydrogen and helium).
[Volume = 4/3 X π X R3, if R = 7 X 1010 cm, V = 1.44 X 1033 cubic centimeters (cc).
Divide the mass of 2 X 1033 g by volume to get 1.4 g/cc.
Unit 51
1. 435 km/sec [Four days is 96 hours or 345000 seconds. The average distance between
Sun and Earth is 1.5 X 108 km. Divide distance by time to get speed: 435 km/sec.
Unit 52
1. The next house is about 6.1 meters (19.9 feet) away. Here’s how I did it: The second
floor window sill of the nearby house is exactly 4.42 meters (14.5 feet) above the ground
at the base of the house. Call this measurement “b.” (Measurements like this are can
easily be made my dropping a weighted string from the window sill and then measuring
the length of string needed to just touch the ground.). Standing against the outside wall of
my house, I measure the angle from the ground directly under the window to the window
sill in question. Call this angle “A.” (This is a bit harder, but can be done with a simple
device known as a quadrant:
http://www.mscd.edu/~printit/pdf_files/AstronomicalQuadrant.pdf). In this case I find that the
angle is 36 degrees. Call the distance between the buildings D.” The proper trig
8
formulation here is D=b/tan(A); D=4.42m/tan 36, which equals 6.08 m. So the house is
about 6 meters away.
Unit 53
1. γ =1.15 [Lorentz factor γ = 1/√(1-V2/C2 . For V = 0.5C, that’s 1/√(1-.52/1) = 1/√(.75)
= 1.15.
Unit 54
1. 3.33 X 105 or 333,000. [A watt is one joule per second, so the question is how many
times will 3 X 10-19 go into 10-14? Just divide: 10-14/3*10-19 = 33,333.
Unit 55
1. 14,500K [Wien’s Law from Unit 23, T=2.9 X 106 K nm/λmax. Using 200nm for λmax
gives (2.9/200) * 106 K = 14500K.
Unit 56
1. 40 and 16. [(MA + MB)=a3/P2, so the combined mass, in terms of the Sun’s mass, is
103/52 = 40 in the first example and 43/22 = 16 in the second.
Unit 57
1. 1.72 X 106km for Alpha Cent A, which is about 24 percent greater than the Sun’s
diameter of 1.39 X 106 km; 1.22 X 106km for Alpha Cent B, or about 88 percent that of
the Sun. [Using the formula from Unit 10, L = d X (A/57.3), where d is the known
distance and A is the angle subtended in degrees, the diameter is L in whatever units you
are using for d. This is easiest if you change d to km. Change the angles for the two stars
from arcseconds to degrees by dividing by 3600. Thus .0085 arcsec is 2.36 X 10-6
degrees; and .0060 arcseconds is 1.67 X 10-6 degrees. One light year is about 9.5 X 1012
km, so 4.4 ly is 4.18 X 1013 km. For Alpha Cent A, this is L = 4.18 X 1013 km X (2.36 X
10-6/57.3) = 1.72 X 106km.
Unit 58
1. Betelgeuse is the red giant; 40 Eridani B is a white dwarf.
9
Unit 59
1. Graphic adapted from the Book
Unit 60
1. [ For a dust grain falling in nebular cloud from R1 to R2
M = mass of Sun: 1.99 X 1030 kg
G= 6.67*10-11m3kg-1s-2
g = GM/R2
PE = mgR where R is the distance of the mass m from the center of the gravitating mass
M
By Conservation of Energy, Kinetic Energy = change in Potential Energy
KE=(PE1-PE2)=1/2mv2
KE=(mg1R1-mg2R2)=1/2mv2 where the subscripts signify the initial (1) and secondary (2)
positions
m(g1R1-g2R2)=1/2mv2
2(g1R1-g2R2)=v2
v=√(2(g1R1-g2R2))
gR = GM/R2 Where gR signifies the acceleration of gravity at some radius R
=========
R1= 1 light year = 9.4608 X1015 m
R2= 106 km 1000000000 m
=========
10
g1= 1.48294E-12 ms-2
g2= 132.733 ms-2
=========
v= 167.5 ms-1 .
Unit 61
1. about 5.7 X 1010 years, or 57 billion years, which is about 5.5 times the expected
lifetime of the Sun. [From Unit 58, Luminosity L=M3.5 where M is the star’s mass. Both
L and M are expressed in terms of the Sun’s mass and luminosity. In this Unit we have
the lifetime of a main sequence star as t=(M/L) X 1010 years. Substituting the L from Unit
58, we get t = (M/M3.5) X 1010 years. If the mass in question is 0.5, then 0.53.5 = 0.088,
and t=(.5/0.088) X 1010 years = 5.7 X 1010 years.
Unit 62
1. 1.4 X 10-6 kg/l [Volume increases as the cube of the radius, so if the red giant Sun has
100 times the current radius, it will have 1003 (106 or one million) times the current
volume. Density is mass divided by volume, so assuming mass doesn’t change, the Sun’s
density at red giant stage will be one millionth what it is now. Currently it is 1.4 kg/l, so
in the red inflated stage, it will be 1.4 X 10-6 kg/l.
