Emerging Spacecraft Technologies and Applications

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Transcript Emerging Spacecraft Technologies and Applications

(Re) Orientation

1 - Introduction2 - Propulsion & ∆V3 - Attitude Control &

instruments

4

- Orbits

& Orbit Determination

LEO, MEO, GTO, GEOSpecial LEO orbitsOrbit TransferGetting to OrbitGPS

5 - Launch Vehicles6 - Power & Mechanisms

Engin 176

7 - Radio & Comms8 - Thermal / Mechanical

Design. FEA

9 - Reliability10 - Digital & Software11 - Project Management

Cost / Schedule

12 - Getting Designs Done13 - Design Presentations

Sporadic Events: •Mixers •Guest Speakers •Working on Designs •Teleconferencing

5.1

Review of Last time

Attitude Determination & Control

Feedback Control Systems descriptionSimple simulationAttitude StrategiesThe simple lifeEight other approaches and variationsDisturbance and Control forcesDesign build & test an Attitude Control System Set point

Design Activity

Team designations Control Actuator Algorithm Plant (satellite) Engin 176 Sensor 5.2

LEO vs. GEO Orbit

• LEO: 1000 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 5.3

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L E O

• • • • • • • •

LEO & GEO Pros / Cons

Pros 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

• • • • • • • •

Cons 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 G E O

• • • •

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

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Orbit Ground Rules

• #1: Orbits are Dynamic.

– There are no w 2 “ stationary and potential energy - centripetal acceleration vs. gravitational acceleration. I.e. r + Mg/r 2 ” orbits. They are a dynamic balance of kinetic = constant and w a v/r so v 2 a l/r => w a l/ r 3/2 => t a r 3/2 also, v = max @ perigee, min @ apogee • #2: Orbit plane bisects earth balance centripetal acceleration.

because gravity is radial and it must • #3: Orbits are not tracks velocity fully determines orbit - knowing instantaneous position and • #4: Orbit plane wrt sun changes slowly if at all.

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Orbits are Dynamic.

Ground Rule #1 means...

Engin 176 • Orbit period is shorter for lower orbits since: w 2 r + Mg/r 2 = constant w a and v/r so v 2 a 1/r => w a 1/ r 3/2 => t a r 3/2 • For an eliptic orbit, V is maximum @ perigee, min @ apogee • Geosynchs don ’ t just “ hang ” there 5.6

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 5.7

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Orbits are not tracks

Ground Rule #3 means...

• Changing Orbit: where new & old orbit intersect, change V to that appropriate to new orbit

A

• If present and desired orbit don ’ t intersect: join them via an intermediate that does

1 B 2 3

• 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 + ª = ª 5.8

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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.

5.9

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Getting to (LEO) Orbit

• #1: Raise altitude from 0 to 300 km (this is the easy part) – Energy = mgh = 100kg x 9.8 m/s 2 x 300,000 m = 2.94 x 108 kg m ∆V = (E)1/2 = 1715 m/s 2 / s 2 [=W-s = J] = 82 kw-hr = 2.94 x 106 m2/s2 per kg • #2: Accelerate to orbital velocity, 7 km/s (the harder part) – ∆V (velocity) – ∆V (altitude) – ∆V (total) = 7000 m/s (80% of V, 94% of energy) = 1715 m/s = 8715 m/s Engin 176 5.10

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/s 2 ∆V = (E)1/2 = 1686 m/s x (300,000 m - 10,000 m) (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) – ∆V (∆H) = 6800 m/s (97% of ground ∆V, 99% of energy) = 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 5.11

Common

µ

spacecraft 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.

5.12

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∆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) ª + ª = ª • Typical orbit maintenance: 100 to 300 m/s (sun synch or geosynch, per year) Engin 176 5.13

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 /r 3 )

r

= 0 where m = 398 601.2 km3 / sec2 (gravitational parameter for Earth) r = r E r E + h (mean radius of Earth plus altitude) = 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 5.14

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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-e 2 ) / (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 ellipse parabola hyperbola e = 0 0< e < 1 e = 1 e > 1 a = radius a > 0 a = ∞ 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).

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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.

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Orbit Description

- 4

Orbital Characteristics

From the 6 orbital elements other characteristics about the spacecraft orbit can be derived.

Perigee radius (closest approach), r p Apogee radius (farthest distance), r Orbital period, T = 2π (a 3 / m ) 1/2 a = a (1-e) = a (1+e) For a circular orbit, r p = r a = r cir The constant orbital velocity can be found as: V cir = ( m / r cir ) 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 5.17

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).

r cir r cir = r E + h = 6378.145 + 300 km = 6678.145 km The circular radius is also the semi-major axis, The orbital period is a = r cir = 6678.145 km T = 2π (a 3 / µ ) 1/2 T = 2π[ (6678.145 km) 3 / (398 601.2 km 3 /s 2 ) ] 1/2 T = 5431.02 s = 90.52 minutes The circular velocity is V cir = ( µ / r cir ) 1/2 = [ (398 601.2 km 3 /s 2 ) / (6678.145 km) ] 1/2 = 7.726 km/s (~ 15018 mph) Engin 176 5.18