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Space Shuttle Mission to Hubble Telescope March 1, 2002. Initial Space Shuttle Orbit ... Related Project: Real-time International Space Station Tracker ... – PowerPoint PPT presentation

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Title: Burton W. Jones


1
Burton W. Jones
  • Distinguished
  • Teacher
  • and
  • MAA Member

2
  • Gauss
  • and
  • Gauss Again

3
These famous names
  • Newton Legendre
  • Euler Gauss Lagrange
  • Laplace Bessel
  • might all be found on a list of

4
Great Astronomers!!!
5
Why Astronomy?
  • Celestial (orbital) mechanics the mathematics
    of things in orbit

6
My introduction to orbital mechanics
  • From the good Dr. Gauss himself, in a most
    unexpected way
  • The Method of Least Squares

7
From Edwards and Penneys Calculus
  • The great German mathematician Carl Friedrich
    Gauss (1777-1855) invented the method of least
    squares when he was 18 years old. A short time
    later, he used it to determine the orbit of the
    first-discovered asteroid Ceres, initially
    observed on the first day of the nineteenth
    century, but lost from sight a few weeks later.

8
From Larson and Hostetlers Calculus
  • The method of least squares was introduced by
    the French mathematician Adrien-Marie Legendre
    (1752-1833).

9
Whats a curious fellow to do?
  • Summarische Übersicht der zur Bestimmung der
    Bahnen der beiden neuen Hauptplaneten angewandten
    Methoden
  • Summary Survey of the Methods for Determining the
    Orbits of the Two New Major Planets

10
What its like to read Gauss(Quotes from
Summarische Übersicht)
11
What its like to read Gauss(Quotes from
Summarische Übersicht)
  • One can easily convince oneself
  • From this it is easy to conclude
  • From this the following three equations follow
    easily
  • It is clear
  • Further, one easily recognizes
  • I allow myself this easily understood expression
    for the sake of brevity

12
Learning orbital mechanics from Gausss
Summarische Übersicht
13
Learning orbital mechanics from Gausss
Summarische Übersicht
  • Method of successive humiliations
  • What I didnt learn (least squares)
  • What I later learned (least squares)
  • Where it led me (a whole new direction in my
    research, with some interesting impacts on my
    teaching!)

14
Todays TalkOrbital mechanics applications in
the standard undergraduate mathematics curriculum
15
Orbit basics
E (central angle)
satellite
?
t t0
f (polar angle)
16
I. Calculus I
Analysis of Keplers Equation
17
Analysis of Keplers Equation
  • Problem 1. Given e0.5, M2.5, solve k(E)0.
  • (I say, Blah blah blah Newtons method.
    Students reach for their calculators. )

18
Analysis of Keplers Equation
  • Problem 2. Are there other solutions?
  • (Three really clever students put their
    calculators in graphing mode.)

19
Analysis of Keplers Equation
  • Problem 3. Same questions
  • for e 1.5, M 0.2.

20
Keplers Equation E-0.5 sin(E)-2.50
E-1.5 sin(E)-0.20
k(E)
k(E)
E
E
21
Consider Keplers Equation E - e sin(E)
M 0, 0
  • Must there be a solution?
  • If so, must the solution be unique?
  • Why NASA cares about these questions
  • (Students look in vain for the right button on
    their calculator.)

  • 22
    k(E) E - e sin(E) M 0
  • Observe
  • k is continuous
  • k(0) 0, k(M1)0 (there is a root)
  • k '(E) 1 - e cos(E) 0 (root is unique)
  • Calculus books are jammed with functions, but
    rarely look at families of functions
  • A valuable point of view in real world
    applications

  • 23
    II. Linear Algebra
    • Textbook problem
    • Consider the bases
    • B(2,1,1), (2,-1,1), (1,2,1) and
    • B(3,1,-5), (1,1,-3), (-1, 0,2).
    • Find the transition matrix from B to B.
    • My students always ask

    24
    • Why in the world would I ever want to express
      vectors in any basis other than
    • B(1,0,0), (0,1,0), (0,0,1)?

