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

ft/s = 7.3 km/s•! Super-circular velocities

–! Earth to Moon: 24,550 mph = 36,000 ft/s = 11.1 km/s

–! Escape: 25,000 mph = 36,600 ft/s = 11.3 km/s

•! 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( )

!xI t( ) = vx t( )!yI t( ) = vy t( )

where rI t( ) = xI2 t( ) + yI2 t( )

vx 0( ) = 7.5, 8, 8.5 km/svy 0( ) = 0x 0( ) = 0y 0( ) = 6,378 km = R

Dynamic Equations Example:Initial Conditions at Equator

18

19

Equatorial Orbits Calculated with Inverse-Square-Law Model

Work“Work” is a scalar measure of change in energy

With constant force,In one dimension

W12 = F r2 ! r1( ) = F"r

With varying force, work is the integral

W12 = FT drr1

r2

! = fxdx + fydy + fzdz( )r1

r2

! , dr =dxdydz

"

#

$$$

%

&

'''

In three dimensionsW12 = F

T r2 ! r1( ) = FT"r

20

Conservative Force!! Assume that the 3-D force

field is a function of position

F = F r( )

Iron FilingsAround Magnet

Force Emanating from Source

!! The force field is conservative if

FT r( )drr1

r2

! + FT r( )drr2

r1

! = 0

… for any path between r1 and r2 and back

21

Gravitational Force is Gradient of a Potential, V(r)

Gravity potential, V(r), is a function only of position

Fg = !m µE

rI3 rI =

""r

m µE

r#$%

&'(!

""rV r( )

22

Gravitational Force FieldGravitational force field

Gravitational force field is conservative because

… for any path between r1 and r2 and back

Fg = !m µE

rI3 rI

23

!!rV r( )drI

r1

r2

" # !!rV r( )drI

r2

r1

" =

# m µE

rI3 rI drI

r1

r2

" + m µE

rI3 rI drI

r1

r2

" = 0

Potential Energy in Gravitational Force Field

Potential energy, V or PE, is defined with respect to a reference point, r0

!PE !V r2( )"V r1( ) = " m µ

r2+V0

#$%

&'(+ m µ

r1+V0

#$%

&'(= "m µ

r2+m µ

r1

PE r0( ) =V r0( ) =V0 ! "U0( )

24

Kinetic EnergyApply Newton’s 2nd Law to the definition of Work

drdt

= v; dr = vdt F = m dvdt

W12 = FT drr1

r2

! = m dvdt

"#$

%&'T

vdtt1

t2

!

= 12m d

dtvTv( )dt

t1

t2

! = 12m d

dtv2( )dt

t1

t2

!

= 12mv2

t1

t2

= 12m v2 t2( )! v2 t1( )"# $%

= 12m v2

2 ! v12"# $% ! T2 !T1 ! &KE

Work = integral from 1st to 2nd timeT (= KE) is the kinetic energy of the point mass, m

25

Total Energy in Point-Mass Gravitational Field

!! Potential energy of mass, m, depends only on the gravitational force field

T ! KE = 1

2mv2

!! Total energy, E, is the sum of the two:

!! Kinetic energy of mass, m, depends only on the velocity magnitude measured in an inertial frame of reference

V = PE = !m µr

E = PE + KE

= !m µr+ 1

2mv2

= Constant

26

Interchange Between Potential and Kinetic Energy in a Conservative System

!m µr2+m µ

r1

"#$

%&'= 12mv2

2 ! 12mv1

2"#$

%&'

PE2 ! PE1 = KE2 ! KE1P1

P2

27

E 2 !E1 = 0

!m µr2+ 12mv2

2"#$

%&'! !m µ

r1+ 12mv1

2"#$

%&'= 0

Specific Energy…Energy per unit of the satellite’s mass

E S = PES + KES

= 1m

! mµr

+ 12mv2"

#$%&'

