Astronomy Projects For Calculus And Differential Equations

Astronomy Projects for Calculus and Differential Equations
Transit of Venus June 2012
Farshad Barman
Mathematics Department
Portland Community College
Rock Creek Campus
Fall 2012

2
Table of Contents
Introduction 3
Mathematical Introduction for Instructors 4
Notes for Assigning the Projects 14
The Projects 16
Martian Project for Calculus I 17
Mercury Project for Calculus II 23
Halley’s Comet Project for Calculus III 43
The Curiosity Mars Rover Project for Differential Equations 59
A Star Orbiting Sagittarius A* Black Hole Project for Differential Equations 69
Halley’s Comet Project for Differential Equations 78

3
Introduction
Portland Community College granted me a Professional Leave during the fall term of 2012.
Part of my work for this sabbatical was to write up the astronomy projects that I have been
working on, and have been giving to my students in calculus and differential equations, for
the last few years. This document is a collection of these projects, their solutions, and sample
student reports, written in such a way that any instructor who is interested can assign them to
his or her students.
I became interested in astronomy in 2004 while reading an article about the transit of Venus
in June of that year. While investigating the mathematics of this event, I realized that there is
a wealth of mathematical applications in astronomy that will benefit our students in calculus
and differential equations. I have also realized that unlike most other standard projects in
math textbooks, projects in astronomy, and the subject of astronomy in general, create quite a
bit more interest in students, and are a great motivational tool for a deeper appreciation of the
mathematical concepts.
I have rewritten these projects so that they will be self explanatory for students and
instructors. I have also included a mathematical introduction for the instructor who will be
assigning these projects. This introduction includes all the mathematics and all the
information that the instructor will need to feel comfortable when assigning these projects.
I would like to thank Portland Community College for giving me this opportunity.

4
Mathematical Introduction for Instructors
Spiral Galaxy M101

5
I will review here, very briefly, all the laws, mathematics, and background information
necessary for instructors who will assign these projects to their students.
You may choose to skip this introduction and go directly to the projects, since they are
self explanatory. You can come back and read this introduction if you need to.
Modeling orbital locations using Kepler’s Laws:
Johannes Kepler proposed his laws of planetary motion in 1609 and 1619. His laws,
which are true for any celestial object orbiting a much bigger celestial object, state that
(Figure 1):
1) A planet revolves around the Sun in an elliptic orbit with the Sun at one focus.
2) The line joining the Sun to a planet sweeps out equal areas in equal times.
3) The square of orbital period is directly proportional to the cube of semi-major axis.
Figure 1. Kepler’s second law says the shaded areas are equal
The second law says that the planet moves faster when it is closer to the Sun, and slower
when farther away, so that the two shaded areas in Figure 1 will be equal. We will not
use the third law explicitly in the projects.
The closest point of orbit around the Sun is called perihelion, while the farthest point is
called aphelion. These two points are called perigee and apogee for the Moon orbiting
the Earth, while they are called periapsis, and apoapsis in general (helio and geo are
Greek for sun and earth). We will let semi-major axis be a, and semi-minor axis be b. In
astronomy, distances in our solar system are given in astronomical units (AU) which is
the semi-major axis of the Earth (about 93 million miles).
Planet moves faster
Planet moves slower
Aphelion
Perihelion
x
y
a.e
2a
2b

6
The ellipse’s eccentricity, the measure of its elongation, is e and is given by:
2
2
1
a
b
e −
= .
This relationship can be solved for b to give:
2
1 e
a
b −
= .
Eccentricity is between 0 and 1. For a circular orbit 0
=
e , and for a very elongated orbit
e is close to 1.
The distance from the center of the ellipse to either focal point is e
a ⋅ . We will let the
planet be at perihelion at 0
=
t , and the orbital period be op in Earth years.
Here is the summary of the above information:
a is the semi-major axis in AU.
2
1 e
a
b −
= is the and the semi-minor axis in AU.
e is the eccentricity of the ellipse.
e
a ⋅ is the distance from the center of ellipse to either focal point in AU.
op is orbital period in years.
Planet at perihelion at 0
=
t .
The following table gives eccentricity, semi-major axis, and orbital period of several
planets and Halley ’s Comet, used in the following projects:
Planet or comet e a (AU) op (y)
Mercury 0.206 0.387 0.241
Earth 0.0167 1.0 1.0
Mars 0.0937 1.524 1.88
Halley’s Comet 0.97 19.34 76
With the coordinate system in Figure 1, and the planet at perihelion at 0
=
t , the
parametric equation for the x and the y coordinates of the planet is given by:
⋅
=
⋅
−
⋅
=
)
sin(
)
(
)
cos(
)
(
E
b
t
y
e
a
E
a
t
x
Elliptic orbit. (1)
The term e
a ⋅ shifts the orbital ellipse left to place the right focal point (where the Sun is)
at the origin. The variable E , called eccentric anomaly, is given by Kepler’s Equation:
)
sin(
2
E
e
E
op
t
⋅
−
=
π
.

7
This transcendental equation in E, comes from Kepler’s second law, and is not easy to
solve for E. Note that for a circular orbit 0
=
e , op
t
E π
2
= , a
b = , and equation (1)
becomes the more familiar equation of a circle with period op:
⋅
=
⋅
=
op
t
a
t
y
op
t
a
t
x
π
π
2
sin
)
(
2
cos
)
(
Circular orbit
Fortunately there is an explicit solution for Kepler’s Equation, found by Friedrich Bessel
in 1824, and fortunately for us, it involves plenty of opportunities to get students in
calculus to practice their skills in differentiation, integration, and power series with it.
Bessel’s solution of Kepler’s Equation is given by the following power series:
⋅
⋅
⋅
+
+
+
+
=
⋅
+
=
∞
=
op
t
e
J
op
t
e
J
op
t
e
J
op
t
op
t
n
n
e
n
J
op
t
E
n
n
π
π
π
π
π
π
6
sin
3
)
3
(
2
4
sin
2
)
2
(
2
2
sin
)
(
2
2
2
sin
)
(
2
2
3
2
1
1
, (2)
where )
(x
Jn is the Bessel function of order n. Bessel functions are transcendental
functions, and are themselves given by power series:
∞
=
+
+
+
−
=
0
2
2
2
)!
(
!
)
1
(
)
(
i
n
i
n
i
i
n
n
i
i
x
x
J .
Now if we expand the Bessel functions in equation (2) above, we get the following
expanded expression, which has been written without simplifying or reducing the
fractions so we can see the pattern.
⋅
⋅
⋅
+
⋅
⋅
⋅
−
⋅
⋅
+
⋅
⋅
−
⋅
⋅
+
⋅
⋅
⋅
−
⋅
⋅
+
⋅
⋅
−
⋅
⋅
+
⋅
⋅
⋅
−
⋅
⋅
+
⋅
⋅
−
⋅
⋅
+
=
op
t
e
e
e
op
t
e
e
e
op
t
e
e
e
op
t
E
π
π
π
π
6
sin
2
!
5
!
2
)
3
(
2
!
4
!
1
)
3
(
2
!
3
!
0
)
3
(
3
2
4
sin
2
!
4
!
2
)
2
(
2
!
3
!
1
)
2
(
2
!
2
!
0
)
2
(
2
2
2
sin
2
!
3
!
2
2
!
2
!
1
2
!
1
!
0
1
2
2
7
7
5
5
3
3
6
6
4
4
2
2
5
5
3
3
1
1
(3)

8
But fear not. There are simple approximations for eccentric anomaly, E, presented in
equation (3). First note that when 0
=
e , equation (3) reduces to
op
t
E
π
2
= , as mentioned
above. When eccentricity, e, is small ( 21
.
0
≤
e ), which is true for all the planets in our
solar system, we need only the first term of the series above, which is e
e
=
⋅
⋅
⋅ 1
1
2
!
1
!
0
1
2
, to
keep error in orbital position to less that about 4.1 %. For a little more accuracy, the next
significant term is
2
2
!
2
!
0
)
2
(
2
2 2
2
2
e
e
=
⋅
⋅
⋅ . When eccentricity is large, such as that of Halley’s
Comet ( 97
.
0
=
e ), we need about 50 terms of the series in equation (2).
Here are the approximations for E. The first two and the last are used in the following
projects.
2
sin
)
(
2
2
4
sin
2
2
sin
2
2
sin
2
2
50
1
2
⋅
+
≅
+
⋅
+
≅
⋅
+
≅
≅
= op
t
n
n
e
n
J
op
t
E
op
t
e
op
t
e
op
t
E
op
t
e
op
t
E
op
t
E
n
n π
π
π
π
π
π
π
π
Totally ignores Kepler’s Second Law
For planets with 21
.
0
≤
e (all the planets
in our solar system)
A little more accurate than above
For Halley’s Comet ( 97
.
0
=
e )
We will plug in the first and the second approximations above for E in equation (1) to get:
⋅
≅
⋅
−
⋅
≅
op
t
b
t
y
e
a
op
t
a
t
x
π
π
2
sin
)
(
2
cos
)
(
(4)
⋅
+
⋅
≅
⋅
−
⋅
+
⋅
≅
op
t
e
op
t
b
t
y
e
a
op
t
e
op
t
a
t
x
π
π
π
π
2
sin
2
sin
)
(
2
sin
2
cos
)
(
(5)

9
Parametric equations (4) and (5) are the equations used in the following projects that will
follow for Calculus I (The Martian Project) and Calculus II (The Mercury Project).
Differentiating these equations to find orbital velocity will involve the chain rule, while
integrating to find area swept will involve integration using substitution.
If we plug in the last approximation for E in equation (1), we get:
⋅
+
⋅
≅
⋅
−
⋅
+
⋅
≅
=
=
2
sin
)
(
2
2
sin
)
(
2
sin
)
(
2
2
cos
)
(
50
1
50
1
op
t
n
n
e
n
J
op
t
b
t
y
e
a
op
t
n
n
e
n
J
op
t
a
t
x
n
n
n
n
π
π
π
π
(6)
The approximation in equation (6) is good for orbits with large eccentricities (when e is
close to 1), such as Halley’s Comet in the project for Calculus III.
Figure 2 below shows the significance and the difference between the arguments op
t
π
2
and E, for circular and elliptic orbits, respectively. The argument op
t
π
2 is called mean
anomaly, while E, as mentioned above, is called eccentric anomaly. The figure on the
left shows a circular orbit with the angle swept by the line connecting the Sun to the
planet progressing linearly with time, while the figure on the right shows this angle for an
elliptic orbit progressing non-linearly with time to account for Kepler’s second law,
slowing the planet as it moves away from the Sun and speeding it up as it gets closer.
Figure 2. The angle swept by the line connecting the Sun to a planet for a circular and
an elliptic orbit
op
t
E
π
2
=
∞
=
+
=
1
2
sin
)
(
2
2
n
n
op
t
n
n
ne
J
op
t
E
π
π

10
Modeling orbital locations using Newton’s Laws:
Students in Differential Equations will solve two of Newton’s laws directly without using
Kepler’s Laws to find the location of planets, with more accuracy.
Newton’s second law of motion, written in vector form is:
a
m
F = , (7)
where F is force in Newton’s (N), and a is acceleration in meters per seconds squared
( 2
/ s
m ). Newton’s law of gravitational force is:
2
r
GMm
F −
= ,
where
kg
s
m
G
⋅
×
= −
2
3
11
10
67
.
6 , is the Universal Gravitational Constant, M is the mass of
the larger celestial object in kilograms (kg), m is the mass of the smaller celestial object
in kilograms, and r is the distance between the celestial objects in meters (m). This
equation in vector form is:
r
r
r
GMm
F ⋅
−
= 2
, (8)
where r is the vector connecting the larger celestial object to the smaller, in meters, and
2
r
r is the unit vector in that direction.
We will equate the two forces in (7) and (8), and will cancel the common m factor:
r
r
r
GM
a ⋅
−
= 2
. (9)
Now if the larger celestial object (the Sun) is fixed at the origin, and the smaller (planet,
comet, or spacecraft) is orbiting it, and its coordinates at time t are )
(t
x and )
(t
y , then
for the orbiting object j
t
y
i
t
x
r )
(
)
( +
= , and j
t
y
i
t
x
a )
(
)
( ′
′
+
′
′
= . We will plug these
into equation (9) to get:
( )
( )3
2
2
)
(
)
(
)
(
)
(
)
(
)
(
t
y
t
x
j
t
y
i
t
x
GM
j
t
y
i
t
x
+
+
−
=
′
′
+
′
′ .
We will equate the x and the y components of the vectors on the left and the right to get:

