at737 satellite orbits and navigation 1. at737 satellite orbits and navigation2 newton’s laws...

AT737

Satellite Orbits and Navigation 1

AT737 Satellite Orbits and Navigation 2

Newton’s Laws

1. Every body will continue in its state of rest or of uniform motion in a straight line except insofar as it is compelled to change that state by an impressed force.

2. The rate of change of momentum is proportional to the impressed force and takes place in the line in which the force acts.

3. Action and reaction are equal and opposite.


Newton's Second Law is the familiar

where F is force, m is mass, a is acceleration, v is velocity, and t is time.

dt

dvmmaF

Newton’s Laws (continued)


Newton’s Law of Universal Gravitation

The force of attraction between two point masses m1 and m2 separated by a distance r is

where G is the Newtonian (or universal) gravitation constant (6.67259 x 10-11 N m2 kg-

2).

221

r

mGmF


Circular Orbit Example

CentripetalForce 2

2

r

mGm

r

mv e GravitationalForce

-23 14 s m10 3.986005eGm

32

2 4r

GmT

e

Period v

rT

2 The NOAA satellites

orbit at about 850 km above the surface (r = 7228 km) and therefore have a period of about 102 minutes.


Kepler’s Laws1. All planets travel in elliptical paths with the sun

at one focus.

2. The radius vector from the sun to a planet sweeps out equal areas in equal times.

3. The ratio of the square of the period of revolution of a planet to the cube of its semimajor axis is the same for all planets revolving around the sun.

The same laws apply if we substitute satellite for planet and earth for sun, but the proportionality constant is different.


Ellipse Geometry

a = semimajor axis = eccentricity (0-1) = true anomalyr = radius

cos1

)1( 2

a

r Equation of an Ellipse


Kepler’s Equation

3

2

sin)(

a

Gm

Tn

eettnM

e

p

M = Mean anomalyn = mean motion constanttp = time of perigeal passagee = eccentric anomaly = eccentricity

Angles M, e, and θ are zero at perigee.

cos1

coscos

cos1

coscos

e

e

e

NOTE: All angles in radians.


Right Ascension & Declination

= declination = right ascension

Need a coordinate system to orient orbital plane in

space


Orientation Angles

i = inclination angle = argument of perigee = right ascension of ascending node

i < 90° progradei > 90° retrograde


Classical Orbital Elements

Element Symbol

Semimajor axis* a

Eccentricity

Inclination i

Argument of perigee o

Right ascension of ascending node

o

Mean anomaly** Mo

Epoch time to*Two-line elements give orbits per day instead of a**ESA uses true anomaly instead of mean anomaly


Sources of Orbital Elements

NOAA Satellite Information System

http://noaasis.noaa.gov/NOAASIS/ml/quicklook.html(current TBUS and TLEs for GOES and NOAA satellites)

T.S. Kelso’s CelesTrak sitehttp://celestrak.com(TLEs for a lot of satellites—still in business in spite of Space Track)

New Government Sitehttp://www.space-track.orgEstablished by Public Law 108-136, Section 913


Keplerian Orbits

Viewed from space, Keplerian orbits are constant and simple.

Viewed from a point rotating with the earth, Keplerian orbits are complex.


Orbit Perturbing Forces

Force Source

Nonspherical gravitational fieldNonspherical, nonhomogeneous Earth

Gravitational attraction of other bodies

Sun, moon, planets

Radiation pressure Solar radiation

Particle flux Solar wind

Lift and drag Residual atmosphere

Electromagnetic forces

Interaction of electrical currents in the satellite with Earth’s magnetic field


Perturbation Equations

2

2

2 sin312

11

r

rJ

r

GmU eee

i

a

rJnn

dt

dM ee 22/322

2 sin2

311

2

31

i

a

rJn

dt

d ee cos12

3 222

2

i

a

rJn

dt

d ee 2222

2 sin2

521

2

3

U is the gravitational potential energy

(g = -U)

ree = equatorial radius of Earth = 6,378,137 m

J2 = 1.08263 x 10-3

a, , and i are unperturbed

Anomalistic mean motion constant


…and more equations

nT

2

Anomalistic period—the time from perigee to moving perigee

The reciprocal of this is what you get in two-line elements in place of the semimajor axis

dtd

nT

2~ Synodic or nodal period—

the time from ascending node to ascending node


Where is that satellite?A step-by-step calculation guide

1. Find the orbital elements of the satellite you are interested in.

2. Update the variable elements (M, , & ) to the time (t ) that you are interested in:

M = Mo + (t – to)(dM/dt), etc.

3. Use Kepler’s equation to calculate the true anomaly ().

4. Use ellipse equation to calculate r, the distance of the satellite from the center of the earth.


Calculations continued…

5. Calculate the argument of latitude: +

(measures angular distance from equator).

6. Calculate latitude: = sin-1(sin sin i)

7. Calculate the right ascension of the satellite at time t:

cos

cossintan 1 i

sat

= right ascension of ascending node as calculated in step 2


Calculations completed

8. Calculate the right ascension of Greenwich (the prime meridian) at time t:

Greenwich = 99.965° + 360.985645 twhere t is the time in days (and fraction) between time t and 0000 UTC 1 January 2000.

9. Calculate the longitude of the satellite: = sat - Greenwich

Homework

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