= ! µr+ 12v2

P1

P2

28

Angular Momentum of a Particle (Point Mass)

h = r !mv( ) = m r ! v( ) = m r ! !r( )29

Angular Momentum of a Particle

•! Moment of linear momentum of a particle–! Mass times components of the

velocity that are perpendicular to the moment arm

•! Cross Product: Evaluation of a determinant with unit vectors (i, j, k) along axes, (x, y, z) and (vx, vy, vz) projections on to axes

r ! v =i j kx y zvx vy vz

= yvz " zvy( )i + zvx " xvz( ) j + xvy " yvx( )k

h = r !mv( ) = m r ! v( )

30

Cross Product in Column Notation

r ! v =i j kx y zvx vy vz

= yvz " zvy( )i + zvx " xvz( ) j + xvy " yvx( )k

r ! v =

yvz " zvy( )zvx " xvz( )xvy " yvx( )

#

$

%%%%%

&

'

(((((

Column notation

Cross product identifies perpendicular components of r and v

31

=0 !z yz 0 !x!y x 0

"

#

$$$

%

&

'''

vxvyvz

"

#

$$$$

%

&

''''

Angular Momentum Vector is Perpendicular to Both Moment

Arm and Velocity

h = mr ! v = m

yvz " zvy( )zvx " xvz( )xvy " yvx( )

#

$

%%%%%

&

'

(((((

= m0 "z yz 0 "x"y x 0

#

$

%%%

&

'

(((

vxvyvz

#

$

%%%%

&

'

((((

= m!rv

32

Specific Angular Momentum Vector of a Satellite

hS =

mmr ! v = r ! v = r ! !r

… is the angular momentum per unit of the satellite’s mass, referenced to the center of attraction

Perpendicular to the orbital plane

33

Acceleration is

a t( ) = !v t( ) = !!r t( ) = ! µ

r2 t( )rI t( )r t( )

"

#$%

&'= ! µ

r3 t( ) r t( )

… or

!!r + µ

r3r = 0

Equations of Motion for a Particle in an Inverse-Square-Law Field

34

Cross Products of Radius and Radius Rate

Then

… because they are parallel

r ! !!r + µ

r3r"

#$%&'= 0

r ! r = 0 !r ! !r = 0

ddtr ! !r( ) = !r ! !r( ) + r ! !!r( ) = r ! !!r( )Chain Rule for Differentiation

35

Specific Angular Momentum

Consequently

r ! !!r + µ

r3r"

#$%&'= r ! !!r( ) + µ

r3r ! r( )

hS = Constant

hS ! h = r ! "r( ) (Perpendicular to the plane of motion)

Orbital plane is fixed in inertial space

= ddtr ! !r( ) = dhS

dt= 0

36

0

Eccentricity Vector is a Constant of Integration

With triple vector product identity (see Supplement)

!!r + µ

r3r!

"#$%&' h = !!r ' h+ µ

r3r ' h = 0

e = Eccentricity vector Constant of integration( )

!!r ! h = " µ

r3r ! h = " µ

r3r ! r ! !r( )

!!r ! h = " µ

r3r ! r ! !r( ) = " µ

r2!rr " r!r( ) = µ d

dtrr

#$%

&'(

Integrating

!!r ! h( )dt" = !r ! h = µ r

r+ e#

$%&'(

37

Significance of Eccentricity Vector

!r ! h" µ r

r+ e#

$%&'(

)*+

,-.

T

h = 0 because !r ! h" µ rr+ e#

$%&'(

)*+

,-.= 0

! !r " h( )T h# µrTh

r# µeTh = 0

!"µeTh = 0

0 0

!! e is perpendicular to angular momentum,

!! which means it lies in the orbital plane!! Its angle provides a reference direction

for the perigee38

General Polar Equation of a Conic Section

rT !r ! h" µ r

r+ e#

$%&'(

)*+

,-.= 0

39

1st term is angular momentum squared

rT !r ! h( ) = hT r ! !r( ) = hTh = h2

Then

h2 ! µ rTrr

+ rTe"#$

%&'= 0 h2 = µ r + rTe( ) = µ r + recos!( ) = 0

r = h2 µ1+ ecos!

r = p1+ e cos!

