《全球定位系统原理》（英文版）Lecture 5 Satellite Orbits

12. 540 Principles of the Global Positioning System Lecture 05 Prof. Thomas Herring 02/20/02 12.540Lec05

02/20/02 12.540 Lec 05 1 12.540 Principles of the Global Positioning System Lecture 05 Prof. Thomas Herring

Satellite Orbits Treat the basic description and dynamics of satellite orbits Major perturbations on GPS satellite orbits Sources of orbit information SP3 format from the International gPs service Broadcast ephemeris message Accuracy of orbits and health of satellites Logistics: Who can attend lecture on Fridays at11-12:30? 02/20/02 12.540Lec05

02/20/02 12.540 Lec 05 2 Satellite Orbits • Treat the basic description and dynamics of satellite orbits • Major perturbations on GPS satellite orbits • Sources of orbit information: – SP3 format from the International GPS service – Broadcast ephemeris message • Accuracy of orbits and health of satellites • Logistics: Who can attend lecture on Fridays at 11-12:30?

Dynamics of satellite orbits Basic dynamics is described by f=Ma where the force, F, is composed of gravitational forces, radiation pressure(drag is negligible for GPS), and thruster firings(not directly modeled) Basic orbit behavior is given by 酩=GMr 02/20/02 12.540Lec05

02/20/02 12.540 Lec 05 3 Dynamics of satellite orbits • Basic dynamics is described by F=Ma where the force, F, is composed of gravitational forces, radiation pressure (drag is negligible for GPS), and thruster firings (not directly modeled). • Basic orbit behavior is given by Ý r Ý = −GM e r 3 r

Simple dynamics GMe=μ=3986006X106m3s2 The analytical solution to the central force model is a Keplerian orbit. For gps these are elliptical orbits Mean motion, n, in terms of period P is given by 2丌 For gPs semimajor axis a- 26400km 02/20/02 12.540Lec05

02/20/02 12.540 Lec 05 4 Simple dynamics • G M e = µ = 3986006x10 8 m 3 s-2 • The analytical solution to the central force model is a Keplerian orbit. For GPS these are elliptical orbits. • Mean motion, n, in terms of period P is given by • For GPS semimajor axis a ~ 26400km n = 2 π P = µ a 3

Solution for central force model This class of force model generates orbits that are conic sections. We will deal only with closed elliptical orbits The orbit plane stays fixed in space One of the foci of the ellipse is the center of mass of the body These orbits are described Keplerian elements 02/20/02 12.540Lec05

02/20/02 12.540 Lec 05 5 Solution for central force model • This class of force model generates orbits that are conic sections. We will deal only with closed elliptical orbits. • The orbit plane stays fixed in space • One of the foci of the ellipse is the center of mass of the body • These orbits are described Keplerian elements

Keplerain elements: Orbit plane Satellite perigee 00 Verna equator equinox Node Greenwich i Inclination Q2 Right Ascension of ascending node o Argument of perigee v True anomaly 02/20/02 12.540Lec05

02/20/02 12.540 Lec 05 6 Keplerain elements: Orbit plane Node i ω Ω ν Z θ 0 Greenwich Vernal equinox Satellite perigee equator i Inclination Ω Right Ascension of ascending node ω Argument of perigee ν True anomaly

Keplerain elements in plane Satellite Apogee a ae vEv Perigee Focus Center of Mass a semimajor axIs V True anomaly b semiminor axis E Eccentric anomaly e eccentricity M Mean anomaly 02/2002 12.540Lec05

02/20/02 12.540 Lec 05 7 Keplerain elements in plane a Focus Center of Mass ae Satellite Apogee Perigee b E ν r a semimajor axis b semiminor axis e eccentricity ν True anomaly E Eccentric anomaly M Mean anomaly

Satellite motion The motion of the satellite in its orbit is given by M(1)=n(t-70) E(t=M(t+esin e(t) vt)=tan e sin e(t/(1-ecose(t) (cos e(t-e/(1-ecose(t)) To is time of perigee 02/20/02 12.540Lec05

02/20/02 12.540 Lec 05 8 Satellite motion • The motion of the satellite in its orbit is given by • T o is time of perigee M ( t ) = n ( t − T0 ) E ( t ) = M ( t ) + esin E ( t ) ν( t ) = tan −1 1 − e 2 sin E ( t)/ ( 1 − ecos E ( t)) (cos E ( t ) − e)/ ( 1 − ecos E ( t)) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥

True anomaly Difference between true anomaly and mean anomaly fore0.001-0.100 -005 -0.1 -0.15 -025 15 3.5 02/20/02 12.540Lec05

02/20/02 12.540 Lec 05 9 True anomaly 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 x 104 -0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 Difference between true anomaly and Mean anomaly for e 0.001-0.100

Eccentric anomaly Difference between eccentric anomaly and mean anomaly 0.15 fore0.001-0.100 0.1 -0.05 -0.15 -0.25 15 2.5 3.5 02/20/02 12.540Lec05

02/20/02 12.540 Lec 05 10 Eccentric anomaly 0 0.5 1 1.5 2 2.5 3 3.5 4 x 104 -0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 Difference between eccentric anomaly and Mean anomaly for e 0.001-0.100