ASEN 5050 SPACEFLIGHT DYNAMICS Intro to STK, More 2-Body Prof. Jeffrey S. Parker University of Colorado – Boulder Lecture 6: The Two Body Problem

Announcements Friday: no class at 9:00. STK Lab 1 ITLL 1B10, 2B10, or Visions Lab – or on your own computer. Groups of 1-3. Intro to STK today Alan will be in ITLL 2B10 to help answer questions from 1-3. Due Friday 9/26 at 9:00 am Homework #2 is due Friday 9/12 at 9:00 am D2L or under my door (ECNT 418). I’ll have Alan pick them up around 9:05 or 9:10 so don’t be late! Homework #3 is due Friday 9/19 at 9:00 am Concept Quiz #5 will be available at 10:00 am, due Monday morning at 8:00 am. Reading: Chapters 1 and 2 Lecture 6: The Two Body Problem

Quiz 4 Lecture 6: The Two Body Problem

Quiz 4 Lecture 6: The Two Body Problem

Quiz 4 Lecture 6: The Two Body Problem

Quiz 4 Lecture 6: The Two Body Problem

Challenge #3 We examined Pluto’s and Neptune’s orbits last time. Question: since Pluto sometimes travels interior to Neptune’s orbit, could they ever collide? If so, what sort of order of duration do we need to wait until it may statistically happen? Years? Millennia? Eons? Lecture 6: The Two Body Problem

Derivation of Kepler’s Equation If t is given: If n is given: M = n(t - tp) Another Useful Relation: Lecture 6: The Two Body Problem

Solving Kepler’s Equation Given M, solve for E What would a plot of this look like? Circular orbit? Elliptical? Lecture 6: The Two Body Problem

Solving Kepler’s Equation Given M, solve for E To see what we have to do, here are a few plots of this relationship: ~ parabolic Near-circular Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.1 and M = 100 deg (1.745 rad) What is E that produces M=100? Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.1 and M = 100 deg (1.745 rad) Subtract M from all values Find zero-crossing What is E that produces Value=0? Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.1 and M = 100 deg (1.745 rad) Step 1: Initial guess (E0 = M) Step 1: Value = -5.64 deg Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.1 and M = 100 deg (1.745 rad) Step 2: Modify guess and find the zero Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.1 and M = 100 deg (1.745 rad) Newton-Raphson Derivative line Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.1 and M = 100 deg (1.745 rad) Newton-Raphson Derivative line Guess #2 Lecture 6: The Two Body Problem

Solving Kepler’s Equation Newton Raphson Method applied to Kepler’s Equation Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.1 and M = 100 deg (1.745 rad) Guess 1: E0 = 100 deg Error 1: -5.64 deg Guess 2: E1 = 105.546 deg Error 2: 0.0263 deg Guess 3: E2 = 105.521 deg Error 3: 0.0000 deg Guess 4: E3 = 105.521 deg Error 4: 0.0000 deg Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.1 Any tricks or challenges? No. Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.95 Any tricks or challenges? Not really. Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.95, M = 180 deg Iteration History: E0 = 180.0000 deg Error = 0.0000 deg Lecture 6: The Two Body Problem

Solving Kepler’s Equation Let’s say e = 0.95, M = 300 deg Iteration History: E0 = 300.0000 deg Value = 47.1386 deg E1 = 210.2122 deg Error = -62.3980 deg E2 = 244.4787 deg Error = -6.4014 deg E3 = 249.0209 deg Error = -0.1562 deg E4 = 249.1375 deg Error = -0.0001 deg E5 = 249.1376 deg Error = -0.0000 deg 2 1 3 Lecture 6: The Two Body Problem

Solving Kepler’s Equation Iteration History: E0 = 350.0000 deg Value = 9.4518 deg E1 = 203.3066 deg Error = -125.1577 deg E2 = 270.1472 deg Error = -25.4220 deg E3 = 295.6314 deg Error = -5.2939 deg E4 = 304.6185 deg Error = -0.5873 deg E5 = 305.8945 deg Error = -0.0111 deg E6 = 305.9195 deg Error = -0.0000 deg Let’s say e = 0.95, M = 350 deg Lecture 6: The Two Body Problem

Solving Kepler’s Equation How many iterations does it take? e = 0 Lecture 6: The Two Body Problem

Solving Kepler’s Equation How many iterations does it take? e = 0.5 Lecture 6: The Two Body Problem

Solving Kepler’s Equation How many iterations does it take? e = 0.9 Lecture 6: The Two Body Problem

Solving Kepler’s Equation How many iterations does it take? e = 0.95 Lecture 6: The Two Body Problem

Solving Kepler’s Equation How many iterations does it take? e = 0.99 Not good! Lecture 6: The Two Body Problem

