Download Study Guide for 1ST Astronomy Exam

Study Guide for 2ND Astronomy Exam
Chapters 4 & 5 from AstronomyNotes
The successful student will be able to…
 Describe the essentials of the geocentric model of the Universe of Aristotle and Ptolemy.
o The position and motion of the Earth
o The nature of terrestrial and celestial matter
o The role of epicycles in the Ptolemaic model of planetary motion.
 Where is a planet on its epicycle when it goes retrograde? Why does it go retrograde only when it is on
that part of its epicycle?
o The role of equants in the Ptolemaic model of planetary motion.
 How was the equant used to explain why there are only 179 days between the Fall Equinox and Spring
Equinox, but 186 days between the Spring Equinox and the Fall Equinox?
 Describe the essentials of the heliocentric model of the Universe of Copernicus.
o The position and motion of the Earth
 List all the motions of the Earth; Rotation on a tilted axis and revolution about the Sun with the time scales
and degree of tilt.
o The calculation of true orbital periods.
 Saturn is in opposition on 2013/04/28 and again on 2014/05/10. There are 378 days between these two
oppositions. Knowing that Earth takes 365.241 days to complete a 360  orbit around the Sun, estimate the
time required for Saturn to complete one 360 orbit around the Sun. (Hint: First find the number of
degrees Earth moves around the Sun in the 378 days between oppositions using a proportion, then find the
number of degrees that Saturn moves through its orbit between oppositions using subtraction. (See
handout from class) and, finally, use a proportion to estimate how long Saturn takes to complete oe 360 
orbit. Ans: The Earth moves 372.6 around its orbit in 378 days. Saturn moves 12.6  around its orbit in
that time. Satrun takes about 10,800 days or 29.6 year.)
o The calculation of distances of the planets from the Sun.
 Identify the right triangles that Copernicus constructed to
determine the relative distances of the planets from the Sun for
inferior and superior planets. See figure to the right.
 Describe how Copernicus could use planetary configurations to
determine the magnitude of the shaded angles in the figure to the
right.
o The true nature of retrograde motion.
 Completely describe the causes of retrograde motion for a
superior planet.
o The reason that superior planets brighten during retrogression.
 Completely describe why superior planets only brighten when in retrograde motion. Include an illustrative
sketch.

 State the flaws of the original heliocentric model of Copernicus.
 Describe Kepler’s three laws of planetary motion and state how the first two laws were contrary to the previously held ideas
of Aristotle and Ptolemy.
 Correctly use the terms ellipse, focus, semi-major axis, perihelion, aphelion, eccentricity in youir answer to
this questions.
 Perform simple mathematical problems using Kepler’s 1st and 3rd Laws similar to the homework problems.
 See example problems in italics below.
o Calculate the period of an object orbiting the Sun given its semi-major axis in AU.
 1685 Toro is an Apollo asteroid that orbits the Sun. Its semi-major axis is 1.37 AU What is its orbital
period in days? . (Ans: 1.60 years = 585.7 days)
o Calculate the semi-major axis of an object orbiting the Sun given its orbital period in years.
 The asteroid 1981 Midas was discovered on March 6, 1973 by Charles T. Kowal at Palomar Observatory.
Its orbital period is 864.541 d. What is its orbital semi-major axis? (Ans: 1.78 AU)
o Calculate the semi-major axis of an object orbiting the Sun given its perihelion and aphelion distances in AU.
 4257 Ubasti (1987 QA) is an Apollo asteroid and Near-Earth object discovered on August 23, 1987 by
Jean Mueller at Palomar Observatory. Its perihelion and aphelion distances are 0.876 AU and 2.42 AU
respectively. What is Ubasti’s orbital semi-major axis?
o Calculate the orbital eccentricity of an object orbiting the Sun given its perihelion and aphelion distances in AU.
 6063 Jason (1984 KB) is an Apollo asteroid discovered on May 27, 1984 by Carolyn and Eugene
Shoemaker at Palomar. Its perihelion and aphelion distances are 0.525 AU and 3.91 AU respectively.
What is Jason’s eccentricity? (Ans: 0.763
o
Calculate the perihelion distance of an object orbiting the Sun given its semi-major axis and orbital eccentricity.
 4183 Cuno is an Apollo, Mars- and Venus-crosser asteroid. Its semi-major axis and eccentricity are 1.98
AU and 0.636 respectively. What is Cuno’s perihelion distance? (Ans: 0.720 AU)
 Discuss Galileo’s observations of the Sun, Moon, Jupiter and Venus and state how they contradicted the previously held
Aristotelian model of the Universe.
 Describe how each of Galileo’s observations contradicted the Aristotelian geocentric model of the
Universe.
The Universal Law of Gravitation
 Describe the characteristics of gravity in words and in an equation.
 Describe and illustrate by example the nature of the inverse square law as it applies to gravity.
o Just like home work #5 problems #10 & 11.
 State the significance of the low value for G.
Measuring a Body’s Mass using Orbital Motion
 Describe what an astronomer needs to observe to calculate the mass of a distance body using properties
of the distant body’s satellite.
o
A picture of Saturn and its moon Titan appears to the right. What would an
astronomer need to know about Saturn and Titan to calculate Saturn’s mass from the
motion of its satellite Titan?
Orbital and Escape Velocities
 Describe how the orbital velocity of an object in circular orbit depends on the distance from the central object and how the
orbital velocity depends on the mass of the central body.
 Similar to HW #5 Problems #4, 17 & 18.
 Describe how the mass of an orbiting object affects its orbital velocity.
 Remember our class problem of the Moon, spacecraft, astronaut and wrench in orbit around the Earth.
Which went faster to stay in orbit?
 Define escape velocity and state how it depends on the mass and radius of the body that you are escaping from.
 Conceptual questions similar to HW #5 Problems #19 to #21.
Plus…
o
o
Use ratios to compare sizes of astronomical objects.
 See practice problems below
Use a proportion to calculate a scale model of an astronomical object.
 See practice problems below
Practice Ratio and Proportion Problems
1.
How many Earths could fit side by side across the Sun? Look up the radii of the Earth and Sun
Radius of the Earth = 6.378  10 m
6
Radius of the Sun = 6.96  10 m
8
384.4  10 6 m
The Ratio
 109
6.378  10 6 m
109 Earths could fit side by side across the Sun.
2.
How many Earth’s could fit between the Earth and the Moon? Look up the distance to the Moon
Distance to the Moon = 384.4  10 m
6
Diameter of the Earth = 1.276  10 m
7
The Ratio
384.4  106 m
 30.1
1.276  10 7 m
About 30 Earths could fit side by side Between the Earth and the Moon.
3.
How many Earth’s could fit between the Sun and the Earth? Look up the AU
1 AU = 1.5  10 m
11
Diameter of the Earth = 1.276  10 m
7
1.5  1011 m
 11,755
The Ratio
1.276  107 m
Almost 12,000 Earths could fit side by side between the Earth and the Sun.
4.
How many Suns could fit side by side across the solar system? Look up Pluto’s Distance from the Sun (There is an
error in this solution. Mathew Thomas found it. Can you? The correct answer is twice the answer below.)
Pluto’s distance from the Sun = 39 AU = 5.85  10
12
m
Diameter of the Sun = 1.39  10 m
9
The Ratio
5.85  1012 m
 4,209
1.39  109 m
About 4,200 Suns could fit side by side across our solar system.
5.
How many solar systems could fit side by side between the Sun and the next nearest star Proxima Centauri? Look up
the distance to Proxima Centauri
Distance to Proxima Centauri = 4.3 ly = 4.07  10
16
Diameter of the Solar System = 78 AU = 1.17  10
13
m
m
The Ratio
4.07  1016 m
 3,479
1.17  1013 m
About 3,400 of our solar systems could fit side by side between the Sun and the nearest star Proxima Centauri.
6.
How many solar systems could fit side by side across the Milky Way galaxy? Look up the diameter of the Milky Way
galaxy.
Diameter of the Solar System = 78 AU = 1.17  10
m
Diameter of the Milky Way = 100,000 ly = 9.46  10
20
13
The Ratio
m
9.46  10 20 m
 8.09  107  81 Million
13
1.17  10 m
About 81 Million of our solar systems could fit side by side across the Milky Way Galaxy.
Proportion Problems
1.
If the Earth were 1 foot in diameter, how far away would the Sun be?
1 AU
x