Unit 63
1. about 9.3 km/sec. [a 10 percent increase in a 40 million kilometer radius is 4 million
kilometers. So the surface of the star moved outward 4 million km in 5 days. There are
432000 seconds in 5 days, so the speed was 4 X 106 km/432000 sec or about 9.3 km/sec.
Unit 64
1. 25, 1, and 0.09 solar radia respectively at 6000K, 30,000K and 100,000 K. Using the
Stefan-Boltzmann Law and holding luminosity L constant, if temperature varies, the
radius must also vary to compensate. L = 4πR2σT4 which can be rearranged as R2 =
L/4πσT4 . The L/4πσ can be considered constant and removed, so we have R  √(1/T4).
Thus at T=6000 the temperature has doubled and R reduces to one quarter the original
value, or about 25 solar radii.
11
Unit 65
1. 4.7 X 105 kg/l [The Sun’s radius is 695000km, whereas the white dwarf’s is 10000 km.
Volume varies with the cube of the radius, so the white dwarf is (105/6.95 X 105))3 or
about 336000 times less voluminous than the Sun, and the density is corresponding the
same factor larger than that of the Sun. The Sun’s density is about 1.4 kg/l, so the white
dwarf density is about 4.7 X 105 kg/l.
Unit 66
1. 1.2 X 1057 neutrinos [If the star is 2 solar masses and half of that mass is of protons,
then the total mass of protons is one solar mass or 2 X 1030kg. Each proton has a mass of
1.67 X 10-27 kg. If all those protons change into neutrons, releasing one neutrino each
time, then there are 2 X 1030/1.67 X 10-27, or 1.2 X 1057 neutrinos released.
Unit 67
1. Escape velocity for the white dwarf is about 6100 km/s; escape velocity of neutron star
is about 193,000 km/sec
Unit 68
1. 8.9 km [Rs=2GM/c2. G= 6.67 X 10-11 m3/kgs2 . C = 3 X 105 km/sec. (Change to m/sec
by multiplying by 1000: 3 X 1011km/sec.) One solar mass is 2 X 1030 kg so 3 solar
masses are 6 X 1030 kg. Substituting all these values into the equation yields 8.9 km.
Unit 69
1. Along about here when I realized that I will never get finished with this if I have to
keep making sketches, the bright thought occurred to me to check if there were any such
sketches in any of the units Dr. Arny did. Low and behold, chapter 13 had the call for a
sketch, but as far as I can tell none was generated. So for now I am skipping it, too.
Unit 70
1. 1. About 26000 years. [One kpc is 3260 light years, so 8kpc = 26080 light years.
Thus the center of the Galaxy is about 26,000 light years away, and light from this
region takes just over 26,000 years to reach us.
Unit 71
1. 4.4 km/sec. [Set up a simple equation: 220 km/sec times 8 kpc = X km/sec times 400
kpc. Rearranging this gives X = 220 km/sec times (8/400) = 4.4 km/sec .
Unit 72
1. 1429 MHz [From Unit 22 Wavelength (λ) times frequency (ν) = the speed of light, c.
This can be rearranged to yield ν = c/λ. Thus frequency = (3 X 1010 cm/sec)/21 cm =
1.429 X 109 Hz or 1429 MHz. The wavelength is actually closer to 21.1 cm, giving a
frequency of about 1421 MHz.
Unit 73
1. About 222 million years for the “Galactic Year.” [Assume that the Sun has a circular
orbit. Circumference = 2πR where R = 8 kpc or 8 X 3260 light years, which converts to
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about 2.45 X 1017 km. Thus C = 2 X π X 2.45 X 1017 km = 1.54 X 1018 km. At 220 km/s,
it would take 1.54 X 1018km/2.2 X 102 km/sec or 7.0 X 1015 seconds to make one
complete orbit. Since there are about 365 X 24 X 60 X 60, or 31536000 seconds in a
year, the orbital time is 7.0 X 1015 seconds/3.15 X 107 sec/year = 2.22 X 108, or about
222 million years.
Unit 74
1 0.047 z and 14100 km/s velocity [By z = Δλ/ λ = (22-21)/21 = 0.047 as the red shift.
This is about 4.7% the speed of light, or 0.0.47 X 300000 km/s or about 14,100 km/s
recessional velocity.
Unit 75
1. a) 10 minutes; b) 0.1 minute of arc (about 6 seconds of arc). [By Section 10.4, L = d X
A/57.3 where L is the linear diameter, d is the distance and A is the angular diameter.
This can be changed to A = (L X 57.3)/d. L and d must be changed into the same units. L
= 30kpc. In the first case, a), d= 10 Mpc or 10000kpc. Thus A = (30 X 57.3)/10^5 =
0.017 degrees, which is about 10 minutes of arc.
Unit 76
1. 9.8 X 108 or nearly a billion year! [1 Mpc = 3260000 light years; 1 light year = 9.46 X
1012 km; so 1 Mpc = 3.08396 X1019 km. Velocity is given as 1000 km/s. Time equals
distance divided by velocity, so we have (3.08396 X1019 km)/ 1000 km/sec, which yields
3.08396 X 1016 seconds. There are 31536000 seconds in a year. After dividing the total
number of seconds by the number of seconds in a year, we get 9.78 X 108 years.