= h2 µ1+ e cos!

, m or km

! : True Anomaly =Angle from perigee direction, deg or rad

40

Elliptical Planetary Orbits!! Assume satellite mass is negligible

compared to Earth’s mass!! Then

!! Center of mass of the 2 bodies is at Earth’s center of mass

!! Center of mass is at one of ellipse’s focal points

!! Other focal point is “vacant”

Properties of Elliptical Orbits

e =ra ! rpra + rp

=ra ! rp2a

rp = a / 1! e( ) ra = a / 1+ e( )

a =ra + rp2

Semi-major axis is the average of the two

41

Eccentricity can be determined from apogee and perigee radii

Properties of Elliptical Orbits

b = rarp

p = h2 µ = a 1! e2( )

A = ! a b = !a2 1" e2 , m2

!! Semi-latus rectum, p, can be expressed as a function of h or a and e

!! Semi-minor axis, b, can be expressed as a function of ra and rp

!! Area of the ellipse, A, is

42

!rp = 0 and v = rp !! p

E S " E = 12rp !! p( )2

" µrp

= 12h2

rp2 " µ

rp

At the periapsis, rp

Energy is Inversely Proportional to the Semi-Major Axis

43

p = h2 µ = a 1! e2( )

rp = a 1! e( )

E = 12rp

2 µp ! 2µrp( ) = µ2a 1! e( )2

1! e2( )! 2 1! e( )"# $%

= !µ 1! e( )2a 1! e( )

E = ! µ

2a

Classification of Conic Section Orbits

44

Orbit Shape Eccentricity, e Energy, E! Semi-Major Axis, a

Semi-Latus Rectum, p

Circle 0 < 0 > 0 a

Ellipse 0 < e < 1 < 0 > 0 a(1– e2)

Parabola 1 0 Undefined("!)

2rp

Hyperbola >1 > 0 < 0 a(1– e2)

!! Specific total energy, E, is inversely proportional to the semi-major axis

12v2 = µ

r+E

v = µ 2r! 1a

"#$

%&'

Velocity is a function of radius and specific energy

E = ! µ

2a

“Vis Viva (Living Force) Integral”

!! Velocity is a function of radius and semi-major axis

v = 2 µ

r+E!

"#$%&

45

vp = µ 2rp! 1a

"

#$%

&'= µ 2

a 1! e( ) !1a

"#$

%&'

= µa1+ e( )1! e( )

Velocity at periapsis

Maximum and Minimum Velocities on an Ellipse

46

va = µ 2ra! 1a

"#$

%&'= µ

a1! e( )1+ e( )

Velocity at apoapsis

vp = µ 2rp! 1a

"

#$%

&'

Hyperbola (a < 0)

Velocities at Periapses and Infinity of Parabola and Hyperbola

47

vp =2µrp

Parabola (a "" ")

v! = 0

v! = " µa

Relating Time and Position in Elliptical Orbit

Rearrange and integrate angular momentum over time and angle

h = µp = r2 d!dt

= 2 d Ellipse Area( )dt

48

dtto

t f

! = r2

µpd"

"o

" f

! = p3

µd"

1+ ecos"( )2"o

" f

!

… but difficult to integrate analytically

Anomalies of an Elliptical OrbitAngles measured from last periapsis

49

! (or " ): True AnomalyE (or # ): Eccentric Anomaly

M : Mean Anomaly

50

Kepler’s Equation for the Mean AnomalyM = E ! esinE

cosE =ae+ p ! r( ) e

a= a ! r

ae

r = a 1! cosE( )!r = !E aesinE

From the diagram,

51

Relationship of Time to Eccentric Anomaly

E = ! µ2a

= 12!r2 + h

2

r2

"#$

%&'! µr

where h2 = µa 1! e2( )

then r2 !r2 = µ 2r! 1a

"#$

%&' r

2 ! a 1! e2( )()*

+,-

leading to a3

µ1! ecosE( )2 dE

dt"#$

%&'

