Orbital Mechanics Space System Design, MAE 342, Princeton University Robert Stengel Copyright 2016 by Robert Stengel. All rights reserved. For educational use only. http://www.princeton.edu/~stengel/MAE342.html Conic section orbits Equations of motion Momentum and energy Kepler’s Equation Position and velocity in orbit 1 Orbits 101 Satellites Escape and Capture (Comets, Meteorites) 2
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Orbital Mechanics!Space System Design, MAE 342, Princeton University!
Robert Stengel
Copyright 2016 by Robert Stengel. All rights reserved. For educational use only.http://www.princeton.edu/~stengel/MAE342.html
Conic section orbitsEquations of motion
Momentum and energyKepler’s Equation
Position and velocity in orbit
1
Orbits 101 Satellites Escape and Capture
(Comets, Meteorites)
2
Two-Body Orbits are Conic Sections
3
Classical Orbital ElementsDimension and Time
Orientation
a : Semi-major axise : Eccentricity
t p: Time of perigee passage
! :Longitude of the Ascending/Descending Nodei : Inclination of the Orbital Plane
" : Argument of Perigee
4
Orientation of an Elliptical Orbit
5
!!iirrsstt PPooiinn"" ooff AArriieess
Orbits 102(2-Body Problem)
•! e.g., –! Sun and Earth or–! Earth and Moon or–! Earth and Satellite
•! Circular orbit: radius and velocity are constant•! Low Earth orbit: 17,000 mph = 24,000
•! Near escape velocity, small changes have huge influence on apogee
6
!! Particle of fixed mass (also called a point mass) acted upon by a force changes velocity with !! acceleration proportional to and in direction of force
!! Inertial reference frame !! Ratio of force to acceleration is the mass of the
particle: F = m a
Newton’s 2nd Law
ddt
mv t( )!" #$ = mdv t( )dt
= ma t( ) = F
F =
fxfyfz
!
"
%%%%
#
$
&&&&
= force vectorm ddt
vx t( )vy t( )vz t( )
!
"
####
$
%
&&&&
=
fxfyfz
!
"
####
$
%
&&&&
7
Equations of Motion for a Particle
dv t( )dt
= !v t( ) = 1mF =
fx mfy m
fz m
!
"
####
$
%
&&&&
=
axayaz
!
"
####
$
%
&&&&
Integrating the acceleration (Newton’s 2nd Law) allows us to solve for the velocity of the particle
v T( ) = dv t( )dt
dt0
T
! + v 0( ) = a t( )dt0
T
! + v 0( ) = 1mFdt
0
T
! + v 0( )
vx T( )vy T( )vz T( )
!
"
####
$
%
&&&&
=
ax t( )ay t( )az t( )
!
"
####
$
%
&&&&
dt0
T
' +
vx 0( )vy 0( )vz 0( )
!
"
####
$
%
&&&&
=
fx t( ) mfy t( ) mfz t( ) m
!
"
####
$
%
&&&&
dt0
T
' +
vx 0( )vy 0( )vz 0( )
!
"
####
$
%
&&&&
3 components of velocity
8
Equations of Motion for a Particle
dr t( )dt
= !r t( ) = v t( ) =!x t( )!y t( )!z t( )
!
"
####
$
%
&&&&
=
vx t( )vy t( )vz t( )
!
"
####
$
%
&&&&
Integrating the velocity allows us to solve for the position of the particle
r T( ) = dr t( )dt
dt0
T
! + r 0( ) = vdt0
T
! + r 0( )
x T( )y T( )z T( )
!
"
####
$
%
&&&&
=
vx t( )vy t( )vz t( )
!
"
####
$
%
&&&&
dt0
T
' +
x 0( )y 0( )z 0( )
!
"
####
$
%
&&&&
3 components of position
9
Spherical Model of the Rotating Earth
RE =xoyozo
!
"
###
$
%
&&&E
=cosLE cos'E
cosLE sin'E
sinLE
!