11
( )
( )
+
−
=
′
′
+
−
=
′
′
3
2
2
3
2
2
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
t
y
t
x
t
GMy
t
y
t
y
t
x
t
GMx
t
x
.
This is a system of non linear second-order differential equations in two unknowns, )
(t
x
and )
(t
y , that we will solve numerically using a computer algebra system (MAPLE).
Before doing that, however, we will need to turn this into a system of four first-order
differential equations in four unknowns. Let )
(
)
( t
x
t
vx
′
= , and )
(
)
( t
y
t
vy
′
= . We will
also need four initial conditions, one for each unknown variable. Let
0
0
0 )
0
(
,
)
0
(
,
)
0
( x
x v
v
y
y
x
x =
=
= , and yo
y v
v =
)
0
( to get:
( )
( )
=
+
−
=
′
=
=
′
=
+
−
=
′
=
=
′
0
3
2
2
0
0
3
2
2
0
)
0
(
,
)
(
)
(
)
(
)
(
)
0
(
),
(
)
(
)
0
(
,
)
(
)
(
)
(
)
(
)
0
(
),
(
)
(
y
y
y
y
x
x
x
x
v
v
t
y
t
x
t
GMy
t
v
y
y
t
v
t
y
v
v
t
y
t
x
t
GMx
t
v
x
x
t
v
t
x
(10)
The system of equations (10) is the one used to model the orbits of the Earth, Mars, the
Rover Curiosity spacecraft, the star S0-2 orbiting the black hole at the center of our
Milky Way galaxy, and Halley’s Comet.
If you are interested in changing the projects to include other planets and/or different
dates, the initial conditions for all the planets, and other celestial objects, at a given time
can be found on JPL’s HORIZONS system website:
http://ssd.jpl.nasa.gov/horizons.cgi
In order to get the xy coordinates and velocities, the settings should be:
Current Settings
Ephemeris Type [change] : VECTORS
Target Body [change] : Mars [499]
Coordinate Origin [change] : Solar System Barycenter (SSB) [500@0]
Time Span [change] : Start=2011-11-26, Stop=2011-11-30, Step=1 d
Table Settings [change] : defaults
Display/Output [change] : default (formatted HTML)

12
A short review of the heliocentric coordinate system:
In most of the projects that follow we will place the x-axis along the major axis of the
elliptic orbit, with the Sun at the right focus, and perihelion on the positive x-axis. This
makes the problem simple without loss of generality. In the project Curiosity’s Orbit to
Mars, however, we are dealing with two planets, and a spacecraft traveling from one to
the other, and need to use the standard astronomical coordinate system called the
heliocentric coordinate system. Here is a short review of this coordinate system.
The three dimensional solar coordinate system, called the heliocentric coordinate system
is shown in Figure 3. The Sun is at the origin, and the xy-plane is the plane of Earth's
orbit. The Earth orbits the Sun, and rotates about its axis counterclockwise as seen from
the positive z-axis.
The Earth's rotation axis (north-south pole line) is in the yz-plane, tilted from the z-axis
by about 23 toward the positive y-axis. The Earth is on the positive y-axis on Winter
Solstice, when the North Pole is tilted away form the Sun (approx. Dec. 21st
), and on the
negative y-axis on Summer Solstice when the North Pole is tilted into the Sun (approx.
June 21st
). The Earth is on the x-axis on the Equinoxes.
Figure 3. The heliocentric coordinate system
Top view of the orbits of Earth and Mars are shown in Figure 4. The Earth's perihelion,
measured clockwise from the positive x-axis, is at about 103 (approx. Jan. 3rd
). Mars's
perihelion is at about 336 .
z
x
Autumn Equinox Winter Solstice
y
Spring Equinox
Summer Solstice

13
Figure 4. Top view of the orbits of Earth and Mars
Mars’s orbital plane is tilted relative the xy-plane (Earth’s orbital plane) by less than 2 .
We will, however, ignore this small tilt, and treat the problem in Curiosity’s Orbit project
as a two-dimensional problem.
x
y
Mars's orbit
Earth's orbit
Mar's Perihelion
at about 336
Earth's Perihelion
at about 103

14
Notes for assigning the projects
These projects have been written such that if a student is new to Calculus II or III or
Differential Equations, and has not done the projects in previous classes, he or she will be
comfortable with the new project in astronomy. There is, therefore, some overlap and
repetition in the first few steps or problems in these projects.
The Martian Project is for students in Calculus I (differential calculus), and is to be
handed out early in the term. Students can read and start working on the project early in
the term, but they would need to know the chain rule as applied to trigonometric
functions when they get to step 2). The project report will be due toward the end of the
term.
The Mercury Project is for students in Calculus II (integral calculus), and is broken up
into eight weekly problems, m0 through m7, to be collected, graded and handed back.
Students will collect these problems, and will write a project report toward the end of the
term. Since each m problem builds on the previous ones, they should be corrected and
handed back promptly. Problems m0 and m1 are very simple, but help students review
parametric equations. Problem m5 needs integration using substitution, and should be
handed out when students have learned that skill. Here is a suggested due date for each
problem for a ten-week term:
Problem Week of class problem due Skills needed to solve the H problem
m0 Second
m1 Third
m2 Fourth
m3 Fifth
m4 Sixth
m5 Seventh Integration by substitution
m6 eighth
m7 ninth
Project Due tenth
The Halley’s Comet Project is for Calculus III (sequences and series), and similar to the
Mercury project, is broken up into five weekly problems called the H problems, to be
collected, graded and handed back. Students will collect these problems, and will write a
project report toward the end of the term. Since each H problem builds on the previous
ones, they should be corrected and handed back promptly. The due date for each problem
should be set by the instructor to insure that the students have learned the topic necessary
for solving the problem. Here is a suggested due date for each problem for a ten-week
term (if series solution to differential equations is not covered, H5 can be skipped):

15
Problem Week of class problem due Skills needed to solve the H problem
H1 Second
H2 Third
H3 Fourth Power Series, Taylor Polynomials
H4 Fifth
H5 Sixth Series solution of Differential
Equations
The three projects, Curiosity’s Path to Mars, A Star Orbiting Sgr A*, and Halley’s
Comet, are for students in Differential Equations. They are all based on modeling the
orbits using Newton’s Laws. The instructor can choose one, and hand it out early in the
term. Students can start working on the project as soon as they have learned system of
differential equations, and converting a second-order differential equation into a system
of two first-order equations. They should also be familiar with solving non-linear
differential equations with a computer algebra system. The project report will be due
toward the end of the term.

16
The Projects
The Cassini spacecraft takes an angled view toward Saturn’s South Pole showing the
rings and the planet casting shadows on each other

17
Martian Project
Calculus I
Spirit and Opportunity Mars rovers send pictures home from Mars
On Christmas Day, 1642, the year Galileo died, there was born a male infant so tiny that,
as his mother told him in later years, he might have been put into a quart mug, and so
frail that he had to wear a bolster around his neck to support his head. This unfortunate
creature was entered in the parish register as “Isaac sonne of Isaac and Hanna Newton”.
There is no record that the wise men honored the occasion, yet this child was to alter the
thought and habit of the world.
James Newman

18
Introduction:
Johann Kepler in 1609 discovered that planets orbit the Sun in elliptic orbits, and that
their orbital velocity is not constant but varies. The following summarizes Kepler’s first
two laws (see the Figure at the end of this handout):
1) The planets orbit the Sun in Elliptic orbits with the Sun at one of the focal points.
2) The line joining the Sun to a planet sweeps out equal areas in equal times.
His second law, simply said, means that planets slow down when they are farther from
the Sun, and speed up when they are closer. Since the line joining the Sun to the planet is
shorter when the planet is closer, the length of the orbit covered by the planet in a given
interval of time would be larger to make the areas swept equal.
Kepler did not have the physics or the mathematical tools to prove his own discovery,
and it was left for the genius of Sir Isaac Newton to do that, in 1665, using his second law
of motion ( F = ma). The 23-year old was a student at the University of Cambridge when
an outbreak of the Plague forced the university to close down for 2 years. Those two
years were to be the most creative in Newton’s life. He conceived the law of gravitation,
the laws of motion, differential calculus, and the proof of Kepler’s Laws.
Mathematics of Orbits:
An ellipse is described by the length of the semi-major axis a , and the length of the
semi-minor axis b ( refer to the Refresher on Parametric Equations sheet at the end of this
handout). The ellipse’s eccentricity, the measure of its elongation, is e and is given by:
2
2
1
a
b
e −
= .
This relationship can be solved for b to give:
2
1 e
a
b −
= .
Eccentricity is between 0 and 1. For a circular orbit 0
=
e , and for a very elongated orbit
e is close to 1. The distance from the center of the ellipse to either focal point is e
a ⋅ .
Note that when b
a = , we have e = 0, and the ellipse is a circle. Our planets have
eccentricities of 0.009 (Neptune) to 0.206 (Mercury).
The point of the orbit closest to the Sun is called perihelion, and the point farthest is
called aphelion. To simplify the calculations for this project, without loss of generality,
we will place the origin at the focal point where the Sun resides, the x-axis along the
major axis. The center of the ellipse is then at )
0
,
( e
a ⋅
− . We will also let time 0
=
t

19
when and the planet is at perihelion. With these assumptions, the parametric equations of
the orbit of a planet are:
⋅
=
⋅
−
⋅
=
)
2
sin(
)
(
)
2
cos(
)
(
op
t
b
t
y
e
a
op
t
a
t
x
π
π
or:
⋅
−
=
⋅
−
⋅
=
)
2
sin(
1
)
(
)
2
cos(
)
(
2
op
t
e
a
t
y
e
a
op
t
a
t
x
π
π
(1)
Where op is the orbital period in Earth years.
Note that when 0
=
e , the above equations turns into the parametric equations of a circle
with center at the origin and radius equal to a.
Although equation (1) models the shape of the orbit correctly, it does not account for
Kepler’s second law (In fact it has total disregard for orbital velocity). To account for
that, we can add a term to the arguments of the cosine and sine functions. This is an
approximation to an otherwise difficult problem, but is a very good one for 2
.
0
<
e :
⋅
+
−
=
⋅
−
⋅
+
⋅
=
)
2
sin(
2
sin
1
)
(
)
2
sin(
2
cos
)
(
2
op
t
e
op
t
e
a
t
y
e
a
op
t
e
op
t
a
t
x
π
π
π
π
(2)
Equations (1) and (2) give the position of a planet as a function of time in years. The x-
and y-components of orbital velocity are given by:
=
=
dt
t
y
d
t
v
dt
t
x
d
t
v
y
x
)
(
)
(
)
(
)
(
(3)
And finally the orbital velocity as a function of time is given by Pythagoras’s Theorem.
)
(
)
(
)
( 2
2
t
v
t
v
t
v y
x +
= . (4)
The orbital constants for Mars are given in the following table:
Semi-major axis in (AU) Eccentricity Orbital Period (years)
a e op
1.524 0.0934 1.88
AU is an Astronomical Unit, which is Earth’s semi-major axis (the mean distance from
the Sun to Earth), and is about 93 million miles.