Unit 77
1. 100 times farther away [Brightness falls off by the inverse square of the distance. Thus
for the quasar’s light to be diminished 10,000 times, it has to be √10000 or 100 times
farther away.
Unit 78
1. 2.8889 X 1041 kg, 3.17779X 1041 kg, 3.46668X 1041 kg. This represents an increase of
about 2.89 X 1040 kg 14.5 billion solar masses per Kpc. [M<R = (V2 X R)/G. Here the
values for R are 10,11, 12… kiloparsecs. One Kpc = 3.08396 X 1019m, and G = 6.672 X
10-11 m3/kgs2 . Substituting the values in and solving, we have the mass at 2.8889X 1041
kg, 3.17779X 1041 kg, 3.46668X 1041 kg, respectively at R = 10, 11 and 12 Kpc.
Unit 79
1. A Hubble Constant of 600 km/sec per Mpc is much higher than the currently accepted
value, and would imply that the Universe is expanding much faster than current
estimates. The greater expansion rate would mean that the Universe would have taken
less time to expand to its current size, and hence that the age of the Universe is much less
than the currently accepted value of 13-14 billion years. This would be a problem, since it
gives the age of the Universe at less than 2 billion years, which is not as old as rocks
found on Earth or the estimated age of the Sun. This of course was a source of great
13
concern for cosmologists, since it makes no sense for the Earth and Sun to be older than
the Universe itself!
Unit 80
1. 1. When the Solar System was formed, the universe was about 2/3rd its current age,
and we might expect about 2/3rd its current size. (This can be confirmed inTable 79.1
where we can see that 5 billion years ago the universe was about 0.67 its current size.)
Light from the CMB would have a wavelength about 2/3rd its current wavelength because
the wavelength is stretched by the expansion of space. By Wien’s law, the temperature
would be about 4 K (3/2 what we currently see or 1.5  2.7 K). At that time we would
have been seeing CMB light coming from regions that were nearer to us than what we see
today. We would expect the kinds of structure to have been similar, but the specific
patterns would have been different and the angular sizes larger (since the CMB would
have been “closer”).
Unit 81
1. a) The mass of the solar system is essentially the mass of the Sun, 2 X 1030 kg. If the
density of the Universe was 0.1kg/l, then the volume occupied would be (2 X 1030)/(0.1
kg/l), which is 2 X 1031 liters or 2 X 1028 cubic meters. This is not much bigger than the
size of the Sun today. b) After 180 seconds the universe would have been 180 times
larger if it were expanding at a constant speed, however it is slowing down, so we might
guess that it is 50 times bigger. With that guess, we would find a density of matter about
503 ≈ 100,000 times lower, or about .000001 kg/l, , implying a volume of 100,000 times
greater. Thus it would be 2 X 1036 liters or 2 X 1033 cubic meters. (Note: detailed
calculations of the young expanding universe indicate that the density of matter declines
by a factor of about 30 each time its age grows about 10 times bigger.)
Unit 82
1. About 16 percent less. Critical density, ρc, is proportional to H2 by the equation ρc =
3H2/8πG. As H2 changes, so does the critical density. Thus a change in H from 71 to 65
km/s/Mpc is (712-652)/712, which is about 0.162 or 16 percent drop in critical density.
Unit 83
1. The mass of bacteria would equal the mass of the Earth in just more than 132 hours.
The mass of the Earth is 5.97 X 1024 kg, whereas the mass of the bacterium is 10-15 kg, so
the ratio is 5.97 X1024/10-15 = 5.97 X 1039. So the question is how many hours will it take
for the number of bacteria to reach about 6 X 1039, assuming that the number doubles
every hour. After the first hour there will be two bacteria, after two there will be 4, after 3
hours there will be 8 bacteria and so on. You can work it out that way, or use the
following equation, where ln is the natural logrithm: Number of hours to reach mass of
Earth = ln(5.97 X 1039 – 10-15)/ln(N) where N is the rate of change per hour. In this case
the number doubles ever hour, so N = 2.
Unit 84
1. The Haircut Equation. HC= Np*Fn*Fd*Fw, where Np is the total population of a
given area in question; Fn is the fraction of people in that population who need a haircut
at any given time; Fd is the fractional delay time in “inverse” days representing the length
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of time on average a person delays getting a haircut; and Fw respresenting the fraction of
the time that weather factors present an excuse for not getting a haircut. (Other
complicating factors could be considered, but let’s stop here.). Given 260 million people
in the US as Np, estimate Fn as ¼; Fd as 1/7 (that is, imagine that the average person
delays getting a haircut for 7 days; and estimate Fw as 1/21, meaning that bad weather
might prevent someone from getting a haircut every 3 weeks or so. Plugging in the
numbers we get: HC = 260 million X ¼ X 1/7 X 1/21 = .442 million or about 442,000
haircuts every day in the US. (Remember, this is just a wild guess!)
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