PHYS-1500 PHYSICAL MODELING
FALL 2006
9/15/2006, Class 6: More Runge-Kutta Method
NAME ________ANSWERS______________
A FALLING MOON
Suppose the Moon could be stopped in its orbit. If so, it would fall to Earth. Answer the
following questions:
1. How much time would elapse from the time the Moon stopped until it struck Earth? Express
your answer in convenient units. For example, if the answer is between one minute and one
hour, express the answer in minutes, if between one hour and one day, express the answer in
hours, etc.
412180 s = 114.5 hr = 4.77 days
2. At what relative speed would the Moon and Earth collide?
About 10,000 m/s
Useful information:
24
Mass of Earth:
Mass of Moon:
ME = 5.98 ×10 kg
22
MM = 7.36 ×10 kg
Radius of Earth:
Radius of Moon:
RE = 6.37 ×10 m
6
RM = 1.74 ×10 m
6
8
Average Separation (Center to Center):
r = 3.82 ×10 m
In terms of the reduced mass  and the center to center distance x Newton’s second law can be
written:
Fa
ME M M
ME M M d 2x
G

M E  M M dt 2
x2
2
ME  M M
d x
2  G
dt
x2
1
This problem can be solved with the same program that was used to solve problem 4, on page 85
of the textbook, provided you set it up to model a system of two bodies of different masses. If
you did not, below is a picture of a spreadsheet that is set up for bodies of different masses. You
can find a copy of this spreadsheet in sheet 2 of an Excel file called r-k.xls at the course web site
under “Activities” in “Class 5”.
Once you have the program, you just have to enter the correct masses and initial separation.
2
LARGE ANGLE PENDULUM
A simple pendulum executes simple harmonic motion for small amplitudes. For larger
amplitudes, the mathematics get very complicated. However, the spreadsheet that you set up for
simple harmonic motion, can handle the large amplitude situation after a few modifications. The
picture below shows a suitably modified spreadsheet that describes a pendulum of length 1.0 m.
Set up a comparable spreadsheet. You can modify the spreadsheet that you constructed for class
5, or you can start with one that is available on line. It is sheet 1 of the r-k.xls file that is referred
to above.
The changes are in the constants in rows 4 and 5, and the acceleration term. The acceleration
term in cell I10 is highlighted and appears in the formula bar at the top of the picture. The key
change is to make the acceleration proportional to the sine of the displacement rather than the
displacement. For the settings given in the picture, the analytic solution and the numerical
solution should agree almost perfectly. As 0 is increased, the two solutions will begin to
disagree. However, it is the analytic solution that is no longer correct.
1. The period of the 1.0 m pendulum is about 2.0 s, for small amplitudes. Increase 0 until the
period of the pendulum is 10% larger than the 2.0 s predicted by the analytic solution (i.e. its
period will be 2.2 s). What is that value of 0?
1.2 to 1.3 rad
2. Sketch the numerical solution for 0 = 3.14.
3
4
3
Angle (rad)
2
1
x(anal)
0
-1
0
2
4
6
8
10
12
x
-2
-3
-4
t (s)
3. Sketch the numerical solution for 0 = 3.15. Briefly explain what the graph means.
12
10
Angle (rad)
8
6
x(anal)
4
x
2
0
-2
0
2
4
6
8
10
12
-4
t (s)
 d 
4. Now set 0 = 0, and set 
 = 0.1 rad/s. The two graphs should agree quite well. Then set
 dt 0
 d 

 = 6.26 rad/s. Sketch the graph below.
 dt 0
4
4
3
Angle (rad)
2
1
x(anal)
0
-1
0
2
4
6
8
10
12
x
-2
-3
-4
t (s)
 d 
5. Set 
 = 6.27 rad/s. Sketch the graph below. Briefly explain what the graph means.
 dt 0
25
Angle (rad)
20
15
x(anal)
x
10
5
0
0
2
4
6
8
t (s)
5
10
12