20
The Project:
Your task in this project is to calculate the location and the orbital velocity of Mars for
the simple (and inaccurate) model given by equation (1), and the better approximation
model given by equation (2). You will make a table and plot the orbital velocity for 1.88
year (one Martian year) for the two models and will compare them. Use three decimal
places in all your numerical results. Here are the steps you can take to arrive at the
result:
A) The simple model:
0) Calculate the average orbital velocity of Mars by noting that Mars travels the
circumference of its elliptic orbit in 1.88 year. The following is a simple approximate
equation for circumference of an ellipse (there is no simple exact formula):
2
2
2
2
b
a
C
+
≅ π
Average orbital velocity is then this distance C divided by time op for Mars. The units
will be AU/y. All your calculations for the instantaneous velocity in the following steps
should orbit this average velocity.
1) Find the x and y locations of Mars for time increments of 0.188 year from 0
=
t to
88
.
1
=
t for the simple model of equation (1). You should have 11 points. Make a graph
of the elliptic orbit and indicate the locations of Mars for the 11 time calculations with
times indicated on each point.
2) Find )
(t
vx and )
(t
vy for model (1) in terms of op
e
a and
,
, . Do not plug in numerical
constants at this time. You should find the derivatives by hand using the derivative
rules we have learned in this class. Write a statement here for each step describing how
you found the derivative by using the derivative rules. For example:
[ ]
3
2
)
( qx
kx
e
d
e
c
dx
d
x
g ⋅
+
⋅
=
[ ] [ ]
3
2
qx
kx
e
d
dx
d
e
c
dx
d
⋅
+
⋅
= derivative of sum rule
[ ] [ ]
3
2 3
2
qx
dx
d
e
d
kx
dx
d
e
c qx
kx
⋅
+
⋅
⋅
= multiplicative constant, and chain rules
⋅
⋅
⋅
=
3) Find )
(t
v for model (1) using equation (4) in terms of the constants op
e
a and
,
, . You
should be able to simplify this expression greatly using trigonometric identities.
4) Plug in values for the constants op
e
a and
,
, in )
(t
v and find numerical values for
orbital velocity for the time increments mentioned in step 1). Tabulate and graph this
function.

21
B) The more accurate model:
5) Find the x and y locations of Mars for time increments of 0.188 year from 0
=
t to
88
.
1
=
t for the more accurate model of equation (2). You should have 11 points. Make
a graph of the elliptic orbit and indicate the locations of Mars for the 11 time calculations
with times indicated on each point.
6) Find )
(t
vx and )
(t
vy for model (2) in terms of op
e
a and
,
, . Do not plug in numerical
constants at this time. You should find the derivatives by hand using the derivative
rules we have learned in this class. Write a one line statement here for each step
describing how you found the derivative by using the derivative rules as in step 2) above.
7) The expressions in step 6) will be too complicated to find )
(t
v for this model as we
did in step 3). To find )
(t
v for model 2), plug the constants op
e
a and
,
, into equations
for )
(t
vx and )
(t
vy , and find numerical values for each velocity component with the
same time increments as in step 1), and then find velocity using equation (4).
8) Calculate the orbital velocity now by using equation 4) for every time data point you
have for the components of velocity in step 6). Tabulate and graph this function.
Your Report
Present all the mathematics and the calculations for both models.
Present the locations and the orbital velocities for each model in separate tables. Each
table should have four columns (for t, x, y, and v).
Make a graph of orbital velocity as a function of time for each model. Choose a scale
that will show the differences in velocities well.
Finally make separate graphs of the elliptic orbit and indicate the locations of Mars for
the 11 time calculations with time and orbital velocity indicated for each point. Make
sure that the graphs are large enough to cover one whole graph paper each.
Your report should then have two tables with 4 columns each, two ellipses with location
of Mars and its velocity indicated on these points, and two graphs of velocity vs. time.
Your report should be complete and easy to understand by a mathematician who
has not seen this handout and has not been to our class.
Your report should include:
I) A cover sheet.
II) A short and complete statement of the problem in your own words. Do not attach any
part of this handout to your report.
III) All your calculations.
IV) All the graphs and tables.
V) A short conclusion of what this project has contributed to your cosmic consciousness.

22
Refresher on Parametric Equations of Conic Sections:
Parametric equation of a circle a
r = center
at (0,0), period π
2 :
=
=
)
sin(
)
(
)
cos(
)
(
t
a
t
y
t
a
t
x
Parametric equation of an ellipse, major
axis 2a, minor axis 2b, center at (0,0),
period π
2 :
=
=
)
sin(
)
(
)
cos(
)
(
t
b
t
y
t
a
t
x
As above, but shift center to )
,
( k
h :
+
=
+
=
k
t
a
t
y
h
t
a
t
x
)
sin(
)
(
)
cos(
)
(
,
( k
h :
+
=
+
=
k
t
b
t
y
h
t
a
t
x
)
sin(
)
(
)
cos(
)
(
As above, but change period to B
+
=
+
=
k
B
t
a
t
y
h
B
t
a
t
x
)
2
sin(
)
(
)
2
cos(
)
(
π
π
+
=
+
=
k
B
t
b
t
y
h
B
t
a
t
x
)
2
sin(
)
(
)
2
cos(
)
(
π
π
axis a
2 , minor axis b
2 , eccentricity e ,
center at )
0
,
( e
a ⋅
−
−
=
⋅
−
=
)
2
sin(
1
)
(
)
2
cos(
)
(
2
B
t
e
a
t
y
e
a
B
t
a
t
x
π
π
y
2b
2a
a.e
x
Aphelion
Perihelion
Planet moves slower
Planet moves faster

23
The Mercury Project
Calculus II
Einstein’s theory of general relativity showed why Mercury’s perihelion shifts very
slowly around the sun. This was a powerful factor motivating the adoption of general
relativity.
This term we will study the orbit of Mercury, its position as a function of time, and
Kepler’s Second Law of planetary motion. I will hand you weekly problems, which I call
m problems. You will hand these problems back to me, they will be graded, and handed
back to you. You will collect these problems and will summarize the results at the end of
the term in a project report.

24
Problem m0
two laws (See Figure):
2) The line joining the Sun to a planet sweeps out equal areas in equal time.
the Sun, and speed up when they are closer.
The ellipse’s semi-major axis is a, while the semi-minor axis is b. The eccentricity, the
measure of its elongation, is e and is given by 2
2
1 a
b
e −
= , which can be solved for b
to give 2
1 e
a
b −
= . Eccentricity is between 0 and 1. For a circular orbit 0
=
e , and for
a very elongated orbit e is close to 1. The distance from the center of the ellipse to either
focal point is e
a ⋅ .
Orbital values for the planet Mercury are:
years
241
.
0
,
206
.
0
,
AU
1
,
AU
387
.
0 2
=
=
−
=
=
op
e
e
a
b
a
AU is astronomical unit, which is the average distance from the Sun the Earth.
The area of an Ellipse is given by
ab
A π
= .
y
2b
2a
a.e
x
Aphelion
Perihelion
Planet moves slower
Planet moves faster

25
In this, and all the subsequent m problems, please round your answers to four decimal
places, unless otherwise mentioned, and include units for the results, where applicable.
Find the area of the orbital ellipse of Mercury:
A =……………………………………………
Every 1/20th
of the orbital period (op/20), the line from the Sun to Mercury sweeps
exactly 1/20th
of the area A you found above. This is true regardless of where Mercury is.
Fill in the table the areas swept by the line from the Sun to Mercury. These should be all
the same numbers, and equal to 1/20th
of the area you found above.
Time interval Exact Area swept in 1/20th
of op
20
11
to
20
10
20
5
to
20
4
.
20
to
0
op
t
op
t
op
t
op
t
op
t
t
=
=
=
=
=
= ………………………
……………………….
………………………
Note: In order to make calculations in the m problems easier with your calculator or
MAPLE program, it is essential to store the formulas with variable names, and then store
all the numerical values into variable names, before you attempt to evaluate the formulas
in these problems.
Here is an example you will see in m5. Calculate −
−
=
op
op
ab
A
πβ
α
β
π 4
sin
)
(
4
4
1
for 20
/
,
0 op
=
= β
α and also for 20
/
5
,
20
/
4 op
op =
= β
α :
> A1:=abs(a*b/4*(4*Pi*(bet-alp)/op1-sin(4*Pi*bet/op1)));
:=
A1
1
4
a b −
4
π ( )
−
bet alp
op1
sin 4
π bet
op1
> a:=0.387; e:=0.206; b:=a*sqrt(1-e^2); op1:=0.241; alp:=0; bet:=op1/20;
area:=evalf(A1);
:=
a 0.3870 :=
e 0.2060 :=
b 0.3787 :=
op1 0.2410 :=
alp 0 :=
bet 0.0121
:=
area 0.0015
> alp:=4*op1/20;bet:=5*op1/20;
area:=evalf(A1);
:=
alp 0.0482 :=
bet 0.0603
:=
area 0.0230

26
Problem m1
Refer to the back of this m1 handout for a refresher on parametric equations of conic
sections.
a) Write the implicit equation of a circle with radius a centered at the origin.
…………………………………………………….
b) Write the parametric equation of a circle with radius a centered at the origin with
parameter t , and a period of π
2 . Your answer will involve sine and cosine functions.
c) Write the parametric equation of a circular orbit with radius a centered at )
,
( k
h with
parameter t , and a period of π
2 . The planet’s position at 0
=
t should be at )
,
( k
a
h +
d) Find the location of this planet, in exact form, at:
......
..........
..........
..........
..........
..........
..........
:
2
......
..........
..........
..........
..........
..........
..........
:
2
/
3
......
..........
..........
..........
..........
..........
..........
:
......
..........
..........
..........
..........
..........
..........
:
2
/
......
..........
..........
..........
..........
..........
..........
:
4
/
......
..........
..........
..........
..........
..........
..........
:
0
π
π
π
π
π
=
=
=
=
=
=
t
t
t
t
t
t

27
e) Write the parametric equation of a circular orbit with radius a centered at the origin
with parameter t , and an orbital period of op . Your answer will involve sine and
cosine functions.
f) Write the parametric equation of an elliptic orbit with major axis a
2 along the x-axis,
minor axis b
2 along the y-axis. The ellipse is centered at )
0
,
0
( with parameter t , and
an orbital period op . The planet’s position at 0
=
t should be at )
0
,
(a
g) Shift the ellipse in f) left so that the origin is at the right focal point. Note that the
distance from center to each focal point is e
a⋅ , where e is the eccentricity of the ellipse.
Write the equation for this orbit. Your equations should be in terms of e
b
a ,
, and op :
h) The orbit of Mercury has the following values:
years
241
.
0
206
.
0
AU
1
,
AU
387
.
0 2
=
=
−
=
=
op
e
e
a
b
a
AU is an “astronomical unit” which is the average distance from the Sun to the Earth ( a
for Earth). If Mercury is at perihelion at 0
=
t , find the location of this planet at the
given times below. Put your answer in ordered pairs. Perihelion is when the planet is
closest to the Sun (for our problem this is )
0
,
( e
a
a ⋅
− )

28
......
..........
..........
..........
..........
..........
..........
:
20
11
......
..........
..........
..........
..........
..........
..........
:
20
10
......
..........
..........
..........
..........
..........
..........
:
20
5
......
..........
..........
..........
..........
..........
..........
:
20
4
......
..........
..........
..........
..........
..........
..........
:
20
......
..........
..........
..........
..........
..........
..........
:
0
op
t
op
t
op
t
op
t
op
t
t
=
=
=
=
=
=
i) Graph the elliptic orbit and locate the above locations on your graph, and attach your
graph. Use a graphing software such as GRAPH, WINPLOT, or MAPLE, with a window
of AU
AU 5
.
0
to
5
.
0
− in both directions, and a scale of AU
1
.
0 . Connect the origin
to the above points and shade the three slices, one from
20
to
0
op
t
t =
= , one from
20
5
to
20
4 op
t
op
t =
= , and one from
20
11
to
20
10 op
t
op
t =
= .
This is an example of how you can plot the orbit of a planet and place the planet's positions on the orbit using MAPLE.
For this example a=1.5 AU, b=1.2 AU, op=3 years, e=0.6. The two locations were calculated for t = 0.15 year and
t = 0.25 year.
> with(plots):
> f:=t->a*cos(2*Pi*t/op1)-a*e; g:=t->b*sin(2*Pi*t/op1);
:=
f →
t −
a cos 2
π t
op1
a e
:=
g →
t b sin 2
π t
op1
> a:=1.5: b:=1.2: e:=0.6: op1:=3:
> p1:=plot([f(t),g(t),t=0..3],x=-3..3,y=-2..2,scaling=CONSTRAINED,
xtickmarks=[-1,1],ytickmarks=[-1,1]):
p2:=pointplot({[f(.15),g(.15)],[f(.25),g(.25)]},symbol=CIRCLE,
color=black,scaling=CONSTRAINED):
display({p1,p2});