    25
    • Why in the world would I ever want to express
      vectors in any basis other than
    • B(1,0,0), (0,1,0), (0,0,1)?
    • Answer Things that are very complicated in one
      coordinate system may be simple in another!

    26
    • We can choose our basis to make the Earths
      rotation simple to describe

    27
    • or we can choose our basis to make the
      satellites motion easy to describe

    z'
    y'
    x'
    28
    • Or we could choose the basis so that the
      elliptical orbit has an easy polar
      representation

    29
    • But we really need all three! So
    • Express vectors in the most convenient basis,
      then use transition matrices to convert as
      necessary between the various bases!

    z
    z'
    y'
    y''
    z''
    ?
    x''
    y
    x'
    x
    30
    • First transition matrix
    • basis to
    • basis

    z'
    y'
    y''
    z''
    x''
    Wp
    x'
    31
    • Next transition matrix

    z
    z'
    y'
    i
    y
    x'
    x
    32
    • Final transition matrix

    z
    y
    RA
    x
    33
    This completes the conversion from
    coordinates to coordinates!
    z
    y
    x
    34
    Now we can easily work with satellite positions
    in one coordinate system, then change them to
    another!
    35
    Vector Time (GMT) 2005/024/133200.000
    Vector Time (MET) N/A Weight (LBS)
    404236.9 M50 Cartesian
    M50 Keplerian
    -----------------------------------
    -------------------------------- X
    4205964.20 a
    6731367.37 meter Y -1135248.32 meter
    e .0009492 Z
    5124138.27 i
    51.32425 XDOT 448.028837
    Wp 48.90129 YDOT 7575.227356 meter/sec
    RA 274.58915 deg ZDOT 1315.109936
    f 28.49040

    M 28.43854
    36
    III. Differential Equations
    • Space Shuttle Mission to Hubble Telescope March
      1, 2002

    Initial Space Shuttle Orbit
    Transfer Orbit
    Hubble Orbit
    ?v
    37
    • Vector Time (GMT) 2002/060/113232.000
    • Vector Time (MET) 000/001030.000
    • Weight (LBS) 254068.0
    • M50 Cartesian
      M50 Keplerian
    • -----------------------------------
      --------------------------------
    • X 429037.29 a
      6700526.03 meter
    • Y -5724011.02 meter e
      .0395379
    • Z 3066426.36 i
      28.47537
    • XDOT 7923.139316 Wp
      53.88105
    • YDOT 93.941694 meter/sec RA
      174.32026 deg
    • ZDOT -476.018849 f
      44.89957

    38
    • SHUTTLE TRAJECTORY DATA
    • Lift off time (UTC) 2002/060/112202.000
    • Maneuvers contained within the current ephemeris
      are as follows
    • IMPULSIVE TIG (GMT) M50 DVx(FPS) LVLH
      DVx(FPS)
    • IMPULSIVE TIG (MET) M50 DVy(FPS) LVLH
      DVy(FPS)
    • DT M50
      DVz(FPS) LVLH DVz(FPS)
    • -----------------------------------------------
      -------------------------
    • 060/120713.533 -99.2
      133.3
    • 000/004511.533 80.6
      -0.1
    • 000/000128.577 -38.0
      -0.0

    39
    • Vector Time (GMT) 2002/060/121157.821
    • Vector Time (MET) 000/004955.821
    • Weight (LBS) 250747.5
    • M50 Cartesian M50
      Keplerian
    • -----------------------------
      --------------------------------
    • X 2861810.80 a
      6769217.50 meter
    • Y 5503917.12 meter e
      .0277775
    • Z -3128507.70 i
      28.48130
    • XDOT -6827.375539 Wp
      52.94227
    • YDOT 817.354008 meter/sec RA
      174.16154 deg
    • ZDOT -1143.785242 f
      197.83838