"
###
$
%
&&&R
Spherical model of earth s surface, earth-fixed (rotating) coordinates
LE : Latitude (from Equator), deg!E : Longitude (from Prime Meridian), degR : Radius (from Earth's center), deg
Earth's rotation rate, ! , is 15.04 deg/hr10
Non-Rotating (Inertial) Reference Frame for the Earth
Celestial longitude, !C, measured from First Point of Aries on the
Celestial Sphere at Vernal Equinox
!C = !E +" t # tepoch( ) = !E +" $t
11
Transformation Effects of Rotation
RE =cos!"t sin!"t 0#sin!"t cos!"t 0
0 0 1
$
%
&&&
'
(
)))R I =
cos!"t sin!"t 0#sin!"t cos!"t 0
0 0 1
$
%
&&&
'
(
)))
xoyozo
$
%
&&&
'
(
)))I
rE =cosLE cos!E
cosLE sin!E
sinLE
"
#
$$$
%
&
'''R + Altitude( ); rI =
cosLE cos!CcosLE sin!CsinLE
"
#
$$$
%
&
'''R + Altitude( )
Transformation from inertial frame, I, to Earth’s rotating frame, E
Location of satellite, rotating and inertial frames
Orbital calculations generally are made in an inertial frame of reference
12
Gravity Force Between Two Point Masses, e.g., Earth and Moon
Magnitude of gravitational attraction
G : Gravitational constant = 6.67 !10"11Nm2 /kg2
m1 : Mass of 1st body = 5.98 !1024 kg for Earthm2 : Mass of 2ndbody = 7.35 !1022 kg for Moonr : Distance between centers of mass of m1 and m2, m
F = Gm1m2
r2
13
Acceleration Due To Gravity
F2 = m2a1on2 =Gm1m2
r2
a1on2 =Gm1
r2!µ1r2
At Earth’s surface, acceleration due to gravity is
ag ! goEarth =
µE
Rsurface2 = 3.98 !10
14 m3 s2
6,378,137m( )2= 9.798m s2
µ1 = Gm1 Gravitational parameter of 1st mass
14
“Inverse-square Law”
Gravitational Force Vector of the Spherical Earth
Force always directed toward the Earth’s center
rIrI
=
xyz
!
"
###
$
%
&&&I
xI2 + yI
2 + zI2=
cosLE cos'I
cosLE sin'I
sinLE
!
"
###
$
%
&&&
Fg = !m µE
rI2rIrI
"
#$%
&'= !m µE
rI3
xyz
(
)
***
+
,
---I
(vector), as rI = rI
(x, y, z) establishes the direction of the local vertical
15
Equations of Motion for a Particle in an Inverse-Square-Law Field
dv t( )dt
= !v t( ) = 1mFg = ! µE
rI2rIrI
"
#$%
&'= ! µE
rI3
xyz
(
)
***
+
,
---
Integrating the acceleration (Newton’s 2nd Law) allows us to solve for the velocity of the particle
v T( ) = dv t( )dt
dt0
T
! + v 0( ) = a t( )dt0
T
! + v 0( ) = 1mFdt
0
T
! + v 0( )
vx T( )vy T( )vz T( )
!
"
####
$
%
&&&&
= 'µE
x rI3
y rI3
z rI3
!
"
####
$
%
&&&&
dt0
T
( +
vx 0( )vy 0( )vz 0( )
!
"
####
$
%
&&&&
3 components of velocity
16
Equations of Motion for a Particle in an Inverse-Square-Law Field
dr t( )dt
= !r t( ) = v t( ) =!x t( )!y t( )!z t( )
!
"
####
$
%
&&&&
=
vx t( )vy t( )vz t( )
!
"
####
$
%
&&&&
As before; Integrating the velocity allows us to solve for the position of the particle
r T( ) = dr t( )dt
dt0
T
! + r 0( ) = vdt0
T
! + r 0( )
x T( )y T( )z T( )
!
"
####
$
%
&&&&
=
vx t( )vy t( )vz t( )
!
"
####
$
%
&&&&
dt0
T
' +
x 0( )y 0( )z 0( )
!
"
####
$
%
&&&&
3 components of position
17
Dynamic Model with Inverse-Square-Law Gravity
No aerodynamic or thrust forceNeglect motions in the z direction
msatellite << mEarth
!vx t( ) = !µE xI t( ) rI 3 t( )!vy t( ) = !µE yI t( ) rI 3 t( )