29
r = ,
center at (0,0), period π
2 :
=
=
)
sin(
)
(
)
cos(
)
(
t
a
t
y
t
a
t
x
period π
2 :
=
=
)
sin(
)
(
)
cos(
)
(
t
b
t
y
t
a
t
x
,
( k
h :
+
=
+
=
k
t
a
t
y
h
t
a
t
x
)
sin(
)
(
)
cos(
)
(
,
( k
h :
+
=
+
=
k
t
b
t
y
h
t
a
t
x
)
sin(
)
(
)
cos(
)
(
+
=
+
=
k
B
t
a
t
y
h
B
t
a
t
x
)
2
sin(
)
(
)
2
cos(
)
(
π
π
+
=
+
=
k
B
t
b
t
y
h
B
t
a
t
x
)
2
sin(
)
(
)
2
cos(
)
(
π
π
axis a
2 , minor axis b
center at )
0
,
( e
a ⋅
−
−
=
⋅
−
=
)
2
sin(
1
)
(
)
2
cos(
)
(
2
B
t
e
a
t
y
e
a
B
t
a
t
x
π
π
y
2b
2a
a.e
x
Aphelion
Perihelion
Planet moves slower
Planet moves faster

30
Problem m2
The following figure shows the orbit of a planet around the Sun. The point of the orbit
closest to the Sun is called perihelion, and the point farthest is called aphelion. To
simplify the calculations for this problem, without loss of generality, we will place the
origin at the focal point where the Sun resides, the x-axis along the major axis. The length
of the major axis is a
2 , and that of the minor axis is b
2 . The center of the ellipse is then
at )
0
,
( e
a ⋅
− . We will also let time t equal zero when and the planet is at perihelion.
With these assumptions, the parametric equations of the orbit of a planet are:
=
⋅
−
=
)
2
sin(
)
(
)
2
cos(
)
(
op
t
b
t
y
e
a
op
t
a
t
x
π
π
(1)
Where op is the orbital period in Earth years, and 2
1 e
a
b −
= .
Note that when 0
=
e , then a
b = , and the above equations turns into the parametric
equations of a circle with center at the origin.
Equation (1) does not account for Kepler’s second law (In fact it has total disregard for
orbital velocity). To account for that, we can add a term to the arguments of the cosine
and sine functions. This is an approximation to an otherwise difficult problem, but is a
very good one for 21
.
0
≤
e :
y
2b
2a
a.e
x
Aphelion
Perihelion
Planet moves slower
Planet moves faster

31
⋅
+
⋅
=
⋅
−
⋅
+
⋅
=
)
2
sin(
2
sin
)
(
)
2
sin(
2
cos
)
(
op
t
e
op
t
b
t
y
e
a
op
t
e
op
t
a
t
x
π
π
π
π
(2)
Equations (1) and (2) (Models 1 and 2) give the position of a planet as a function of time
in years. The values of op
e
b
a and
,
,
, for Mercury are given in problem m1.
Find the locations for Mercury for the following times for the two models above. You
calculated the first model’s locations in problem m1, and can just copy them here. Refer
to the Note in m0 to make your calculations easier.
Model 1 Model 2
........
..........
..........
..........
..........
:
20
11
........
..........
..........
..........
..........
:
20
10
........
..........
..........
..........
..........
:
20
5
........
..........
..........
..........
..........
:
20
4
.........
..........
..........
..........
..........
:
20
.......
..........
..........
..........
..........
:
0
op
t
op
t
op
t
op
t
op
t
t
=
=
=
=
=
=
........
..........
..........
..........
..........
:
20
11
........
..........
..........
..........
..........
:
20
10
........
..........
..........
..........
..........
:
20
5
........
..........
..........
..........
..........
:
20
4
.........
..........
..........
..........
..........
:
20
.......
..........
..........
..........
..........
:
0
op
t
op
t
op
t
op
t
op
t
t
=
=
=
=
=
=
Graph the elliptic orbit and locate the planet locations for model 2, as you did for model 1
in problem m1, with the same viewing window and scales. Connect the Sun to the above
points and shade the three slices, one from
20
to
0
op
t
t =
= , one from
20
5
to
20
4 op
t
op
t =
= , and one from
20
11
to
20
10 op
t
op
t =
= .

32
Problem m3
In problems m3 through m6 we will work on finding the area swept by a line connecting
the Sun to a planet using geometry and integral calculus. The graph in Fig. 1 is given by
the parametric equation:
=
=
)
(
)
(
t
g
y
t
f
x
Fig. 1 Fig 2
Fig 3 Fig 4
a) Find the area OCA in Fig 2 in terms of g
f , and α only.
Area OCA = …………………………………………....
D C
B
A
O
β
=
t
B
D
O
α
=
t
O
A
C

33
b) Find the area of the triangle ODB in Fig. 3 in terms of g
f , and β .
Area ODB = …..…………………………………………
c) If the area DCAB in Fig. 4 is A1, find the area A of the slice OAB in terms of
β
α,
,
, g
f and A1 (think of adding and subtracting areas of triangles to A1).
Area OAB = …………………………………………….………………………. Eq. (1)
The above equation gives the area swept by a line connecting the Sun to a planet, if the
functions )
(t
f and )
(t
g are the parametric equations for the orbit of that planet. We will
program this equation for Mercury to find areas swept in problem m4.

34
Problem m4
A planet’s elliptic orbit has major axis a
2 along the x-axis, minor axis b
2 along the y-
axis, eccentricity e , orbital period op , and the Sun at the right focal point and the planet
at perihelion at 0
=
t . There are two models that predict the position of this planet.
Model 1:
⋅
=
⋅
−
⋅
=
)
2
sin(
)
(
)
2
cos(
)
(
op
t
b
t
y
e
a
op
t
a
t
x
π
π
(1)
Model 2:
⋅
+
⋅
=
⋅
−
⋅
+
⋅
=
)
2
sin(
2
sin
)
(
)
2
sin(
2
cos
)
(
op
t
e
op
t
b
t
y
e
a
op
t
e
op
t
a
t
x
π
π
π
π
(2)
Write the formula for the area A swept by the line connecting the Sun to the planet from
α
=
t to β
=
t you found in m3 (Eq. (1) in m3). We will call it AS for area swept.
AS = ………………………………………………………………………..
Program this equation in MAPLE, or your calculator, to find the area swept for the two
models, as follows. First note that MAPLE is case-sensitive, while your TI calculator
may not be. Following the note in m0, let )
(
1 t
f and )
(
1 t
g equal to )
(t
x and )
(t
y
functions for model 1, and )
(
2 t
f and )
(
2 t
g be equal to )
(t
x and )
(t
y functions for
model 2. Use function notation for these functions.
Let the area swept be called ASM1 and ASM2 (for area swept model 1, and area swept
model 2). Let the area DCAB in m3, which we called 1
A , be called A1M1 and A1M2
(for 1
A model 1, and 1
A model 2). These are expressions, not functions. Use alp and bet
for alpha and beta.
Do not declare any numeric values for any constants or variables at this stage.

35
Your program in MAPLE will look like this (note that in some versions of MAPLE op is
reserved, so call it op1):
> restart; interface(displayprecision = 4): Digits := 20:
> f1:=t->a*cos(2*Pi*t/op1)-a*e;
g1:=t->b*sin(2*Pi*t/op1);
f2:=t->a*cos(2*Pi*t/op1+e*sin(2*Pi*t/op1))-a*e;
g2:=t->b*sin(2*Pi*t/op1+e*sin(2*Pi*t/op1));
> ASM1:=A1M1+ .......;
ASM2:=A1M2+ .......;
Your calculator functions and expressions will look like this ( →
sto is the store key):
2
2
1
1
1
1
)
(
1
)
/
2
sin(
)
(
1
)
/
2
cos(
asm
sto
m
a
asm
sto
m
a
t
g
sto
op
t
b
t
f
sto
e
a
op
t
a
→
+
→
+
→
∗
→
∗
−
∗
π
π
Save this MAPLE program, or functions and expressions in your calculator. We will find
formulas for 1
A (A1M1, and A1M2) for the two models in m5, and will input values for
the variables including alpha and beta in m6.

36
Problem m5
In problem m3 you wrote the formula for the areas swept by a line from the Sun to a
planet. Model 1 is simple but inaccurate, model 2 is more complicated but more accurate.
The only missing part of the equations is area 1
A , which we will find in this problem .
Given a parametric curve
=
=
)
(
)
(
)
(
)
(
t
g
t
y
t
f
t
x
Area 1
A between this curve and the x-axis from α
=
t to β
=
t (area DCAB in the figure
above), as we will see in class, is given by:
dt
t
f
dx
t
g
y
ydx
A )
(
),
(
,
1
′
=
=
=
β
α
.
′
=
β
α
dt
t
f
t
g )
(
)
( Eq. (1)
The absolute value sign above is to insure positive areas.
D C
B
A
O
β
=
t
α
=
t

37
Write the integral formulas for 1
A for model 1 (that is starting with equation (1) in m4,
find )
(t
f ′ and then plug in )
(t
f ′ and )
(t
g in equation (1) above, but do not integrate
here). Use chain rule to find the derivative of )
(t
f , and show your steps. Pull all the
multiplicative constants out of the integral and simplify the integrand. We will integrate
this on next page.
1
A (model 1)
=……………………………………………………………………………………Eq. (2)

38
Your next task is to integrate the integral equation for 1
A for model 1 (Eq. (2) you found
above) by hand. Start with equation (2) above, use the double angle identity to convert
the sine squared to a square-less cosine, and integrate using substitution. Show all your
work here. The result for 1
A should have no integral sign and should be in terms of
op
b
a and
,
,
,
, β
α and should be simplified.
1
A (Model 1):
=……………………………………………………………………………… Eq. (3)
It is not easy to find 1
A for model 2 as we did for model 1. We will leave the integral
formula for 1
A for model 2 as is in Eq. (1) above, but will replace )
(t
f and )
(t
g with
)
(
2 t
f and )
(
2 t
g .
1
A (Model 2) ′
=
β
α
dt
t
f
t
g )
(
)
( 2
2 Eq. (4)

39
Problem m6
In problem m5 you found formulas for area 1
A for model 1 (Eq. (3) in m5) and for model
2 (Eq. (4) in m5). We will now find numerical values for 1
A and, finally, the areas swept
by the line connecting the Sun to Mercury.
Add to your MAPLE program, or calculator functions and expressions you wrote in m4,
new lines to define A1M1 and A1M2, using equations (3) and (4) in m5. These are
expressions, not functions.
Now you can declare numerical values for op
e
b
a ,
,
, and α and β . You can now
change alpha and beta to change the intervals and get corresponding values for the areas.
Find numerical values for the three time intervals given in problem m0 for the
expressions for 1
A for model 1 and model 2, and list the areas in the following table. The
values of the orbit of Mercury and the intervals are given in m0 and repeated here.
20
11
,
20
10
20
5
,
20
4
20
,
0
,
years
241
.
0
206
.
0
AU
1
,
AU
387
.
0 2
op
op
or
op
op
or
op
op
e
e
a
b
a
=
=
=
−
=
=
β
α
Model 1 area 1
A (A1M1) Model 2 area 1
A (A1M2)
......
..........
..........
..........
:
20
11
to
20
10
.......
..........
..........
..........
:
20
5
to
20
4
.
..........
..........
..........
..........
:
20
to
0
op
t
op
t
op
t
op
t
op
t
t
=
=
=
=
=
=
......
..........
..........
..........
:
20
11
to
20
10
.......
..........
..........
..........
:
20
5
to
20
4
.
..........
..........
..........
..........
:
20
to
0
op
t
op
t
op
t
op
t
op
t
t
=
=
=
=
=
=
And finally, find the areas swept for model 1 and model 2 in the following table:

40
Model 1 area swept by line connecting the Sun
to Mercury (ASM1)
Model 2 area swept by line connecting the Sun
to Mercury (ASM2)
......
..........
..........
..........
:
20
11
to
20
10
.......
..........
..........
..........
:
20
5
to
20
4
.
..........
..........
..........
..........
:
20
to
0
op
t
op
t
op
t
op
t
op
t
t
=
=
=
=
=
=
......
..........
..........
..........
:
20
11
to
20
10
.......
..........
..........
..........
:
20
5
to
20
4
.
..........
..........
..........
..........
:
20
to
0
op
t
op
t
op
t
op
t
op
t
t
=
=
=
=
=
=
According to Kepler’s Second Law, the areas above must be the same, but neither of the
above models is exact. Model 2, however, should be better than model 1.

