(Re) Orientation 1 - Introduction 2 - Propulsion & ∆V 3 - Attitude Control & instruments 4 - Orbits & Orbit Determination LEO, MEO, GTO, GEO Special LEO orbits Orbit Transfer Getting to Orbit GPS 5 - Launch Vehicles 6 - Power & Mechanisms 7 - Radio & Comms 8 - Thermal / Mechanical Design. FEA 9 - Reliability 10 - Digital & Software 11 - Project Management Cost / Schedule 12 - Getting Designs Done 13 - Design Presentations Another night of F=MA? Education is the process of realizing that you don’t know what you thought you knew... Sporadic Events: •Mixers •Guest Speakers •Working on Designs •Teleconferencing Engin 176

Review of Last time Attitude Determination & Control Design Activity Feedback Control Systems description Simple simulation Attitude Strategies The simple life Eight other approaches and variations Disturbance and Control forces Design build & test an Attitude Control System Or in a word, F=MA Design Activity Team designations Mission selections Homework - ACS for mission Plant (satellite) Set point Error Control Algorithm Sensor Disturbances Actuator Engin 176

LEO vs. GEO Orbit LEO: 1000 km GEO: 36,000 km Low launch cost/risk Short range Global coverage (not real time) Easy thermal environment Magnetic ACS Multiple small satellites / financial “chunks” Minimal propulsion GEO: 36,000 km Fixed GS Antenna Constant visibility from 1 satellite Nearly constant sunlight Zero doppler Engin 176

LEO & GEO Pros / Cons Pros Cons LEO GEO Low launch cost High launch reliability Low radiation (except poles and SAA) Short slant range RF, imaging, probes Global coverage from polar Doppler ranging & GPS Easy thermal for 0°C < T < 20°C Financial small chunks Atmospheric drag Tracking antennas Doppler compensation Short, infrequent, irregular contacts Requires autonomy No global perspective Lots of clutter Frequent battery cycling Except sun-synch Low temp hard to achieve LEO Large Fixed Antennas are cheap Constant visibility from 1 satellite Nearly constant solar illumination Zero Doppler Need large antenna 1/3 second R/T delay Large orbit maintenance ∆V Large insertion ∆V; complex launch Thermal Averaging difficult Exposure to solar weather Poor Polar Visibility Finances chunky GEO Engin 176

Orbit Ground Rules #1: Orbits are Dynamic. There are no “stationary” orbits. They are a dynamic balance of kinetic and potential energy - centripetal acceleration vs. gravitational acceleration. I.e. w2r + Mg/r2 = constant and w a v/r so v2 a l/r => w a l/ r3/2 => t a r3/2 also, v = max @ perigee, min @ apogee #2: Orbit plane bisects earth because gravity is radial and it must balance centripetal acceleration. #3: Orbits are not tracks - knowing instantaneous position and velocity fully determines orbit #4: Orbit plane wrt sun changes slowly if at all. Engin 176

Orbits are Dynamic. Ground Rule #1 means... Orbit period is shorter for lower orbits since: w2r + Mg/r2 = constant and w a v/r so v2 a 1/r => w a 1/ r3/2 => t a r3/2 For an eliptic orbit, V is maximum @ perigee, min @ apogee Geosynchs don’t just “hang” there Engin 176

Ground Rule #2 means... Must be geosynch @ the equator Orbit plane bisects earth Ground Rule #2 means... Must be geosynch @ the equator Orbit planes & inclination are fixed Orbits are not tracks - knowing instantaneous position + velocity fully determines the orbit Orbit plane wrt sun changes slowly Launch “windows”: the orbit plane always includes the launch site and the earth’s center. These might need to have an orientation wrt e.g. the sun or have a specific time of day at launch, or be aligned to a target Engin 176

Orbits are not tracks Ground Rule #3 means... Changing Orbit: where new & old orbit intersect, change V to that appropriate to new orbit If present and desired orbit don’t intersect: join them via an intermediate that does Do V changes where V is a minimum (at apogee) Orbit determination: requires a single simultaneous measurement of position + velocity. GPS and / or ground radar can do this. A 1 2 3 B ª + = Engin 176

Orbit plane wrt sun changes slowly Ground Rule #4 means... Polar, sun synch dawn - dusk orbit… isn’t, after a few months. Geosynch satellites… don’t remain geosynch on their own. Engin 176

Getting to (LEO) Orbit #1: Raise altitude from 0 to 300 km (this is the easy part) Energy = mgh = 100kg x 9.8 m/s2 x 300,000 m = 2.94 x 108 kg m2/ s2 [=W-s = J] = 82 kw-hr = 2.94 x 106 m2/s2 per kg ∆V = (E)1/2 = 1715 m/s #2: Accelerate to orbital velocity, 7 km/s (the harder part) ∆V (velocity) = 7000 m/s (80% of V, 94% of energy) ∆V (altitude) = 1715 m/s ∆V (total) = 8715 m/s Engin 176

Getting to (LEO) Orbit - from an airplane Airplane is at 10 km altitude and 200 m/s airspeed #1: Raise altitude from 10 to 300 km Energy = mgh = 100kg x 9.8 m/s2 x (300,000 m - 10,000 m) ∆V = (E)1/2 = 1686 m/s (98% of ground based launch ∆V) (or 99% of ground based launch energy) #2: Accelerate to orbital velocity, 7 km/s from 0.2 km/s airplane ∆V (velocity) = 6800 m/s (97% of ground ∆V, 99% of energy) ∆V (∆H) = 1686 m/s (98% of ground ∆V, 99% of energy) ∆V (total, with airplane) = 8486 m/s ∆V (total, from ground) = 8715 m/s Engin 176