2  6,378 km  1 ft
x  2  6,378 km   1 AU  1 ft
x
1 AU  1 ft
2  6,378 km 
1.5  108 km  1 ft
2  6,378 km 
x  11,759 ft
x
x  2.23 miles
If the Earth were 1 foot in diameter, the Sun be 2.23 miles away.
2.
If an AU were shrunk to 1 foot, how far away would the next nearest star be?
4.23 ly
x

1 AU 1 ft
x  1 AU  4.23 ly  1 ft
4.23 ly  1 ft
1 AU
4.23  63,240 AU  1 ft
x
1 AU
x  267,505 ft
x
x  50.7 miles
If the AU were 1 foot in diameter, the nearest star be 50.7 miles away.
3.
If an AU were shrunk to 1 foot, what would be the diameter of the Milky Way galaxy?
100,000 ly
x

1 AU
1 ft
x  1 AU  100,000 ly  1 ft
100,000 ly  1 ft
1 AU
100,000  63,240 AU  1 ft
x
1 AU
x  6.324  109 ft
x
x  1.20  10 6 miles
If the AU were 1 foot in diameter, the Milky Way Galaxy would be 1.20 million miles across.
4.
If an AU were shrunk to 1 foot, what would be the diameter of the Earth?
2  6,378 km  
x
1 AU
1 ft
x  1 AU  2  6,378 km   1 ft
2  6,378 km  1 ft
x
1 AU

2  6,378 km   1 ft
x
1.5  108 km
x  8.5  10 5 ft
x  .001 inches
If the AU were 1 foot in diameter, the Earth would be 0.001 inch across.
5.
If a light year were shrunk to 1 foot, what would be the distance to the nearest large galaxy, M31?
2.5  10 6 ly
x

1 ly
1 ft
x  1 ly  2.5  10 6 ly  1 ft
x
2.5  10 6 ly  1 ft
1 ly
x  2.5  10 6 ft
x  473 miles
If the light year were 1 foot in diameter, the nearest large galaxy, M31, would be 473 miles away.
6.
If the Milky Way galaxy were shrunk to 1 foot in diameter, what would be the diameter of the observable Universe?
27.4  109 ly
x

100,000 ly
1 ft
x  100,000 ly  27.4  109 ly  1 ft
x
27.4  109 ly  1 ft
100,000 ly
x  27.4  10 4 ft
x  51.9 miles
If the Milky Way Galaxy were 1 foot in diameter, the diameter of the visible universe would be 51.9 miles.