41
Problem m7
In problem m6 you found the approximate areas swept by a line from the Sun to Mercury
for two models. Model 1 is simple but inaccurate, model 2 is more complicated but more
accurate. You found the exact areas swept during these intervals (1/20th
of the area of the
orbital ellipse) for Mercury in problem m0. Fill in the areas for both models from the
second table in m6 here, compare to the values in problem m0 and find the percent errors
for each interval and fill in the error columns. Note that percent error is:
100
exact
exact
e
approximat
error
% ×
−
=
Exact area swept in 1/20th
of an orbital period (op/20) from m0 :…………………………
Model 1 area swept by line
connecting the Sun to Mercury
for time intervals :
% error Model 2 area swept by line
connecting the Sun to Mercury
for time intervals:
% error
........
..........
:
20
11
,
20
10
.........
..........
:
20
5
,
20
4
........
..........
:
20
,
0
op
op
op
op
op
………..
………..
……….. ........
..........
:
20
11
,
20
10
.........
..........
:
20
5
,
20
4
........
..........
:
20
,
0
op
op
op
op
op
……..…
……..…
……..….

42
Writing Your Project Report
You are now ready to present your scientific work on Kepler’s Second Law for Mercury.
Here is a guideline for your presentation for the results of problems m0 through m7.
a) Please do not attach or refer to any of the m problems in your report. Write your
report as if someone who does not know anything about the m problems, and has
never been to our class, but knows math, is reading your report. You are writing
your report for an OUTSIDER.
b) You do not need to present all the preliminary steps in m1. Present the main ideas of
the two models, planet locations for the intervals we have worked with, the area formulas,
numerical values for the areas, and the differences in the accuracy of the two models.
Present all the tables, and graphs that are relevant to understanding these main ideas.
Your report:
1) Introduction: Summarize Kepler’s Laws (m0) and the two models that we have been
working with (m2). Present the orbital values (a, b, e, op) for Mercury (m0). Summarize
what you will be doing in this project.
2) Project Report: Present the two models and planet locations you found for each
model in m1 and m2, with tables and graphs. Present the equations for the areas swept
by a line from the Sun to Mercury by starting with a figure similar to Fig. 4 in m3, and
starting with Area OAB in m3. You can then derive and present the area equations for
1
A in m5 for each model (Eqs. (3) and (4) in m5).
Present the exact area that should be swept in 1/20th
of an orbital period (m0). Present
the areas swept for the periods
20
11
,
20
10
and
,
20
5
,
20
4
,
20
,
0
op
op
op
op
op
for model 1 and
model 2 and the errors in a table (m7).
3) Summary: Summarize the results of this project and all that you have learned.

43
Halley’s Comet Project
Calculus III
Comet Halley from Mount Wilson, 1986
"The diversity of the phenomena of nature is so great, and the treasures hidden in the
heavens so rich, precisely in order that the human mind shall never be lacking in fresh
nourishment."
Johannes Kepler
This term we will study Halley’s Comet, its position as a function of time, and Kepler’s
Second Law of planetary motion. I will hand you weekly problems, which I call H
problems. You will hand these problems back to me, they will be graded, and handed
back to you. You will collect these problems and will summarize the results at the end of
the term in a project report.

44
Halley’s Comet Project
Calculus III
This term we will study the orbit of Halley’s Comet and its position as a function of time.
I will hand you weekly problems I will call H problems. We will use power series to
estimate the locations of the comet at various times during the 76 years it takes to orbit
the Sun. You will summarize the results of these problems at the end of the term in a
project report.
Edmond Halley's Comet
In 1705 Edmnnd Halley predicted, using Newton’s newly formulated laws of motion, that
the comets seen in 1531, 1607, and 1682 are all the same comet and would return in
1758 (which was, alas, after his death). The comet did indeed return as predicted and was
later named in his honor. The average period of Halley's orbit is 76 years. Comet Halley
was visible in 1910 and again in 1986. Its next passage will be in early 2062.
Comets, like all planets, orbit the Sun in elliptic orbits, but their orbits are very eccentric
(the major axis is much larger than the minor axis). The point where the comet is closest
to the Sun is called perihelion, and the point where it is the farthest is called aphelion
(see the figure in the refresher sheet attached).
At aphelion in 1948, the comet was 35.25 AU from the Sun, while at perihelion on
February 9, 1986, it was only 0.5871 AU from the Sun. An astronomical unit (AU) is the
semi-major axis for Earth, which is about 93 million miles.
The ellipse’s semi-major axis is a, while its semi-minor axis is b. The eccentricity, the
measure of its elongation, is e and is given by 2
2
1 a
b
e −
= , which can be solved for b
to give 2
1 e
a
b −
= . Eccentricity is between 0 and 1. For a circular orbit 0
=
e , and for
a very elongated orbit e is close to 1. The distance from the center of the ellipse to either
focal point is e
a ⋅ .
We will let 0
=
t designate February 1986. With this convention 20
=
t is February
2006, and 76
=
t is February 2062 when the comet will return to perihelion again.
The orbits of the Earth, Uranus, Neptune and Halley’s
Comet
Close up view of the orbit of Earth and Halley’s Comet

45
r = center
at (0,0), period π
2 :
=
=
)
sin(
)
(
)
cos(
)
(
t
a
t
y
t
a
t
x
period π
2 :
=
=
)
sin(
)
(
)
cos(
)
(
t
b
t
y
t
a
t
x
,
( k
h :
+
=
+
=
k
t
a
t
y
h
t
a
t
x
)
sin(
)
(
)
cos(
)
(
,
( k
h :
+
=
+
=
k
t
b
t
y
h
t
a
t
x
)
sin(
)
(
)
cos(
)
(
+
=
+
=
k
B
t
a
t
y
h
B
t
a
t
x
)
2
sin(
)
(
)
2
cos(
)
(
π
π
+
=
+
=
k
B
t
b
t
y
h
B
t
a
t
x
)
2
sin(
)
(
)
2
cos(
)
(
π
π
axis a
2 , minor axis b
center at )
0
,
( e
a ⋅
−
−
=
⋅
−
=
)
2
sin(
1
)
(
)
2
cos(
)
(
2
B
t
e
a
t
y
e
a
B
t
a
t
x
π
π
y
2b
2a
a.e
x
Aphelion
Perihelion
Planet moves slower
Planet moves faster

46
Problem H1
a) Write the parametric equation of a circular orbit with radius a centered at the origin
with parameter t , and an orbital period of op . The planet is at )
0
,
(a at 0
=
t .Your
answer will involve sine and cosine functions.
b) Write the parametric equation of an elliptic orbit with major axis a
2 along the x-
axis, minor axis b
2 along the y-axis. The ellipse is centered at )
0
,
0
( with parameter t ,
and an orbital period op . The planet’s position at 0
=
t should be at )
0
,
(a
c) Shift the ellipse in b) left so that the origin is at the right focal point. Note that the
distance from center to each focal point is e
a⋅ , where e is the eccentricity of the ellipse
(see Refresher ). Write the equation for this orbit. Your equations should be in terms of
e
b
a ,
, and op :
d) The orbit of Halley’s Comet has the following values:
years
76
97
.
0
AU
1
,
AU
34
.
19 2
=
=
−
=
=
op
e
e
a
b
a
AU is an “astronomical unit” which is the average distance from the Sun to the Earth ( a
for Earth).

47
Kepler’s Law states that the line connecting the Sun to the planets or comets sweeps
equal areas in equal time. The equation in c) ignores this law and will, therefore, give the
correct orbit, but incorrect locations for Halley’s Comet. We will see in Problem H2
how to find the correct positions. If Halley’s Comet is at perihelion at 0
=
t (Feb. 1986),
find the incorrect location of this planet using the equation in c) at the given times below.
Put your answer in ordered pairs )
,
( y
x and use three decimal places. Perihelion is when
the planet is closest to the Sun (for our problem this is )
0
,
( e
a
a ⋅
− )
time in years Incorrect locations
......
..........
..........
..........
..........
..........
..........
:
2062)
(Feb
76
t
......
..........
..........
..........
..........
..........
..........
:
75
......
..........
..........
..........
..........
..........
..........
:
70
......
..........
..........
..........
..........
..........
..........
:
60
......
..........
..........
..........
..........
..........
..........
:
50
......
..........
..........
..........
..........
..........
..........
:
40
......
..........
..........
..........
..........
..........
..........
:
30
......
..........
..........
..........
..........
..........
..........
:
2006)
(Feb
20
......
..........
..........
..........
..........
..........
..........
:
10
......
..........
..........
..........
..........
..........
..........
:
5
......
..........
..........
..........
..........
..........
..........
:
1
......
..........
..........
..........
..........
..........
..........
:
5
.
0
......
..........
..........
..........
..........
..........
..........
:
1986)
(Feb
0
=
=
=
=
=
=
=
=
=
=
=
=
=
t
t
t
t
t
t
t
t
t
t
t
t
e) Graph the elliptic orbit and locate the above locations on your graph. Use MAPLE,
and attach your graph. This is an example of how you can plot the orbit of a planet and place
the planet's positions on the orbit using MAPLE.
> with(plots):
> f:=t->a*cos(2*Pi*t/op1)-a*e; g:=t->b*sin(2*Pi*t/op1);
> a:=1.5: b:=1.2: e:=0.6: op1:=3:
> p1:=plot([f(t),g(t),t=0..3],x=-3..3,y=-2..2,scaling=CONSTRAINED,
xtickmarks=[-1,1],ytickmarks=[-1,1]):
p2:=pointplot({[f(.15),g(.15)],[f(.25),g(.25)]},symbol=CIRCLE,
color=black,scaling=CONSTRAINED):
display({p1,p2});

48
Problem H2
Johann Kepler in 1609 discovered that planets and comets orbit the Sun in elliptic orbits
and that their orbital velocity is not constant but varies. The following summarizes
Kepler’s first two laws (See Figure):
1) The planets orbit the Sun in elliptic orbits with the Sun at one of the focal points.
His second law simply said means that planets slow down when they are farther from the
Sun, and speed up when they are closer. Since the line joining the Sun to the planet is
shorter when the planet is closer, the length of the orbit traveled by the planet in a given
interval of time would be larger to make the areas swept equal.
For a circular orbit the eccentricity e is zero, but as the orbit gets more eccentric
(elongated), e approaches 1. The point of the orbit closest to the Sun is called
perihelion, and the point farthest is called aphelion. To simplify the calculations for this
problem, without loss of generality, we will place the origin at the focal point where the
Sun resides, the x-axis along the major axis. The length of the major axis is a
2 , and that
of the minor axis is b
2 . The center of the ellipse is then at )
,
0
( e
a ⋅
− . We will also let
time t equal zero when and the planet is at perihelion. With these assumptions, the
parametric equations of the orbit of a planet are:
⋅
=
⋅
−
⋅
=
)
2
sin(
)
(
)
2
cos(
)
(
op
t
b
t
y
e
a
op
t
a
t
x
π
π
or:
⋅
−
=
⋅
−
⋅
=
)
2
sin(
1
)
(
)
2
cos(
)
(
2
op
t
e
a
t
y
e
a
op
t
a
t
x
π
π
(1)
y
2b
2a
a.e
x
Aphelion
Perihelion
Planet moves slower
Planet moves faster

49
Where op is the orbital period in Earth years.
Note that when 0
=
e , the above equations turns into the parametric equations of a circle
with center at the origin.
Equation (1) does not account for Kepler’s Second Law (It assumes an almost constant
velocity). To account for that Kepler developed the following equation called Kepler’s
Equation:
)
sin(
2
E
e
E
op
t
⋅
−
=
π
(2)
For a given time t , you first solve for E from (2) and then plug E in equation (1) instead
of
op
t
π
2
:
⋅
−
=
⋅
−
⋅
=
)
sin(
1
)
(
)
cos(
)
(
2
E
e
a
t
y
e
a
E
a
t
x
(3)
The variable E is called eccentric anomaly, while the expression
op
t
π
2 is called mean
anomaly. Note that for a circular orbit when 0
=
e , these two are the same, but as e gets
closer to 1, these two will be different.
Equation (3) will give the correct position of the comet at a given time t . The only
problem with this is that because equation (2) is an implicit equation in E, and cannot be
solved for E, you must solve for E using a numerical technique. Fortunately your TI
calculator and MAPLE have SOLVE commands to do this for us ( solve (equation in x , x)
for TI and MAPLE)
We will study techniques to approximate E as a function of t in explicit form in
problems H3 and H4. This will give us in t
expression
an
)
( =
t
E which we will then
plug into (3) for E as an expression.
a) For problem H2 let t equal the values in the table below, solve for E from (2) using
the solver command on your calculator or MAPLE (make sure your calculator is in radian
mode). Now use (3) to find the correct locations for Halley’s Comet. Write the location
in ordered pairs )
,
( y
x , and carry your results to three decimal places.