Kepler’s Equation Algorithm 2 in Vallado: Lecture 6: The Two Body Problem

Orbital Elements (Vallado, 1997) Lecture 6: The Two Body Problem

Orbital Elements Now, let’s define our other orbital elements. The inclination, i, refers to the tilt of the orbit plane. It is the angle between and , and varies from 0-180°. Lecture 6: The Two Body Problem

Orbital Elements The right ascension of the ascending node, W, is the angle in the equatorial plane from to the ascending node. The ascending node is the point on the equator where the satellite passes from South to North (opposite for the descending node). The line of nodes connects the ascending and descending nodes. The node vector, , points towards the ascending node and is denoted: The node lies between 0° and 360°. Lecture 6: The Two Body Problem

Orbital Elements The argument of periapse, w, measured from the ascending node, locates the closest point of the orbit (periapse) and is the angle between and . Lecture 6: The Two Body Problem

Orbital Elements The true anomaly, n, is the angle between periapse and the satellite position; thus: ( is positive going away from periapse, negative coming towards periapse.) Lecture 6: The Two Body Problem

Special Cases Elliptical Equatorial Orbits – W is undefined, so we use true longitude of periapse, , This is equivalent to astronomers’ longitude of periapse, , where Lecture 6: The Two Body Problem

Special Cases Circular Orbits – w is undefined, use argument of latitude, u, where: Lecture 6: The Two Body Problem

Special Cases Circular Equatorial – w and W undefined Lecture 6: The Two Body Problem

Two Line Element Sets (Vallado, 1997) Available on class web page. (Can be read by many programs including STK.) , are “Kozai” means. B* is a drag parameter. Lecture 6: The Two Body Problem

Two Line Element Sets Example 1 16609U 86017A 93352.53502934 .00007889 00000 0 10529-3 34 2 16609 51.6190 13.3340 0005770 102.5680 257.5950 15.59114070 44786 Epoch: Dec 18, 1993 12h 50min 26.5350 sec UTC Errors can be as large as a km or more. Lecture 6: The Two Body Problem

Orbital Elements from and (and t) Algorithm 9 in the book First compute the following vectors Compute the energy: p. 112 - 116 Lecture 6: The Two Body Problem

Orbital Elements from and (and t) Test using Example 2-5 in book Also, Lecture 6: The Two Body Problem

STK STK! Systems Tool Kit version 10.0 Lecture 6: The Two Body Problem

Welcome Splash Lecture 6: The Two Body Problem

STK Boot-up Lecture 6: The Two Body Problem

Save Save frequently Save each scenario in its own directory Please note: each scenario includes many files, for each object. If two scenarios are in the same directory and share objects, they may not behave properly. Lecture 6: The Two Body Problem

Lecture 6: The Two Body Problem

A satellite Lecture 6: The Two Body Problem

View Lecture 6: The Two Body Problem

ASEN 5050 SPACEFLIGHT DYNAMICS Coordinate and Time Systems Prof. Jeffrey S. Parker University of Colorado - Boulder Lecture 6: The Two Body Problem

Coordinate Systems Given a full state, with position and velocity known. Or, given the full set of coordinate elements. What coordinate system is this state represented in? Could be any non-rotating coordinate system! Earth J2000 or ecliptic J2000 or Mars, etc. Lecture 6: The Two Body Problem

Coordinate Systems Celestial Sphere Celestial poles intersect Earth’s rotation axis. Celestial equator extends Earth equator. Direction of objects measured with right ascension (a) and declination (d). Lecture 6: The Two Body Problem

Coordinate Systems The Vernal Equinox defines the reference direction. A.k.a. The Line of Aries The ecliptic is defined as the mean plane of the Earth’s orbit about the Sun. The angle between the Earth’s mean equator and the ecliptic is called the obliquity of the ecliptic, e~23.5. Lecture 6: The Two Body Problem

Coordinate Frames Inertial: fixed orientation in space Rotating Inertial coordinate frames are typically tied to hundreds of observations of quasars and other very distant near-fixed objects in the sky. Rotating Constant angular velocity: mean spin motion of a planet Osculating angular velocity: accurate spin motion of a planet Lecture 6: The Two Body Problem

Coordinate Systems Coordinate Systems = Frame + Origin Inertial coordinate systems require that the system be non-accelerating. Inertial frame + non-accelerating origin “Inertial” coordinate systems are usually just non-rotating coordinate systems. Is the Earth-centered J2000 coordinate system inertial? Lecture 6: The Two Body Problem

Useful Coordinate Systems ICRF International Celestial Reference Frame, a realization of the ICR System. Defined by IAU (International Astronomical Union) Tied to the observations of a selection of 212 well-known quasars and other distant bright radio objects. Each is known to within 0.5 milliarcsec Fixed as well as possible to the observable universe. Motion of quasars is averaged out. Coordinate axes known to within 0.02 milliarcsec Quasi-inertial reference frame (rotates a little) Center: Barycenter of the Solar System Lecture 6: The Two Body Problem

