PHYS-1500 PHYSICAL MODELING
FALL 2006
10/20/2006Class 15a: Improved Orbital Motion
NAME _______KEY_____________________
For this exercise, you are to construct an improved spreadsheet model of an artificial satellite
orbiting Earth. The first one that was constructed was simple enough, and seemed to conserve
angular momentum. However, it did not produce truly circular orbits, and energy was conserved
only approximately. This time, conservation of energy and angular momentum will be built into
the program. That will improve the accuracy of the program, without adding a lot of computation
time. The most difficult part of the new program will be the use of polar coordinates. The notes
that have been provided outline how velocity and acceleration are described in polar coordinates.
In polar coordinates, the angular momentum and mechanical energy of a satellite in orbit can be
written,
GMm 1
GMm
L  mr 2
and
E  12 mv 2 
 2 m( r 2  r 2 2 ) 
.
r
r
Since the mass of the satellite is not specified, it is convenient to write the angular momentum
and mechanical energy per mass, both of which are conserved.
L
E 1 2 2  2 GM
 r 2
and
 2 ( r  r  ) 
m
m
r
If E/m, and L/m are known at any point in the orbit, their values will remain the same at all
L/ m
points. Then, the value of  can be found at any value of r from the equation,   2 . In
r
E 1 2
GM
 ( r  r 2 2 ) 
addition, at any value of r, r can be determined since,
, and
m 2
r
E GM 1 2

 2 ( r  r 2 2 ) . Then,
m
r
4 2
2
 E GM  2  2
 E GM  r 
 E GM  ( L/ m)
, and we get,
r2  2 

r


2



2


 m
 m
r 
r  r 2
r 
r2
m
 E GM  ( L / m)
r   2 

r 
r2
m
2
and
 
L/ m
.
r2
Therefore, at any value of r, the velocity is known. That simplifies the calculation of the orbit.
It almost appears that acceleration is not needed, but that is not true. Without an acceleration
term, if r ever became zero, it would never change again, so the program would fail. Therefore,
we need an equation for r . We can get that as follows,
2
GM
GM
GM
GM ( L / m) 2
 L / m
2
2




ar   2  r  r , so
r   2  r   2  r  2    2 
.
 r 
r
r
r
r
r3
When at a value of r, the program will calculate r and r at that point. Then r at the next point
will be estimated by assuming that r changes by an amount r t. The average of the two values
will be used to calculate the change of r. Then,  can be found at both values of r and the
average will be used to calculate the change of Finally, in order to plot the orbit, x and y will
have to be calculated from the equations, x = r cos  and y = r sin .
Below is a copy of a spreadsheet with those formulae entered. Below that are the contents of the
first several rows of the spreadsheet, for columns A through K. Set up your spreadsheet to match
the one below, and then copy row 11 to rows 12 through 211.
G6 =B5*G5
G7 =G5^2/2-B3*B4/B5
A10 0
A11 =A10+$G$3
B10
J3 =MAX(B10:B211)
J4 =MIN(B10:B211)
=B5
B11
J6 =SQRT(B3*B4/B5)
J7 =SQRT(2)*J6
=B10+(C10+E10)*$G$3/2
C10 =SQRT(2*($G$7+$B$3*$B$4/B10)-($G$6/B10)^2)
C11 =IF(2*($G$7+$B$3*$B$4/B11)-($G$6/B11)^2>0,SQRT(2*($G$7+$B$3*$B$4/B11)($G$6/B11)^2)*F10,0)
D10 =-$B$3*$B$4/B10^2+$G$6^2/B10^3
D11 =-$B$3*$B$4/B11^2+$G$6^2/B11^3
E10 =C10+D10*$G$3
E11 =C11+D11*$G$3
F10 =IF(E10<0,-1,1)
F11 =IF(E11<0,-1,1)
G10 0
G11 =G10+(H10+H11)*$G$3/2
H10 =$G$6/B10^2
H11 =$G$6/B11^2
I10 =B10*COS(G10)
I11 =B11*COS(G11)
K11 =AVERAGE(J3:J4)
K12 =-K10
When the spreadsheet is complete, create a graph of y vs. x.
J10 =B10*SIN(G10)
J11 =B11*SIN(G11)
For best performance, choose t so that no more than one orbit is plotted. The program may give
strange results if the satellite is allowed to go around its path too many times. The program starts
the satellite in an orbit a distance R0 from the center of Earth, on the positive x axis, with a
velocity of v0 in the positive y direction. Now repeat the check that you made on the last
program.
1. For R0 = 6.50 ×106 m = 6500 km, calculate the value of v0 that should produce a circular
orbit.
v0 = ___7833.52___m/s___
units
2. Enter the value you just calculated in the spread sheet. Does it produce a circular orbit? List
the maximum and minimum distances of the satellite from the center of Earth. This can be
done easily by using the MAX and MIN functions.
rmax = __6.50 ×106 __m _______
units
rmin = __6.50 ×106 __m ___
units
3. Now set R0 = 7.50 ×106 m = 7500 km, and repeat 1. and 2.
v0 = _7292.61__m/s_
units
rmax = ___7.50 ×106 __m ___
units
rmin = ____7.50 ×106 __m __
units
4. Now set R0 = 8.50 ×106 m = 8500 km, and repeat 1. and 2.
v0 = __6850.21__m/s___
units
rmax = ___8.50 ×106 __m ____
units
rmin = ___8.50 ×106 __m ___
units
Class 15b: Kepler’s Third Law
Check to see if the new program, in polar coordinates, obeys Kepler’s Third Law. For each of the
following starting conditions, find the semi-major axis and the period of the orbit. List them in
the table provided. Then, plot T² as a function of r³. The plot should be a straight line. Is it? Does
it have the correct slope?
(2 ) 2
In theory, the slope should be equal to
= 9.898 ×10-14 s²/m³.
GM E
t(s)
R0 (m)
v0 (m/s)
Semi-major axis (m)
Period (s)
66
7.00 ×106
9000
1.21 ×107 m
13240 s
33
9.00 ×106
6000
7.58 ×106 m
6560 s
50
9.00 ×106
7000
1.01 ×107 m
10050 s
102
9.00 ×106
8000
1.62 ×107 m
20480
200
1.00 ×107
8000
2.53 ×107 m
40000
Kepler's Third Law
1.80E+09
1.60E+09
y = 9.88E-14x - 2.95E+05
Period Squared
1.40E+09
1.20E+09
1.00E+09
8.00E+08
6.00E+08
4.00E+08
2.00E+08
0.00E+00
0
2E+21
4E+21
6E+21
8E+21
1E+22 1.2E+22 1.4E+22 1.6E+22 1.8E+22
Sem i-Major Axis Cubed
The graph is a straight line.
The slope is very close at 9.88 ×10-14 s²/m³.