•Wednesday 4:00 – 4:55pm
•Sundquist Science Center E-109
•Dr. Spencer Buckner
•www.apsu.edu/astronomy
Office: SSC B – 326 (enter through B-332)
Hours: MWF 10:15 – 11:30am
and MWThF 1:30 – 2:30pm
or by appointment
Email: buckners@apsu.edu
Phone: 221-6241
21st Century Astronomy: The Solar System,
4th Edition, by Kay, Palen, Smith and Blumenthal
If you don’t have the book, don’t go buy it since it isn’t required.
Mostly we will just use the problems in the back of the chapters.
There are old editions in the department library you can borrow.
Exams…………45%
Homework…….30%
Projects………..20%
Participation…….5%
90 – 100…A
80 – 89…..B
70 – 79…..C
60 – 69…..D
<60……….F
There will be three one-hour exams during the
semester. The exams will be entirely problems similar
to the homework problems. A formula sheet will be
given out with the exam. Tentative exam dates are
•Wednesday September 23
•Wednesday October 28
•Wednesday December 9
A scientific calculator will be required for the exams
Homework will be assigned from the Student
Questions at the back of each chapter in the 21st
Century Astronomy textbook. Additional problems
from other sources will also be assigned to
supplement the back-of-chapter questions. They will
be due at the beginning of the next class meeting.
First Homework set is due next week: Chapter 1 #46
& 50 plus Supplemental Problems
Solutions to the homework will be posted in the
class D2L shell a day or two after they are turned in.
There will be two projects assigned during the
semester. The first project will be due September 30.
The second project will be due at the final exam
period Wednesday December 9. In addition to a
written report, you will make a short (10-15 minute)
oral presentation on your project.
• First Project List is posted in
Problems in Planetary Astro Stuff
on www.apsu.edu/astronomy and in
the Content section of the class
D2L website.
• NO REPEATS!
• First project presentations will be
Wednesday October 21
How to work a problem
•Step 1: What are you trying to solve for?
•Step 2: What information are you given?
•Step 3: What equation(s) do you need to
solve the problem?
•Step 4: Plug in the numbers and solve the
problem
•Step 5: Check for reasonableness
Most objects viewed through a telescope
have diameters measured in arcseconds. To
calculate the linear diameter of such an
object we can use the Small Angle
Approximation applied to the formula for
arclength (s = rq). The result is the formula
rq
S
where the number 206,265 comes
206, 265
from dividing 360° in arcseconds by 2p. If
Jupiter is observed to be 31.2 arcseconds
wide when it is at a distance of 944 million
kilometers, what is the actual diameter of the
planet?
Answer the first two questions
•Step 1: What are you trying to solve for?
What is the diameter of the planet
Jupiter in kilometers?
•Step 2: What information are you given?
Angular size of the planet (31.2”)
and distance from Earth to Jupiter
( 944 million kilometers)
Step 3: what equation(s) do you need?
Look for example problems in the chapter, in
the appendix, in other textbooks or online.
For this problem, it is given in the statement of
the problem
rq
s
206, 265
Step 4: Plug in numbers and solve
r = 944 x106 km
q = 31.2”
rq
s

206, 265
6
944

10
km   31.2"

206, 265
 142, 791km  143, 000km
Significant figures: the final answer should never have
more significant figures that the least number of
significant figures in the numbers you start with.
Step 5: check for reasonableness
In this case we can check the result with the
value in the back of the textbook which gives
the radius as 71,492 km
An Example for You
Under excellent seeing conditions a good sized
telescope on Earth can resolve angular details as
small as 1 arcsecond. What is the greatest
distance at which you could resolve a typical
person (1.7 meters)?
Example for You Solution
1. What are you trying to find?
Distance at which 1.7 meters subtends an
angle of 1”
2. What information is given?
Angle: 1”
Linear size: 1.7 m
3. What equation is needed?
rq
s
206, 265
First rearrange the equation
rq
206, 265s
s
 r 
206, 265
q
Next, plug and chug
r
206, 265s
q
206, 265 1.7 m

 350, 650  350km
1"