Download Additional Exercises for Chapter 4 Computations of Copernicus and

Additional Exercises for Chapter 4
Computations of Copernicus and Brahe
The fact that any tangent to a circle is perpendicular to the radius to the point of tangency was
verified in the Additional Exercises for Chapter 3. This fact is also a consequence of Proposition
3.4. It will play a role in the first exercise below.
47. Consider Figure 4.1 in the case of Mercury and suppose that Copernicus measured the angle
α to be 22◦ . What estimate for the distance of Mercury from the Sun could he derive from
this measurement? Ans: 0.3746 AU
Take any triangle and let its sides have lengths a, b, and c respectively. Let α be the angle
opposite the side with length a, β the angle opposite the side with length b, and γ the angle
γ
a
β
b
c
α
opposite the side of length c. Then
sin α
sin β
sin γ
=
=
.
a
b
c
This relationship is known as the Law of Sines.
48. Copernicus estimated the distances of the outer planets (Mars, Jupiter, and Saturn) from the
Sun in terms of the distance from the Earth to the Sun as follows. Start with a situation in
which the Sun S, the Earth E, and the planet P are aligned. After a certain time, say a few
months, the Earth and the planet will have moved to, say, positions E1 and P1 respectively.
See the diagram below. Copernicus could estimate the angle θ because he knew the elapsed
time and the period of the outer planet. Knowing the period of the Earth’s orbit, he could
estimate ∠ESE1 . This provided him with an estimate for γ. Since he was able to measure α,
he could estimate β. The formula
sin α
DP =
sin β
gave him an estimate for the distance DP of the planet from the Sun in AUs. Use the Law of
E1
α
β
DP
γ
1 AU
S
P1
E
P
Sines to verify this formula.
49. Suppose that Copernicus knew that the period of the orbit of Mars was about 1.9 years. He
also knew that at a certain time the Earth, Mars, and the Sun were aligned. Let’s suppose
that he observed Mars for two months from that time and that he then measured the angle
α to be 114.9◦ . What estimate for the distance of Mars from the Sun could he have derived
from this? Compare this answer with data from Table 4.2. Ans: γ = 28.4◦ , 1.518 AU
50. One way to confirm the motion of the Earth around the Sun is by observing a shift in the
position of nearer stars (as seen against the firmament of more distant stars). The new star (an
exploding supernova) that appeared in 1572 provided an excellent opportunity. According to
the Greek philosphers, changes in the universe could only occur in realms close to the Earth.
So this new star was thought to be relatively near. Brahe pointed his precision instruments
skyward, but did not detect any change in its position. He knew that this could only mean
that the star had to be far off! Brahe was able to pinpoint the angular position of a star with
1
-th of a degree. If Brahe would have been able to measure a
an accuracy of one minute or 60
change in the position of a star with this accuracy, what could he have concluded about the
distance of the new star from the Earth? [Hint: Review Section 1.7 and focus on the angle
θ.] Ans: At least 0.1 LY.
Galileo and Sums of Odd Numbers
51. In his experiments with inclined planes, Galileo considered sums of consecutive odd numbers
of the form 1 + 3 + 5 + . . . . A look at the figure below confirms that 1 + 3 = 22 , 1 + 3 + 5 = 32 ,
1 + 3 + 5 + 7 = 42 , and that 1 + 3 + 5 + 7 + 9 = 52 . This suggests that the sum of the first k
odd numbers should be equal to k 2 . Expressed in sigma notation this is the equality
k
X
(2i − 1) = 1 + 3 + 5 + · · · + (2k − 1) = k 2 .
i=1
2
Check this equality, for k = 6. Then verify it for any k by using the trick of Gauss from Note
4 at the end of Chapter 4. In the context of Galileo’s observations, this shows that if a ball
rolls down an inclined plane starting from rest, then the total distance it travels is proportional
to the square of the time of the motion.
Parabolas, Circles, and Ellipses
52. Consider the parabolas y = x2 + x − 11 and y = 2x2 − 4x − 7. Note that both of them open
upward. Make use of “completing the square” to find the lowest point and the x-intercepts
for each of the parabolas. What are their y-intercepts? Determine the points where the two
parabolas intersect and then sketch both of their graphs. Express the region that the two
parabolas enclose in set theoretic notation. Ans: Lowest points ( −1
, −45
), (1, −9), points of
2
4
2
2
intersection (1, −9), (4, 9), region {(x, y) | 2x − 4x − 7 ≤ y ≤ x + x − 11}.
53. Make a careful sketch of a unit circle and the x-axis. Consider the angle θ with radian
measure 22 and the corresponding point Pθ . How many full revolutions and what fraction of
a revolution do you need to go around the circle (clockwise or counter clockwise?) so as to
locate this point? Go around the circle and place the point on the circle with care. Use the
location of the point to estimate sin θ and cos θ. Then do the same for θ equal to −11 radians.
Then check all your estimates with a calculator.
2
2
54. Consider the ellipse x52 + y32 = 1 . Show that (4, 0) is a focal point. Consider the circle with
center this focal point and radius 2. Verify that the points of intersection of the circle and
√
the ellipse are ( 15
, ± 34 7).