Common µspacecraft Orbits Remote Sensing: Favors polar, LEO, 2x daily coverage (lower inclinations = more frequent coverage). Harmonic orbit: period x n = 24 hours LEO Comms: Same! - multiple satellites reduce contact latency. Best if not in same plane. Equatorial: Single satellite provides latency < 100 minutes • Sun Synch: Dawn/Dusk offers Constant thermal environment & constant illumination (but may require ∆V to stay sun synch) Elliptical: Long dwell at apogee, short pass through radiation belts and perigee... Molniya. Low E way to achieve max distance from earth. MEO: Typically 10,000 km. TRW Odyssee comms cluster, GPS. Engin 176

∆V Directory Shuttle to 500 km circular: 300 m/s LEO to GTO: 2000+ m/s GTO to GEO: 2000+ m/s LEO Inclination change: 7000 m/s x sin (∆angle) Typical orbit maintenance: 100 to 300 m/s (sun synch or geosynch, per year) ª + = Engin 176

Orbit Description* Sir Isaac Newton used his Second Law of Motion combined with his Law of Gravitation to describe the motion of a small body orbiting about a much larger body. The 2-body equation of motion (∑E=constant) is: r + (m /r3) r = 0 where m = 398 601.2 km3 / sec2 (gravitational parameter for Earth) r = rE + h (mean radius of Earth plus altitude) rE = 6378.145 km (mean radius of Earth) To solve for the position vector, r, we need 6 constants of integration (second order equation in 3 dimensional space). Therefore, we can find the current position and velocity based on a previous known position and velocity. * Thanks to Rob Baltrum Engin 176

Orbit Description - 2 A solution to the 2-body equation of motion is a polar equation of a conic section given by: r = a(1-e2) / (1 + e cos n) The position in the orbital plane, r, depends on the values of a (semi-major axis), e(eccentricity), and n (polar angle or true anomaly). The conic section equation describes 4 major type of orbits: circle e = 0 a = radius ellipse 0< e < 1 a > 0 parabola e = 1 a = ∞ hyperbola e > 1 a < 0 Therefore, if the polar equation for a conic section can be used to describe a spacecraft position in the orbital plane (3 terms - a, e, n) we then only need to describe the orientation of the orbit about the Earth (or central body). Engin 176

The 6 Classical Orbital Elements Orbit Description - 3 The 6 Classical Orbital Elements The first 3 elements describe the type of conic section the orbit represents. The second 3 elements describe the orientation of the orbit with respect to the Earth (or central body). a, semi-major axis - a constant distance ( in kilometers) which describes the size of the orbit. e, eccentricity - a dimensionless constant which describes the shape of the orbit. (elliptical, circular, parabolic, hyperbolic) n, true anomaly - the angle in degrees, measured in the direction and plane of the spacecraft’s motion, between the perigee point to the position vector of the spacecraft at any time. This determines where in the orbit the S/C is at a specific time. i, inclination - the angle in degrees between the angular momentum vector and the unit vector in the Z-direction. This is a measure of how the orbit plane is ‘tilted’ with respect to the Equator. Ω, longitude or right ascension of the ascending node - the angle in degrees from the Vernal Equinox (line from the center of the Earth to the Sun on the first day of autumn in the Northern Hemisphere) to the ascending node along the Equator. This determines where the orbital plane intersects the Equator (depends on the time of year and day when launched). w, argument of perigee - the angle in degrees, measured in the direction and plane of the spacecraft’s motion, between the ascending node and the perigee point. This determines where the perigee point is located and therefore how the orbit is rotated in the orbital plane. Engin 176

Orbital Characteristics Orbit Description - 4 Orbital Characteristics From the 6 orbital elements other characteristics about the spacecraft orbit can be derived. Perigee radius (closest approach), rp = a (1-e) Apogee radius (farthest distance), ra = a (1+e) Orbital period, T = 2π (a3 / m) 1/2 For a circular orbit, rp = ra = rcir The constant orbital velocity can be found as: Vcir = (m / rcir)1/2 Beta angle is the angle in degrees between the Sun vector and the normal to the orbital plane. This angle is critical for power and thermal analysis to determine the amount of time in the the Sun and in eclipse behind the Earth. Engin 176

Orbit Description - 5 Example - A Shuttle Orbit (LEO / low inclination) Shuttle generally launches directly into a circular orbit from Kennedy Space Center (Latitude 28.4°). to an altitude of h=300 km (~ 162 nmi). Since the Shuttle goes directly into its orbit, the inclination is equal to the launch facility latitude, i = 28.4°. The orbit is circular, therefore the eccentricity is zero, e = 0. The altitude is the same throughout the orbit since it is circular (perigee and apogee radius are equal). rcir = rE + h = 6378.145 + 300 km rcir = 6678.145 km The circular radius is also the semi-major axis, a = rcir = 6678.145 km The orbital period is T = 2π (a3/µ)1/2 T = 2π[ (6678.145 km)3/ (398 601.2 km3/s2) ]1/2 T = 5431.02 s = 90.52 minutes The circular velocity is Vcir = (µ / rcir) 1/2 = [ (398 601.2 km3/s2) / (6678.145 km) ] 1/2 = 7.726 km/s (~ 15018 mph) Engin 176