50
......
..........
..........
..........
..........
..........
..........
.......
..........
:
2062)
(Feb
76
t
......
..........
..........
..........
..........
..........
..........
..........
..........
:
75
......
..........
..........
..........
..........
..........
..........
..........
..........
:
70
......
..........
..........
..........
..........
..........
..........
.........
..........
:
60
......
..........
..........
..........
..........
..........
..........
..........
..........
:
50
......
..........
..........
..........
..........
..........
..........
..........
..........
:
40
......
..........
..........
..........
..........
..........
..........
..........
..........
:
30
......
..........
..........
..........
..........
..........
..........
.........
..........
:
2006)
(Feb
20
......
..........
..........
..........
..........
..........
..........
..........
..........
:
10
......
..........
..........
..........
..........
..........
..........
..........
..........
:
5
......
..........
..........
..........
..........
..........
..........
..........
..........
:
1
......
..........
..........
..........
..........
..........
..........
..........
..........
:
5
.
0
......
..........
..........
..........
..........
..........
..........
..........
..........
:
1986)
(Feb
0
comet
the
of
locations
Correct
E
of
Value
years
in
Time
=
=
=
=
=
=
=
=
=
=
=
=
=
t
t
t
t
t
t
t
t
t
t
t
t
b) Plot the orbit and locate these locations as you did in Problem H1.
c) Observe the difference in these locations and that in Problem H1 and summarize with a
short statement.

51
Problem H3
We saw in problem H2 that to find the correct locations of Halley’s Comet we had to
solve the following implicit equation for E (eccentric anomaly):
)
sin(
2
E
e
E
op
t
⋅
−
=
π
, (1)
and then plug the value of E into the orbital equation for Halley’s Comet given by:
⋅
−
=
⋅
−
⋅
=
)
sin(
1
)
(
)
cos(
)
(
2
E
e
a
t
y
e
a
E
a
t
x
(2)
Implicit equations are not very convenient when scientist want to predict the location of
planets and comets in the sky, or want to design spacecraft to land on or fly by these
celestial objects. It is important to find an explicit expression for E as a function of time,
t
n
ression i
some
t
E exp
)
( = , that we can plug directly in the arguments of the cosine
and sine functions in (2).
In this H problem and the next we will study power series that will approximate E as an
explicit function of t . First, we need to study Bessel functions before we can proceed.
Bessel functions, like )
ln(
and
),
cos(
),
sin( x
x
x functions, are called transcendental
functions and can be presented explicitly only by power series. They are written as
.....
),
(
),
(
),
(
),
( 3
2
1
0 x
J
x
J
x
J
x
J
The subscript gives the order of the function (the above are Bessel functions of order 0,
order 1, order 2, order 3, …. ). Bessel function of order k is the solution to the following
differential equation:
...
,
3
,
2
,
1
,
0
,
0
)
( 2
2
2
=
=
−
+
′
+
′
′ k
y
k
x
y
x
y
x . (3)
For example, )
(
2 x
J is the solution to 0
)
2
( 2
2
2
=
−
+
′
+
′
′ y
x
y
x
y
x . In Chapter 7 we will
study differential equations, and in section 8.10 and a later H problem we will learn
techniques to solve these differential equations. The solutions to these differential
equations are given by the power series:

52
.
.
.
2
)!
2
(
!
)
1
(
)
(
2
)!
1
(
!
)
1
(
)
(
2
)
!
(
)
1
(
)
(
0
2
2
2
2
2
0
1
2
1
2
1
0
2
2
2
0
∞
=
+
+
∞
=
+
+
∞
=
+
−
=
+
−
=
−
=
i
i
i
i
i
i
i
i
i
i
i
i
i
i
x
x
J
i
i
x
x
J
i
x
x
J
(4)
1) Write a general power series for a Bessel function of order k .
......
..........
..........
..........
..........
..........
..........
..........
..........
)
( =
x
Jk ……
2) Write the first four terms of the power series of each Bessel functions in (4), in
exact form, and end each with ⋅
⋅
⋅
+ to indicate infinite series. Leave the
denominators in factorial and power form like
5
2
!
3
!
2 to show the patterns (DO NOT
EXPAND THESE INTO LARGE NUMBERS)
=
=
=
=
)
(
)
(
)
(
)
(
3
2
1
0
x
J
x
J
x
J
x
J

53
3) Turn the summations in equation (4) above to partial sums, and choose n for the
upper limit of the sums
=
n
i 0
such that the partial sums will give Taylor polynomials
,
,
, 22
21
20 T
T
T and 23
T for )
(
and
),
(
),
(
),
( 3
2
1
0 x
J
x
J
x
J
x
J , respectively:
n = ………………..
4) Enter the Taylor polynomials )
(
and
),
(
),
(
),
( 23
22
21
20 x
T
x
T
x
T
x
T approximations
for )
(
and
),
(
),
(
),
( 3
2
1
0 x
J
x
J
x
J
x
J , respectively, into a MAPLE worksheet or your
calculator [the command is: sum ( … ,i = 0 .. n); for MAPLE and )
,
0
,
(...., n
i for
TI ]. Plot these four functions on the same set of axes on the window ]
10
,
0
[
∈
x ,
]
1
,
1
[−
∈
y and attach your graphs.
5) MAPLE knows these functions as )
,
( x
k
BesselJ , where k is the order and x the
independent variable. For example )
,
2
( x
BesselJ is )
(
2 x
J . Your calculators
unfortunately don’t have Bessel functions in their catalogue. Use MAPLE to graph
)
(
0 x
J through )
(
3 x
J on the same set of axes and on the same windows as in 4) and
attach the graphs.
6) Write a short statement as to how the partial sum of the series form of Bessel
functions and MAPLE’s Bessel functions compare. Where are they similar, where
are they different.

54
Problem H4
We saw in problems H2 and H3 that to find the correct locations of Halley’s Comet we
had to numerically solve the following implicit equation for E (eccentric anomaly):
)
sin(
2
E
e
E
op
t
⋅
−
=
π
, (1)
and then plug the value of E into the orbital equation for Halley’s Comet given by:
⋅
−
=
⋅
−
⋅
=
)
sin(
1
)
(
)
cos(
)
(
2
E
e
a
t
y
e
a
E
a
t
x
. (2)
In order to avoid having to solve the implicit equation (1) numerically, astronomers and
mathematicians have developed a solution for the eccentric anomaly E (t) as an explicit
function of t , which is a power series form given by:
)
2
sin(
)
(
2
2
)
(
1 op
k
t
k
e
k
J
op
t
t
E
k
k ⋅
⋅
⋅
+
=
∞
=
π
π
. (3)
In (3), )
( e
k
Jk ⋅ is the Bessel function of order k that we studied in H3 with arguments
,....
3
,
2
, e
e
e . Note that )
( e
k
Jk ⋅ itself is a transcendental function and has a power series
expansion.
You will use MAPLE to do this problem. See the note below if you would like to use
your calculator. You can enter this power series as written in (3) into MAPLE using
)
,
( x
k
BesselJ syntax in MAPLE for )
( e
k
Jk ⋅ . Note that k is the order, and x is the
argument, which is e
k ⋅ here.
1)With 97
.
0
=
e for Halley’s Comet, use MAPLE to find the approximate (decimal)
values for the terms )
2
(
),
( 2
1 e
J
e
J , )
4
(
),
3
( 4
3 e
J
e
J , and write 76
/
2 t
π plus the first four
terms of the power series for E(t) in (3), then end with ⋅
⋅
⋅
+ to indicate infinite series.
Leave the term op
t
π
2 as 76
2 t
π , but turn all the coefficients of the sine functions into
decimals.
.........
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
.......
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
)
( =
t
E

55
2) Enter equation (3) in MAPLE using the first 50 terms (k=0..50), using the function
notation for )
(t
E [this will look like E := t->sum (….) ]. Enter the following values of t
in the table below to find the values of )
(t
E . Plug these values of )
(t
E into equation (2)
to find the )
,
( y
x locations of Halley’s Comet and fill the table below:
......
..........
..........
..........
..........
..........
..........
.......
..........
:
2062)
(Feb
76
t
......
..........
..........
..........
..........
..........
..........
..........
..........
:
75
......
..........
..........
..........
..........
..........
..........
..........
..........
:
70
......
..........
..........
..........
..........
..........
..........
.........
..........
:
60
......
..........
..........
..........
..........
..........
..........
..........
..........
:
50
......
..........
..........
..........
..........
..........
..........
..........
..........
:
40
......
..........
..........
..........
..........
..........
..........
..........
..........
:
30
......
..........
..........
..........
..........
..........
..........
.........
..........
:
2006)
(Feb
20
......
..........
..........
..........
..........
..........
..........
..........
..........
:
10
......
..........
..........
..........
..........
..........
..........
..........
..........
:
5
......
..........
..........
..........
..........
..........
..........
..........
..........
:
1
......
..........
..........
..........
..........
..........
..........
..........
..........
:
5
.
0
......
..........
..........
..........
..........
..........
..........
..........
..........
:
1986)
(Feb
0
comet
the
of
location
e
Approximat
E
of
Value
years
in
Time
=
=
=
=
=
=
=
=
=
=
=
=
=
t
t
t
t
t
t
t
t
t
t
t
t
3) Write a short statement as to how this compares with your correct locations you got in
problem H2.

56
*Note: You can use your calculator to do this problem, but since your calculator does not
know Bessel functions, you need to use the power series for )
( e
k
Jk ⋅ :
∞
=
+
+
+
⋅
−
=
⋅
0
2
2
2
)!
(
!
)
(
)
1
(
)
(
i
k
i
k
i
i
k
k
i
i
e
k
e
k
J ,
and plug that in (3) to get:
)
2
sin(
2
)!
(
!
)
(
)
1
(
2
2
)
(
1
0
2
2
op
k
t
k
k
i
i
e
k
op
t
t
E
k
i
k
i
k
i
i
⋅
⋅
+
⋅
−
+
=
∞
=
∞
=
+
+
π
π
.
The calculator will give similar answers to MAPLE if you use the first fifty terms for
both of the series (partial sums). The calculator is, however, excruciatingly slow. If you
do this, it is best to store the numbers in op, and e and enter this equation with op and e
symbols and not numbers. Store the function as f(t), and then enter f(0), f(0.5) f(1), ….
to get values for E(t).

57
Problem H5
In previous H problems we used Bessel functions to model the orbit and the location of
Halley’s Comet. In this last H problem we will actually solve a differential equation to
find the power series of one of these Bessel functions as an example of how the power
series for Bessel functions are derived. I will hand you a guideline to ask you to
summarize the results of problems H1 through H5 into a project report next week
Bessel function of order k, )
(x
Jk as we have seen before, is the solution to the following
differential equation:
...
,
3
,
2
,
1
,
0
,
0
)
( 2
2
2
=
=
−
+
′
+
′
′ k
y
k
x
y
x
y
x .
For example, )
(
2 x
J is the solution to 0
)
2
( 2
2
2
=
−
+
′
+
′
′ y
x
y
x
y
x .
If we let k=0, and choose proper initial conditions, the solution to the initial value
problem :
0
)
0
(
,
1
)
0
(
,
0
2
2
=
′
=
=
+
′
+
′
′ y
y
y
x
y
x
y
x (1)
is the Bessel function of order 0 ( that is )
(
0 x
J ).
a) Solve the above initial value problem (1) using the power series technique. Make sure
you show all your steps and put the final answer in form.
b) Find the interval of convergence of )
(
0 x
J .
c) Assuming that you can solve the differential equation for any Bessel function )
(x
Jk ,
find the interval of convergence of the general Bessel function )
(x
Jk . The form for
)
(x
Jk was found in H3.
d) Graph several Taylor polynomials for )
(
0 x
J until you reach one that looks like a
good approximation to )
(
0 x
J over the interval [-5, 5]. Present the graphs and the Taylor
polynomial that does this approximation
This problem will be graded on the use of good mathematical notation and complete
write up of your work.