Useful Coordinate Systems ICRF2 Second International Celestial Reference Frame, consistent with the first but with better observational data. Defined by IAU in 2009. Tied to the observations of a selection of 295 well-known quasars and other distant bright radio objects (97 of which are in ICRF1). Each is known to within 0.1 milliarcsec Fixed as well as possible to the observable universe. Motion of quasars is averaged out. Coordinate axes known to within 0.01 milliarcsec Quasi-inertial reference frame (rotates a little) Center: Barycenter of the Solar System Lecture 6: The Two Body Problem

Useful Coordinate Systems EME2000 / J2000 / ECI Earth-centered Mean Equator and Equinox of J2000 Center = Earth Frame = Inertial (very similar to ICRF) X = Vernal Equinox at 1/1/2000 12:00:00 TT (Terrestrial Time) Z = Spin axis of Earth at same time Y = Completes right-handed coordinate frame Lecture 6: The Two Body Problem

Useful Coordinate Systems EMO2000 Earth-centered Mean Orbit and Equinox of J2000 Center = Earth Frame = Inertial X = Vernal Equinox at 1/1/2000 12:00:00 TT (Terrestrial Time) Z = Orbit normal vector at same time Y = Completes right-handed coordinate frame This differs from EME2000 by ~23.4393 degrees. Lecture 6: The Two Body Problem

Useful Coordinate Systems Note that J2000 is very similar to ICRF and ICRF2 The pole of the J2000 frame differs from the ICRF pole by ~18 milliarcsec The right ascension of the J2000 x-axis differs from the ICRF by 78 milliarcsec JPL’s DE405 / DE421 ephemerides are defined to be consistent with the ICRF, but are usually referred to as “EME2000.” They are very similar, but not actually the same. Lecture 6: The Two Body Problem

Useful Coordinate Systems ECF / ECEF / Earth Fixed / International Terrestrial Reference Frame (ITRF) Earth-centered Earth Fixed Center = Earth Frame = Rotating and osculating (including precession, nutation, etc) X = Osculating vector from center of Earth toward the equator along the Prime Meridian Z = Osculating spin-axis vector Y = Completes right-handed coordinate frame Lecture 6: The Two Body Problem

Useful Coordinate Systems Earth Rotation The angular velocity vector ωE is not constant in direction or magnitude Direction: polar motion Chandler period: 430 days Solar period: 365 days Magnitude: related to length of day (LOD) Components of ωE depend on observations; difficult to predict over long periods Lecture 6: The Two Body Problem

Useful Coordinate Systems Principal Axis Frames Planet-centered Rotating System Center = Planet Frame: X = Points in the direction of the minimum moment of inertia, i.e., the prime meridian principal axis. Z = Points in the direction of maximum moment of inertia (for Earth and Moon, this is the North Pole principal axis). Y = Completes right-handed coordinate frame Lecture 6: The Two Body Problem

Useful Coordinate Systems IAU Systems Center: Planet Frame: Either inertial or fixed Z = Points in the direction of the spin axis of the body. Note: by convention, all z-axes point in the solar system North direction (same hemisphere as Earth’s North). Low-degree polynomial approximations are used to compute the pole vector for most planets wrt ICRF. Longitude defined relative to a fixed surface feature for rigid bodies. Lecture 6: The Two Body Problem

Useful Coordinate Systems Example: Lat and Lon of Greenwich, England, shown in EME2000. Greenwich defined in IAU Earth frame to be at a constant lat and lon at the J2000 epoch. Lecture 6: The Two Body Problem

Useful Coordinate Systems Synodic Coordinate Systems Earth-Moon, Sun-Earth/Moon, Jupiter-Europa, etc Center = Barycenter of two masses Frame: X = Points from larger mass to the smaller mass. Z = Points in the direction of angular momentum. Y = Completes right-handed coordinate frame Lecture 6: The Two Body Problem

Announcements Friday: no class at 9:00. STK Lab 1 ITLL 1B10, 2B10, or Visions Lab – or on your own computer. Groups of 1-3. Intro to STK today Alan will be in ITLL 2B10 to help answer questions from 1-3. Due Friday 9/26 at 9:00 am Homework #2 is due Friday 9/12 at 9:00 am D2L or under my door (ECNT 418). I’ll have Alan pick them up around 9:05 or 9:10 so don’t be late! Homework #3 is due Friday 9/19 at 9:00 am Concept Quiz #5 will be available at 10:00 am, due Monday morning at 8:00 am. Reading: Chapters 1 and 2 Lecture 6: The Two Body Problem