    40
    Apply Eulers Method to a System of Differential
    Equations
    41
    IV. Calculus III The Mother Lode
    42
    Derivation of Keplers Laws
    43
    Vector Time (GMT) 2005/024/133200.000
    Vector Time (MET) N/A Weight (LBS)
    404236.9 M50 Cartesian
    M50 Keplerian
    -----------------------------------
    -------------------------------- X
    4205964.20 a
    6731367.37 meter Y -1135248.32 meter
    e .0009492 Z
    5124138.27 i
    51.32425 XDOT 448.028837
    Wp 48.90129 YDOT 7575.227356 meter/sec
    RA 274.58915 deg ZDOT 1315.109936
    f 28.49040

    44
    z
    h
    v
    i
    r
    y
    x
    45
    Vector Time (GMT) 2005/024/133200.000
    Vector Time (MET) N/A Weight (LBS)
    404236.9 M50 Cartesian
    M50 Keplerian
    -----------------------------------
    -------------------------------- X
    4205964.20 a
    6731367.37 meter Y -1135248.32 meter
    e .0009492 Z
    5124138.27 i
    51.32425 XDOT 448.028837
    Wp 48.90129 YDOT 7575.227356 meter/sec
    RA 274.58915 deg ZDOT 1315.109936
    f 28.49040

    46
    V. My all time favorite student project
    • Given the position and velocity YDOT, ZDOT of the International Space Station at
      time t0
    • Determine where (azimuth and altitude) and when
      to look for the space station as it flies over
      your location.

    47
    • Solution uses trigonometry, vector calculus,
      linear algebra, and just a dash of computer
      programming.
    • Students see the results of the project in an
      extraordinary way they go outside and look in
      the calculated direction at the calculated time,
      and watch the space station fly by!

    48
    (No Transcript)
    49
    Related Project Real-time International Space
    Station Tracker
    50
    Research Opportunities Outside the Classroom
    Context
    History
    Astronomy
    Mathematics
    51
    Which brings us toGauss Again!(A Current
    Project)
    • Problem Calculate the date of Easter in a given
      year.
    • Certainly of mathematical interest (number
      theory?)
    • Certainly of historical interest
    • Has its roots in astronomy (full moon, vernal
      equinox, etc.)

    52
    But what does it have to do with Gauss?
    • Berechnung des Osterfestes
    • Calculation of the Easter Date
    • published August 1800

    53
    Gausss Easter Algorithm
    • then Easter falls on the 22 d eth of March or
      the d e 9th of April. (For 2000-2099, M24,
      N5)

    54
    Example
    • a 2005 mod 19 10
    • b 2005 mod 4 1
    • c 2005 mod 7 3
    • d 19aM mod 30 4
    • e 2b4c6dN mod 7 1
    • Then Easter falls on the 22 d eth of March,
      or March 27th.

    55
    Gausss explanation of the various steps of the
    algorithm
    56
    Gausss explanation of the various steps of the
    algorithm
    • The analysisdoes not allow itself to be shown
      here in its complete simplicity
    • A further development of this circumstance would
      be too long and drawn out here.
    • which one can easily convince oneself
    • From this it is clear

    57
    A parting gem from Dr. Gauss
    • Handschriftliche Bemerkung at the end of his
      article
    • From Christmas to Easter is on average,
    • however, weekdays.
    • (Absolutely no explanation whatever!)

    58
    773 ???? 950
    • My own observations
    • A complete (Gregorian) calendar cycle consists of
      400 years, or 146,097 days

    59
    773 ???? 950
    • My own observations
    • A complete (Gregorian) calendar cycle consists of
      400 years, or 146,097 days
    • Prime factorization of 146,097 337773
    • Coincidence? Seems unlikely, but

    60
    I have no idea!
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