58
Writing Your Project Report
You are now ready to present your scientific work on Kepler’s Laws as applied to
Halley’s Comet. Here is a guideline for your presentation for the results of problems H1
through H5.
a) Please do not attach or refer to any of the H problems in your report. You can cut and
paste results it they are neatly done, but use your own words. Write your report as if
someone who does not know anything about the H problems, but knows math is
reading your report. Your report should be word processed.
b) You will summarize all the information that you have learned in the H problems in
your report.
Your report:
1) Introduction: Summarize Kepler’s Laws and what the project will present ( the exact
implicit and the approximate explicit equations for E ) just in words ( No equations in
Introduction).
2) Main Presentation : Organize the information in a way that will make sense to an
outside reader. First present Kepler’s first two laws. Next present Halley’s Comet and
its orbital elements (a, e, op, b). Then the implicit equation that Kepler derived, and
finally the approximate explicit form that were later developed, and compare the results.
You also need to present Bessel functions in there as well. Emphasize Power Series for
E(t) and Bessel Functions that we have learned in this class. Include all the tables and
the formulas and the graphs that we have developed in the H problems. The solution to
)
(
0 x
J you found in H5 can be presented here or as an appendix in the back. If you
present it as an appendix in the back, mention here that as an example we will present the
solution for the first Bessel function in Appendix A.
3) Summary: Summarize the results of this project and all that you have learned just in
words. The summary will be just in words with no equations or graphs or tables.

59
Curiosity's Fight Path to Mars
A Project for Differential Equations (Math 256)
On November 25th
, 2011, NASA launched a spacecraft that carried a rover called
Curiosity to Mars. The rover landed at Mars’s Gale crater on August 6th
, 2012. In this
project you will model the 352 million mile (567 million km) flight path of this
spacecraft, and will send it into orbit around the Sun towards Mars.
You are part of the Initial Condition Group at NASA that will work on sending the rover
to Mars. Your task is to model the orbital path, and find the correct initial velocity of the
spacecraft to send it into orbit around the Sun towards Mars. NASA's Curiosity Rover
groups are:
The Launch Group (launches the spacecraft into space out of Earth's gravitational field)
The Initial Condition Group (your group, models orbital path, finds initial velocity)
The Course Correction Group (corrects the course to Mars every few weeks)
The Entry and Landing Group (works on entry into Mars's atmosphere, and landing)

60
Introduction:
two laws (See Figure 1):
interval of time would be larger to make the areas swept equal. The spacecraft carrying
Curiosity rover to Mars will be orbiting the Sun as well as Earth and Mars, and the above
laws applies to it and any other celestial object orbiting the Sun, or any other star.
The closest point of orbit to the Sun is called perihelion (periapsis), and the farthest
point is called aphelion (apoapsis).
Kepler did not have the physics and mathematical tools to prove his own discovery, and it
was left for the genius of Sir Isaac Newton to do that, in 1665, using his second law of
motion and his law of gravitation ( a
m
F ⋅
= , and 2
d
m
M
G
F
⋅
⋅
−
= ). The 23-year old was
a student at the University of Cambridge when an outbreak of the Plague forced the
university to close down for 2 years. Those two years were to be the most creative in
Newton’s life. He conceived the law of gravitation, the laws of motion, differential
calculus, and the proof of Kepler’s law.
Figure 1.
minor
axis =
2b
major axis = 2a
x
Aphelion
(Apoapsis)
Perihelion
(Periapsis)
y
Sun
a

61
Path of a Spacecraft from Earth to Mars
It may seem that the best way to send a spacecraft from Earth to Mars is to choose the
shortest path straight out to Mars, and time it so that the spacecraft and Mars will meet at
the same time and place, as shown in the Figure 2. But since the spacecraft is working
against Sun's gravity, it will need an impractically large amount of fuel.
Figure 2. How not to send a spasceship to Mars
In 1925 German engineer Wolfgang Hohmann proposed that the most efficeint way to
send a spacecraft to Mars (or to any outer planet) is to set the spacecraft in orbit around
the Sun with its perihelion at Earth's orbit, where it is launched, and its aphelion at Mars's
orbit, where it will meet Mars. The spacecraft needs almost no fuel to orbit the Sun,
except for a small amount to give it the correct initail velocity to send it into orbit, and
some course corrections. Figure 3 shows the flight path, with the standard xy coordinate
system for the solar system. See Appendix A for a detailed explanation of the solar
coordinate system.
Figure 3. How to send a spasceship to Mars
Mars at
Launch
Mars at
landing
Orbital
direction
Mars at
landing
Mars at
launch
Earth at
launch
x
y
Earth’s orbit
Mars’s orbit
Spacecraft’s orbit

62
The Mathematics of the Project
Your task in this project is to model the orbit of the two planets, and the spacecraft, and
find the initial velocity of the spacecraft to set it into orbit around the Sun so that it will
meet Mars on August 6th
, 2012.
We will let 0
=
t designate Nov. 25th
, 2011. Counting the number of days to August 6th
,
and dividing it by 365.24 days per year, we can find that 70
.
0
=
t on August 6th
, 2012.
0
=
t is Nov. 25th
, 2011 (Launch)
70
.
0
=
t is August 6th
, 2012 (Landing)
We will use differential equations to model the orbits and locations of Earth, Mars, and
the spacecraft using Newton’s two laws mentioned above.
Newton’s second law of motion in vector form is:
→
→
= a
m
F (1)
where
→
F is the force vector in N (Newtons), and
→
a is the acceleration vector in 2
s
m ,
and m is the mass in kg.
Newton’s law of gravitation in vector form is:
r
r
r
GMm
F
→
→
⋅
−
= 2
(2)
where
kg
s
m
G
⋅
×
= −
2
3
11
10
67
.
6 is the universal gravitational constant, M is the mass of
the larger object (the Sun), and is kg
10
2 30
× , and m is the mass the smaller one (the
planets or the spacecraft). The vector r is the vector connecting the Sun to the orbiting
objects. Note that in (1) and (2), the variables a
F, and r are all functions of time, and
)
(
)
( t
r
t
a ′
′
= .
The following will describe the steps you need to take to do this project.
1) Astronomers use Astronomical Units (AU) for distance, and years instead of seconds.
An Astronomical Unit is the average distance between the Sun and the Earth (about 93
million miles). This is the same as the semi-major axis for the Earth (a in Figure 1
above). There are AU
m
10
598
.
149 9
× , there are y
s
10
15569
.
3 7
× . In order to use
more conventional units that astronomers use, convert the units of universal gravitational
constant from
kg
s
m
⋅
2
3
to
kg
y
AU
⋅
2
3
using the information above. Use at least three decimal
places.

63
2) The motion force in Equation (1) and the gravitational force in Equation (2) are equal.
Equate the right hand sides of equations (1) and (2), and cancel the common factor on the
left and right sides.
3) Using )
(
)
( t
r
t
a ′
′
= , and j
t
y
i
t
x
t
r )
(
)
(
)
( +
= , convert the equation in step 2 above to
an equation involving )
(t
x , )
(t
y and their second derivatives. Note that these are the xy-
coordinates of each orbiting object.
4) When two vectors are equal, their components are equal (this means that when
j
d
i
c
j
b
i
a +
=
+ , then c
a = and d
b = ). Equate the x and the y components on each
side of the equation you got in step 3. This should give you two second-order differential
equations, one involving )
(t
x ′
′ , )
(t
x , and )
(t
y , and the other involving )
(t
y ′
′ , )
(t
x , and
)
(t
y . These equations will not have i and j in them.
5) Let )
(
)
( t
x
t
vx ′
= , and )
(
)
( t
y
t
vy ′
= , and turn each equation into a system of two first-
order differential equations. Collect these into a system of four first-order differential
equations (the order should be )
(
),
(
),
(
),
( t
vy
t
y
t
vx
t
x ). The four unknowns are the x
and the y coordinate, and the velocities in the x and the y directions of each celestial
object orbiting the Sun. This system of four first-order differential equations will
separately model the orbits of Earth, Mars and the spacecraft, if we use the correct initial
conditions for each orbiting object. In other words, if you use the four initial conditions
for Mars in this system of equations, and solve them, you can get the x, and the y
coordinates of Mars, and its velocity in the x and the y directions for a given time.
6) The four initial conditions for this system, for each orbiting object at 0
=
t (Nov. 25th
,
2011) are given below. The letter e is for Earth, m is for Mars, and c is for the curiosity
spacecraft. Note that the initial positions for the spacecraft are the same as Earth's. That
is because when the spacecraft is launched outside of Earth's strong gravitational field, it
is about 200 miles above Earth, which is negligible compared to cosmic distances.
xe(0) = 0.44503 AU
vxe(0) = -5.71113 AU / y
ye(0) = 0.88106 AU
vye(0) = 2.80924 AU / y
xm(0) = -0.81449 AU
vxm(0) = -4.23729 AU / y
ym(0) = 1.41483 AU
vym(0) = -2.11473 AU / y
xc(0) = 0.44503 AU
vxc(0) = Yours to find out AU / y,
yc(0) = 0.88106 AU
vyc(0) = Yours to find out AU / y
vxc(0) = Yours to find out AU / y,

64
7) Use MAPLE, or your calculator, to solve this system numerically as described in a) or
b) below. The initial conditions for Earth and Mars for 0
=
t are given above. You will
have to choose the initial velocity for the spacecraft (vxc(0), and vyc(0)), so that the
spacecraft is launched into orbit around the Sun, and is within 0.08 AU of Mars on
August 6th
( 7
.
0
=
t ).
Hint: First solve the system for Earth and Mars, to make sure that the orbits look
reasonable, and your calculations are correct. Then input reasonable numbers for vxc(0),
and vyc(0) to get a reasonable orbit for the spacecraft. Note that since the spacecraft's
orbit is a bit larger and more elliptic than Earth's, the spacecraft's initial velocity should
be a tad bit bigger than Earth's.
a) With MAPLE use the DEplot( … ) command and the dsolve( … type=numeric)
command. The first one will give you the graph of the orbit, while the second one will
give you the positions at different times after 0
=
t (Nov. 25th
, 2011). Make stepsize
equal to .005, and use scene=[x,y]. You will have to do this three times, one for Earth,
one for Mars, and one for the Curiosity spacecraft. The only difference between these
MAPLE commands will be the initial conditions. Once you solve the Earth's orbit, you
can do copy and paste and edit to get the other two orbits.
See Appendix B for an example of how you can use MAPLE to a system numerically
and plot locations
b) If you use your TI calculator, you can, unfortunately, only get one solution and graph
at a time. You need to set the following setting:
MODE Graph Diff Eq
y = Let x = y1, vx = y2, y = y3, vy = y4, and type your four equations and four initial
conditions in: =
=
′
=
=
′
=
=
′
=
=
′ 4
,
4
,
3
,
3
,
2
,
2
,
1
,
1 yi
y
yi
y
yi
y
yi
y
y = F1 Format Coord RECT
Solution RK
Fields FLDOFF
y = F7 Axes CUSTOM , x Axis y1, y Axis y3
WINDOW t0 = 0, tmax = 0.7, tstep= 0.005, xmin=-2*7/3, xmax=2*7/3, ymin=-2,
ymax=2 (This will give you equal scaling on the axes, as in zoom square)
8) Present the initial velocity for the spacecraft in a table, and find the xy-coordinates of
each celestial object and present that in a second table as shown blow. Use four decimal
places for the initial velocity table, and three decimal places for the xy-coordinates.
vxc(0) vyc(0)

65
t (y) xy-coord. of Earth xy-coord. of Mars xy-coord. of Curiosity
0
0.35
0.70
9) Present the graph of orbits and their locations superimposed on it like the one in
Appendix B. See Appendix B for an example of how to attach these plots so the
positions are superimposed on the orbits.
10) Calculate the distance between Mars and the Curiosity spacecraft at 7
.
0
=
t y. This
distance should be no more than 0.08 AU. Present it in a table as shown below.
Reminder: The distance equation is 2
2
1
2
2
1 )
(
)
( y
y
x
x
d −
+
−
= .
t (y) Distance between Mars and Curiosity spacecraft (AU)
0.7
Your Report
Your report should be complete and easy to understand by a mathematician who
has not seen this paper and has not been to our class.
Your report should include:
I) A cover sheet.
II) A short and complete statement of the problem in your own words.
III) All your math and calculations.
IV) All the graphs and tables (include the tables in steps 8, 9, and 10 above).
V) All the MAPLE worksheets or calculator work (attached as an Appendix to the back
of your report)
VI) A short conclusion of what this project has contributed to your cosmic
consciousness.
VII) The report should be all your own original words, graphs, equations, and tables
(with the possible exception of the s in this handout you are reading now).