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55. Consider a semicircle of radius 5 and inscribe half of the ellipse with semimajor axis 5 and
3
π/3
O
5
3
semiminor axis 3. Compute the shaded area of the accompanying figure.
56. The figure below shows a coordinate system along with the upper half of a circle of radius 5
and the upper half of an ellipse with semimajor axis 5. The origin O is the center of both the
full circle and the full ellipse. The point F = (2, 0) is a focus of the ellipse. The point P0 is
on the circle, the point P is on the ellipse, and the dotted segment P0 P X is perpendicular to
the x-axis. The y-coordinate of P0 is 4 and the x-coordinate of the point X is 3.
y
P0
P
β
0
F
X
N
x
i. Compute the semiminor axis of the ellipse and write down an equation for the ellipse.
ii. Determine the y-coordinate of the point P and use your calculator to find the angle β in
radians.
iii. Compute the area of the elliptical sector N F P.
Kepler’s Battle with Mars
57. An analysis of the data led Kepler to the conclusion that the orbit of Mars is not a circle, but
that it curves inward little by little and then returns to the amplitude of the circle. The figure
below shows what he had in mind. Consider Mars in the position M where the deviation
between its orbit and the circle is largest and let d be this deviation (in AUs). Let O be the
c
M
b
orbit
θ
O
S
center of the orbit and the surrounding circle and let b be the distance from M to O. Let S
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be the position of the Sun and let θ = ∠OM S.
i. Show that if the orbit is an ellipse, then d = b(sec θ − 1). Kepler became aware of this
equality at a time when he did not yet believe the orbit to be ellipical. The fact that it
holds for an elliptical orbit tells us that he was on the right track.
ii. Kepler computed θ to be 5.3◦ . What did this tell him about the eccentricity of the
elliptical orbit of Mars?
Orbital Matters
58. Which of the equations of Section 4.7 confirms that the locations of the two positions of a
planet at which its distance to the Sun are least (the perihelion position) and greatest (the
aphelion position) are both on the major axis? How does it confirm this?
59. Use the data from Table 4.2 to compute for Mercury, the linear eccentricity, the semiminor
axis, and the closest and farthest distances from the Sun (all in AUs). If you were to draw
the elliptical orbit of Mercury at a scale of 1 AU = 20 inches would you be able to visually
distinguish it from a circle? To answer the last question compare Mercury’s ellipse to its
surrounding circle. How far from the center of the ellipse would the Sun be? [Note: Table 4.2
informs us that Mercury’s ellipse is flatter than the ellipses of the other planets.]
60. The following information appeared in the article Up Close and Personal With Mars in the
Science Times Section of the New York Times on Tuesday, August 19, 2003. Minutes before
6 a.m. Eastern time on Aug. 27, 5:51 to be exact, the separation between Mars and Earth
will be precisely 34,646,418 miles or 55,758,006 kilometers. This is asserted to be the closest
that Mars has been in 59,619 years. But in the year 2287, the article goes on to say “Mars
and Earth will cruise even closer”. Use the information in Section 4.7 to show that the
closest possible distance between Earth and Mars is about 33.9 million miles or 54.6 million
kilometers. Can there be a time when this is achieved? What would a “yes” imply about the
relative position of the two major axes of the orbits of the Earth and Mars at such a time?
The closest Mars approaches to Earth were computed by Dr. E. Myles Standish, Jet Propulsion
Laboratory, Pasadena, JPL, and posted to DOME-L, a newsgroup for planetarium professionals,
in response to a question about Mars’s close approach in August of 2003. Standish’s data is based
primarily on the convergence of Earth’s and Mars’s aphelion/perihelion positions. There is no Mars
approach on the record that was closer than the 2003 event, but there will be seven even closer
approaches over the next several hundred years. They are listed below.
AUs
Kilometers
Date
0.37200418
55,651,033
Sept. 8, 2729
0.37200785
55,651,582
Sept. 3, 2650
0.37217270
55,676,243
Sept. 5, 2934
5
0.37225400
55,688,405
Aug. 28, 2287
0.37230224
55,695,623
Sept. 11, 2808
0.37238224
55,707,590
Aug. 30, 2571
0.37238878
55,708,568
Sept. 2, 2366
0.37271925
55,758,006
Aug. 27, 2003
61. Mars was at perihelion on August 27, 2003. How far is it from the Sun today?
62. Consider Mercury in its orbit after an elapsed time of t from its perihelion position.
i. Consider the successive approximation of the angle β. How many steps will always be
enough to insure that β has been computed with an accuracy of four decimal places?