2

= 1

or

µa3dt = 1! ecosE( )dE

52

Integrating the Prior Result …µa3dt

t0

t1

! = 1" ecosE( )dEE0

E1

!µa3

t1 " t0( ) = E " esinE( ) E0E1 = M1 "M 0

t1 ! t0( ) = M1 !M 0

µ a3 or t1 = t0 +

M1 !M 0

µ a3

µa3

P ! 0( ) = E ! esinE( ) 02"

= 2"

P = 2" a3 µ

Orbital Period, P

Time is proportional to Mean Anomaly

Orbital Period

Orbital period is related to the total energy

P = 2! a3

µ= ! "µ2

2E 3 where E < 0 for an ellipse

53

Mean Motion, n, is the inverse of the Period

P = 2! a3

µ!

2!n

where n is the Mean Motion

Position and Velocity in Orbit at Time, tMean Anomaly, given time from perigee passage, tp

54

M t( ) = µa3

t ! t p( )Eccentric Anomaly, E(t), given Mean Anomaly, M(t)

E t( )! esinE t( ) = M t( )

Eo t( ) = M t( ) + esinM t( )Iterate until !Mi < Tolerance

!Mi = M t( )" Ei t( )" esinEi t( )#$ %&!Ei+1 = !Mi 1" ecosEi t( )#$ %&

Ei+1 t( ) = Ei t( ) + !Ei+1

Newton’s method of successive approximation, using M(t) as starting guess for E(t)

Position and Velocity in Orbit at Time, tCalculate True Anomaly, given Eccentric Anomaly

55

! t( ) = 2 tan"1 1+ e1" e

tan E t( )2

#

$%

&

'(

Compute magnitude of radius

r t( ) = a 1! e2( )1+ ecos" t( )

Position and Velocity in Orbit at Time, t

56

Radius vector, in the orbital plane

r t( ) =x t( )y t( )

!

"##

$

%&&=

r t( )cos' t( )r t( )sin' t( )

!

"##

$

%&&

Velocity vector, in the orbital plane

v t( ) =vx t( )vy t( )

!

"

##

$

%

&&= µ

p'sin( t( )e+ cos( t( )

!

"##

$

%&&

see Weisel, Spaceflight Dynamics, 1997, pp. 64-66

!!eexxtt TTiimmee::""llaanneettaarryy DDeeffeennssee

57

##uupppplleemmeennttaall MMaa$$rriiaall

58

First Point of Aries!(Ecliptic Intercept at Right)"

59

Dimension of energy?"

Dimension of linear momentum?"

Dimension of angular momentum?"

Scalar (1 x 1)"

Vector (3 x 1)"

Vector (3 x 1)"60

Sub-Orbital (Sounding) Rockets 1945 - Present

NASA Wallops Island Control Center

CanadianBlack Brant XII

LTVScout

61

MATLAB Code for Flat-Earth Trajectories

Script for Analytic Solution

g = 9.8; t = 0:0.1:40; vx0 = 10; vz0 = 100; x0 = 0; z0 = 0; vx1 = vx0; vz1 = vz0 – g*t; x1 = x0 + vx0*t; z1 = z0 + vz0*t - 0.5*g*t.* t;

Script for Numerical Solution

tspan = 40; % Time span, sxo = [10;100;0;0]; % Init. Cond.[t1,x1] = ode45('FlatEarth',tspan,xo);

Function for Numerical Solution

function xdot = FlatEarth(t,x) % x(1) = vx% x(2) = vz% x(3) = x% x(4) = z g = 9.8; xdot(1) = 0; xdot(2) = -g; xdot(3) = x(1); xdot(4) = x(2); xdot = xdot';end 62

Trajectories Calculated with Flat-Earth Model

!! Constant gravity, g, is the only force in the model, i.e., no aerodynamic or thrust force