66
Appendix A
A short review of the heliocentric coordinate system
The three dimensional solar coordinate system, called the Heliocentric Coordinate
System is shown above. The Sun is at the origin and the xy-plane is the plane of Earth's
orbit. The Earth orbits the Sun, and rotates about its axis counterclockwise as seen from
the positive z-axis. The Earth's rotation axis (north-south pole line) is in the z direction,
but tilted towards the positive y-axis. The Earth is on the positive y-axis on Winter
Solstice when the North Pole is tilted away form the Sun (approx. Dec. 21st
), and on the
negative y-axis on Summer Solstice when the North Pole is tilted into the Sun (approx.
June 21st
). The Earth is on the x-axis on the Equinoxes.
Top view of the orbits of Earth and Mars are shown above. The Earth's perihelion,
measured clockwise from the positive x-axis, is at about 103 (approx. Jan. 3rd
). Mars's
perihelion is at about 336 .
The orbital period of Earth is 1 year, and that of Mars is 1.88 years. The Earth's semi-
major axis is 1 AU, and that of Mars is 1.52 AU.
x
y
z
x
Autumn Equinox Winter Solstice
y
Spring Equinox
Summer Solstice
Mars's orbit
Earth's orbit
Mar's Perihelion
at about 336
∗
∗
Earth's Perihelion
at about 103

67
Appendix B
You will solve the system of four equations in your project numerically using
Maple. The following is an example of how you can solve a system of three Diff. Eqs. in
three unknowns x(t), y(t), and z(t) numerically. You can use the same commands as in
this example, but will have to modify it to reflect your own project.
x'(t) = -2x(t)+3y(t)+z(t)
y'(t) = 2x(t)-6y(t)-2z(t)
z'(t) = x(t)-2y(t)-3z(t)
DEplot gives the graph of the numeric solution.
dsolve gives the numeric solution that can be evaluated at any t.
You can then save the DEplot as p1, and the pointplot of locations as p2, and then
display the graph and the locations of each solution on top of it. When saving plots,
make sure to end the command with a colon (:), and not a semicolon (;). The colon
suppresses the output until you are ready to display the plots.
> with(DEtools): with(plots):
> DEplot({diff(x(t),t)=-2*x(t)+3*y(t)+z(t),
diff(y(t),t)=2*x(t)-6*y(t)-2*z(t),diff(z(t),t)=x(t)-2*y(t)-
3*z(t)},[x(t),y(t),z(t)],t=0..4,[[x(0)=1,y(0)=2,z(0)=-
1]] ,scene=[x,y],arrows=thin,linecolor=red,stepsize=.05);
> sol:=dsolve({diff(x(t),t)=-2*x(t)+3*y(t)+z(t),
diff(y(t),t)=2*x(t)-6*y(t)-2*z(t),diff(z(t),t)=x(t)-2*y(t)-
3*z(t),x(0)=1,y(0)=2,z(0)=-1},
[x(t),y(t),z(t)],type=numeric);
:=
sol proc
proc
proc
proc( ) ... end proc
end proc
end proc
end proc
rkf45_x
> sol(0);
[ ]
, , ,
=
t 0 =
( )
x t 1. =
( )
y t 2. =
( )
z t -1.

68
> sol(0.2);
=
t .2 =
( )
x t 1.22464334242087536 =
( )
y t 1.08667890529896894
, , ,
[
=
( )
z t -.797300639411056334 ]
> sol(.5);
=
t .5 =
( )
x t 1.09121351452654158 =
( )
y t .649786284216056998
, , ,
[
=
( )
z t -.410865740179701122 ]
> sol(1);
=
t 1 =
( )
x t .774048507820265241 =
( )
y t .354922732807339714
, , ,
[
=
( )
z t -.0932091994835609061 ]
> sol(2);
=
t 2 =
( )
x t .349881052241647838 =
( )
y t .127842021374964642
, , ,
[
=
( )
z t .0272897365510226253 ]
> sol(3);
=
t 3 =
( )
x t .151077207063233798 =
( )
y t .0510881785512952366
, , ,
[
=
( )
z t .0205865881299654302 ]
> sol(4);
=
t 4 =
( )
x t .0643413765672093269 =
( )
y t .0212120472193912287
, , ,
[
=
( )
z t .00993510209622652276 ]
> p1:=DEplot({diff(x(t),t)=-2*x(t)+3*y(t)+z(t),
diff(y(t),t)=2*x(t)-6*y(t)-2*z(t), diff(z(t),t)=x(t)-
2*y(t)-3*z(t)}, [x(t),y(t),z(t)], t=0..4,
[[x(0)=1,y(0)=2,z(0)=-1]], scene=[x,y], arrows=thin,
linecolor=red,stepsize=.05):
> p2:=pointplot({[1,2],[1.22,1.09],[1.09,0.65],[.77,0.35],
[.35,0.13],[.15,.05]},symbol=circle,color=black):
> display(p1,p2);

69
A Star Orbiting the Galactic Center Black Hole
A Project For Differential Equations
Infrared image of the stars orbiting the black hole at the center of the Milky Way galaxy.
The stars' orbits (dashed lines) show that they are in the thrall of a very compact object
with the mass of 4.5 million suns. This object is a black hole called Sagittarius A star
(Sgr A*). Colored dots show the stars' positions each year from 1995.7 to 2008.7 (mid
August of each year). The background is a 2008.7 infrared image. Lighter colored dots
are closer to 1995, darker colored dots are closer to 2008.
Information and image from UCLA Galactic
Center, and Scientific American, Dec. 2009

70
Introduction:
two laws (See Figure):
interval of time would be larger to make the areas swept equal. Since we will study a star
orbiting a massive black hole, we can restate these laws as:
1) Stars orbit a black hole in Elliptic orbits with the black hole at one of the focal points.
2) The line joining the black hole to a star sweeps out equal areas in equal time.
Kepler did not have the physics and mathematical tools to prove his own discovery, and it
was left for the genius of Sir Isaac Newton to do that, in 1665, using his second law of
motion and his law of gravitation ( a
m
F ⋅
= , and 2
d
m
M
G
F
⋅
⋅
−
= ). The 23-year old was
a student at the University of Cambridge when an outbreak of the Plague forced the
university to close down for 2 years. Those two years were to be the most creative in
Newton’s life. He conceived the law of gravitation, the laws of motion, differential
calculus, and the proof of Kepler’s law.
minor
axis =
2b
major axis = 2a
x
Apoapsis
Periapsis
y

71
A Star Orbiting a Black Hole
In recent years astronomers have been able to abserve the closest stars orbiting the
massive black hole at the center of our galaxy (the Milky Way Galaxy). This behemeth
black hole is called Sagittarius A Star, or Sgr A* for short, and is about 24,000 light years
from us. All the billions of stars in our galaxy, including the Sun, orbit this black hole.
The stars nearest Sgr A* have orbital periods less than 100 years, while the stars in the
suburbs of the Milky Way have orbital periods of 100 million years or more.
Recent advances in infrared imaging of distant stars have made it possible for
astronomers to see the stars nearest this black hole. The star we will study in this project
is called S0-2. The picture on the cover sheet of this document you are reading shows
several of these stars named S0-1, S0-2, S0-3, S0-5, S0-16, S0-19, and S0-20.
Stars, like planets orbiting the Sun, orbit this black hole in elliptic orbits, but their orbits
are very eccentric (the major axis is much larger than the minor axis). The point where
the star is closest to Sgr A* is called periapsis, and the point where it is the farthest is
called apoapsis (see the above).
We will place the xy-plane on the plane of the star's orbit with periapsis on the positive x-
axis as the above.
Astronomers use Astronomical Units (AU) for distance. An Astronomical Unit is the
average distance between the Sun and the Earth. This is the same as the semi-major axis
for the Earth (a in above). There are AU
m
10
598
.
149 9
× . They also use years instead
of seconds. There are y
s
10
15569
.
3 7
× .
The star S0-2 at periapsis, which occurred at 2002.3 (about mid March 2002), was only
117.7 AU from the Sgr A*, and was moving at 7
.
1689
− AU/y at that point. The negative
velocity indicates the star was moving in the negative y direction at periapsis.
The period of the S0-2's orbit is approximately 14.8 years. You will find this more
exactly yourselves.
We will let 0
=
t designate 2002.3. With this convention, 10
=
t , for example, is 2012.3,
and 8
.
14
=
t is 2017.1 (early 2017) when S0-2 will return to periapsis again
(approximately).
The Mathematics of the Project
In this project we will use differential equations to find the orbit and locations of the star
S0-2 using Newton’s two laws mentioned above.
Newton’s second law of motion in vector form is:
→
→
= a
m
F (1)

72
where
→
F is the force vector in N (Newtons), and
→
a is the acceleration vector in 2
s
m ,
and m is the mass in kg.
Newton’s law of gravitation in vector form is:
r
r
r
GMm
F
→
→
⋅
−
= 2
(2)
where
kg
s
m
G
⋅
×
= −
2
3
11
10
67
.
6 is the universal gravitational constant, M is the mass of
the larger object, and m is the mass the smaller one. The larger mass, Sgr A*, is 4.5
million times that of our Sun, and the mass of our Sun is kg
10
2 30
× . The smaller mass
m is the mass of the star S0-2. The vector r is the vector connecting Sgr A* to the
orbiting star S0-2. Note that in (1) and (2), the variables a
F, and r are all functions of
time.
The following will describe the steps you need to take to do this project.
1) In order to use more conventional units that astronomers use, convert the units of
universal gravitational constant from
kg
s
m
⋅
2
3
to
kg
y
AU
⋅
2
3
using the information above.
2) The motion force in Equation (1) and the gravitational force in Equation (2) are equal.
Equate the right hand sides of equations (1) and (2), and cancel the common term.
3) Using )
(
)
( t
r
t
a ′
′
= , and j
t
y
i
t
x
t
r )
(
)
(
)
( +
= , convert the equation in step 2 above to
an equation involving )
(t
x , )
(t
y and their derivatives. Note that these are the x-y
coordinates of the star S0-2.
4) When two vectors are equal, their components are equal (this means that when
j
d
i
c
j
b
i
a +
=
+ , then c
a = and d
b = ). Equate the x and the y components on each
side of the equation you got in step 3. This should give you two second-order differential
equations, one involving )
(t
x ′
′ , )
(t
x , and )
(t
y , and the other involving )
(t
y ′
′ , )
(t
x , and
)
(t
y . These equations will not have i and j in them.
5) Let )
(
)
( t
x
t
vx
′
= , and )
(
)
( t
y
t
vy
′
= , and turn each equation into a system of two first-
order differential equations. Collect these into a system of four first-order differential
equations.
6) To come up with four initial conditions for this system, note the position and velocity
of the star S0-2 mentioned above under A Star Orbiting a Black Hole. This will actually
give you the four initial conditions that you need for the system in step 5.

73
7) Use MAPLE, or your calculator, to solve this system numerically.
a) With MAPLE use the DEplot( … ) command and the dsolve( … type=numeric)
command. The first one will give you the graph of the orbit, while the second one will
give you the positions at different times after 2002.3. Make stepsize equal to .005, and
use scene=[x,y].
b) If you use your calculator, you need to set the following setting:
MODE Graph Diff Eq
y= type your four equations and four initial conditions in ,.....
1
,
1 yi
y ′
y= F1 Format Coord RECT
Solution RK
Fields FLDOFF
y= F7 Axes CUSTOM
x Axis y?
y Axis y?
8) Find the xy-coordinates of the star for years in table below and fill the table. Use only
three digits after the decimal point for the xy-coordinates. Present the graph of orbit of
S0-2 and its locations superimposed on it. This should look like the picture on the cover
sheet of this instruction document for S0-2, but with different star locations. The last row
in the table will be a time when S0-2 will return to periapsis again. You need to find this
exactly by running MAPLE until you find S0-2 at periapsis. You would also want to
include this time divided by 2 (? /2 row ) to place S0-2 at apoapsis to help you find the
lengths of the major and minor axes.
T Date x-coordinate y-coordinate
0 2002.3
1
2
3
4
5
6
7
? /2 (apoapsis time)
8
9
10
11
12
13
14
? (periapsis time)

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