[Hint: ε6 ≤ 0.00008, ε7 ≤ 0.00002.]
ii. Carry out the computation of β with four decimal place accuracy for the position that
Mercury has exactly 20 days after its perihelion position. Has Mercury travelled more
than one quarter of its orbit by that time or less? Ans: T = (0.2408)(365.2422) = 87.9503
days, β1 = 1.4288, β = 1.6340.
iii. Compute the corresponding r and α. Ans: r = 0.3921 AU
63.
i. Starting from perihelion, a planet (comet or asteroid) will complete one-fourth of its
orbit in the time t = T4 − T2πε , where T is the period and ε the astronomical eccentricity
of the orbit. Verify this formula in two different ways. First, by computing the area At
in Figure 4.35 and using Kepler’s second law, and again by using one of the formulas of
Section 4.7. How far is the planet from the Sun at this time? Conceptually what is the
difference between t and T4 ?
ii. How many days is t for the Earth? What is the difference between T4 and t for the Earth?
How many years is t for Halley? [The period and astronomical eccentricity of Halley are
T = 76 years and ε = 0.967.] Ans: For Halley t = 7.3 years. Surprised?
64. Consider the planet Venus. What evidence do you have that its orbit is closer to being a circle
than that of the other five planets known at the time of Galileo and Kepler. The period of
its orbit is 0.6152 years or 224.7 days and its astronomical eccentricity is ε = 0.0068. Turn
to Figure 4.33 and focus on the angular position α of Venus. Starting from α = 0, how many
days does it take for Venus to revolve through the first 60◦ = π3 ? How many days for the
second 60◦ = π3 , and how many for the third 60◦ = π3 ? Ans: 37.03 days; 74.48 − 37.03 = 37.45
days; 112.35 − 74.48 = 37.87 days. [Different accuracy and roundoff procedures will lead to
different estimates.]
65. The collection of known planets is larger now than at the time of Galileo and Kepler. Beyond
Saturn, the planets Uranus, Neptune and Pluto (in order of increasing distance from the Sun,
as well as the time of their discovery) have been added to the list. Pluto, located in 1930
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by the astronomer Clyde Tombaugh, was the last. In 1951, the astronomer Gerard Kuiper
predicted that other small bodies would be found beyond the orbits of Neptune and Pluto.
Indeed, a number of such objects have been found. (This fact that led to the expulsion of
Pluto from the list of planets in 2007.) One of them was the subject of an article in the New
York Times (the Science Times section) on November 7, 2000. The Times reported that a
team of astronomers had discoved a miniplanet in an orbit close to that of Pluto. Officially
designated 2000 EB173, this object was said to be about 400 miles wide (and hence about
one fourth the size of Pluto) with an orbital period of 243 years. Little additional data about
2000 EB173 was provided in the article. However there was a diagram showing that the orbit
of 2000 EB 173 is beyond that of Neptune in general, but that there is a stretch where the
orbit falls just barely inside that of Neptune. (Pluto’s orbit is similar in this regard.) It is
known that Neptune’s orbit has a semimajor axis of 30.05 AU and that, with an astronomical
eccentricity of 0.011, it is close to being circular. Make use of relevant pieces of the information
to estimate the semimajor axis and astronomical eccentricity of 2000 EB173. Ans: a ≈ 39,
ε ≈ 0.23. [Given the description, only rough estimates make sense.]
Computer Model of Elliptical Orbits Generated by Kepler’s Equations
Go to the link “Computer Model of Elliptical Orbits Generated by Kepler’s Equations” attached
to the website for Chapter 4 and study it. Experiment with the simulations of the Inner and Outer
planets. Notice that the orbit of a comet or asteroid can be simulated by feeding in the relevant
data.
66. Simulate the orbits of the asteroids Ceres, Icarus, and Chiron, the comets Halley, Hale-Bopp,
Hyakutake, and the miniplanet 2000 EB173. (In reference to the latter, the data in Exercise
65 leads to the estimates a = 39 AU and ε = 0.23.) Use “Inner Planets” for Icarus and “Outer
Planets” for the comets and the miniplanet. Notice that Hale-Bopp moves off screen quickly
and vanishes. Estimate how long you would have to wait before you will see it return to the
screen. [Hint: Estimate the distance that the simulated Hale-Bopp is from the Sun when it
goes off screen as well as the time it takes for that to happen. Then use the equations of
Section 4.7 to estimate the corresponding angle β as well as the period T . CHANGE this
problem as follows. Simulate Hale-Bopp by putting in T = 187.5 and ε = 0.99. Set the
simulation by clicking on the yellow disc. Now forget ... and try to determine the orbital
data for the dot moving on the screen. Now measure a − e to be 83 of an inch. Then run the
simulation with trails on. Observe that Hale-Bopp disappears in 13 seconds and (use a ruler
that r = 5 81 inches when it leaves the screen. For α = π2 measure r = 43 inches. Using the
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equation of the ellipse (with x = e) now get: b2 (1 − ε2 ) = 16
as well as a(1 − ε) = 38 . Now,
2
2
compute − ab 2 = 4(1−ε)
. But 1 − ab 2 = ε2 . can solve ... ?]
1−ε
67. Read the NY Times article One Find, Two Astronomers: An Ethical Brawl. What do you
think about the ethical issue that it raises? Use the web to look up the orbital data of “Xena”
- this is an object orbiting our Sun that is bigger than Pluto - and simulate its orbit by using
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the “Outer Planets” page.
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