!! Can neglect motions in the y direction

vx t( ) = vx0

!vz t( ) = !g z positive up( )!x t( ) = vx t( )!z t( ) = vz t( )

vx 0( ) = vx0vz 0( ) = vz0x 0( ) = x0z 0( ) = z0

Dynamic Equations

Initial Conditions

vx T( ) = vx0vz T( ) = vz0 ! gdt

0

T

" = vz0 ! gT

x T( ) = x0 + vx0T

z T( ) = z0 + vz0T ! gt dt0

T

" = z0 + vz0T ! gT 2 2

Analytic (Closed-Form) Solution

63

Trajectories Calculated with Flat-Earth Model

vx 0( ) = 10m / svz 0( ) = 100,150, 200m / sx 0( ) = 0z 0( ) = 0

64

MATLAB Code for Spherical-Earth Trajectories

Script for Numerical Solution R = 6378; % Earth Surface Radius, km tspan = 6000; % seconds options = odeset('MaxStep', 10) xo = [7.5;0;0;R]; [t1,x1] = ode15s('RoundEarth',tspan,xo,options); for i = 1:length(t1) v1(i) = sqrt(x1(i,1)*x1(i,1) + x1(i,2)*x1(i,2)); r1(i) = sqrt(x1(i,3)*x1(i,3) + x1(i,4)*x1(i,4)); end

function xdot = RoundEarth(t,x)% x(1) = vx% x(2) = vy% x(3) = x% x(4) = y mu = 3.98*10^5; % km^2/s^2 r = sqrt(x(3)^2 + x(4)^2); xdot(1) = -mu * x(3) / r^3; xdot(2) = -mu * x(4) / r^3; xdot(3) = x(1); xdot(4) = x(2); xdot = xdot';end

Function for Numerical Solution

65

x2

a2 +y2

b2 = 1

a : Semi - major axis, m or kmb : Semi - minor axis, m or km

x !( ) = a cos !( )y !( ) = b sin !( )

! : Angle from x - axis (origin at center) rad

Equations that Describe Ellipses

a ob

yx

66

Constructing Ellipses

F1P1 + F2P1 = F1P2 + F2P2 = 2a

String, Tacks, and Pen

HypotrochoidArchimedes Trammel

Foci (from center),

x fy f

!

"##

$

%&&

1,2

= ' a2 ' b2

0

!

"##

$

%&&, a2 ' b2

0

!

"##

$

%&&

67

Ellipses

Eccentricity, e = 1! b2

a2

b2

a2= 1! e; b = a 1! e

Semi - latus rectum ("The Parameter"),

p = b2

a, m

68

How Do We Know that Gravitational Force is Conservative?

Because the force is the derivative (with respect to r) of a scalar function of r called

the potential, V(r):

V r( ) = !m µr+Vo = !m µ

rTr( )1 2+Vo

!V r( )!r

=!V !x!V !y!V !z

"

#

$$$

%

&

'''= m µ

r3

xyz

"

#

$$$

%

&

'''= (Fg

This derivative is also called the gradient of V with respect to r 69

Conservation of EnergyEnergy is conserved in an elastic collision, i.e. no

losses due to friction, air drag, etc.“Newton’s Cradle” illustrates interchange of

potential and kinetic energy in a gravitational field

70

Examples of Circular Orbit Periods for Earth and Moon

Period, minAltitude above

Surface, km Earth Moon0 84.5 108.5

100 86.5 1181000 105.1 214.610000 347.7 1905

71

Typical Satellite Orbits

Sun-Synchronous

Orbit

GPS Constellation

26,600 km

72

Geo-Synchronous Ground Track

Geo-Synchronous Ground Track

42, 164 km

Marco Polo-1 & 2

73

Background Math"

a ! b ! c( ) " a i c( )b # a ib( )c= aTc( )b # aTb( )c

Triple Vector Product Identity"

rT !r = r i !r = r dr

dt

Dot Product of Radius and Rate "

74