orbital mech. chapter 1

7/24/2019 Orbital Mech. Chapter 1

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1ME 461 / 561 - Chapter 1

ME 461 / 561 Orbital Mechanics

Chapter 1 - The Two-Body Problem

Professor Christopher D. Hall


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2ME 461 / 561 - Chapter 1

Keplers Laws:

First Law (1609):

The orbit of each planet is

an ellipse with the sun at a focus.

Second Law (1609):

The line joining the planet to the sun

sweeps out equal areas in equal time.

Third Law (1619):

The square of the period of a planet is

proportional to the cube of its mean

distance from the sun.

These laws were formed for motion of planets about the sun, butalso apply to artificial satellites orbiting the sun, planets, or moons

Keplers Laws:

First Law (1609):

The orbit of each planet is

an ellipse with the sun at a focus.

Second Law (1609):

The line joining the planet to the sun

sweeps out equal areas in equal time.

Third Law (1619):

The square of the period of a planet is

proportional to the cube of its mean

distance from the sun.

These laws were formed for motion of planets about the sun, butalso apply to artificial satellites orbiting the sun, planets, or moons

FocusFocus

P2 a3


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3ME 461 / 561 - Chapter 1

Newtons Laws (Principia, 1687):

First Law:

Every body continues in its state of rest or of uniform motion in a

straight line unless it is compelled to change that state by forcesimpressed upon it.

Second Law:

Third Law:

Universal Gravitational Law:

Newtons Laws (Principia, 1687):

First Law:

Every body continues in its state of rest or of uniform motion in a

straight line unless it is compelled to change that state by forcesimpressed upon it.

Second Law:

Third Law:

Universal Gravitational Law:

- ~F

m1 m2

~F

~F id(m~iv)dt

m

M

~r~Fgm=

GMm

r2~r

r


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4ME 461 / 561 - Chapter 1

The N-Body Problem

~Fother =

~FDrag+

~FThrust+

~Fsolarpressure +

~Fperturbe+ . . .

Applying Newtons 2nd Law for the jth particle: Applying Newtons 2nd Law for the jth particle:

~rjn,

~rn ~rjStart

point

Start

pointEnd

point

End

point

Z

Y

X

O ~Fother

~rj

mj

~Fgj

mn

P~F =

id(mij~vj)

dt

~Fgj = GmjnX

k=1

mkr3kj

~rkj

~rn

m2

m1 ~rjn


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5ME 461 / 561 - Chapter 1

The N-Body Problem

~Fother| {z}0

+~Fgj =mj

id(i~vj)

dt =mj

id2(i~rj)

dt2

Gravitational forcebetween earthand satellite

Gravitational forcebetween earthand satellite

Perturbing effects ofsun, moon,

Perturbing effects ofsun, moon,

~rj =

G

nXk=1

mk

r

3

kj

~rkj

~r12 = G(m1+ m2)

r312~r12

| {z }

nXk=3

Gmk~rk2

r3k2 ~rk1

r3k1

| {z }


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6ME 461 / 561 - Chapter 1

The Two-Body Problem

Assumptions

The bodies are spherically

symmetric (we can assumeeach as a point mass).

The only force is thegravitational force, alongthe line joining the centers

of two bodies.

Assumptions

The bodies are spherically

symmetric (we can assumeeach as a point mass).

The only force is thegravitational force, alongthe line joining the centers

of two bodies.

Z

X0

m

X

O0

Y0

Z0

Y

~rm

~rM

~r

M

~r=~rm ~rM

Differential Eq. of therelative motion

Differential Eq. of therelative motion

Nonrotating FrameNonrotating Frame

Inertial FrameInertial Frame

m~rm =

GMm

r3 ~r

~rm =

GM

r3 ~r

M~rM= GMm

r3 ~r~rM= Gmr3 ~r~r= G(M+m)r3 ~r


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7ME 461 / 561 - Chapter 1

G(M+ m) GM

When ( M>>m) i.e. , artificial satellites, space probes, ballistic missiles, When ( M>>m) i.e. , artificial satellites, space probes, ballistic missiles,

Constants (Integrals) of the Motion:

Conservation of mechanical energy

Conservation of angular momentum

Constants (Integrals) of the Motion:

Conservation of mechanical energy

Conservation of angular momentum

Equation of motion of TBP

E=T+ V

PotentialenergyPotentialenergyKinetic

energyKineticenergy

MechanicalenergyMechanicalenergy

~h= ~r ~v

~r+ r3 ~r=~0


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8ME 461 / 561 - Chapter 1

Conservation of Mechanical Energy

Lets start with the EOM:Lets start with the EOM:

Using the fact thatUsing the fact that

Then:Then:

vv+ r2 r= 0

which can be written as:which can be written as:

ddt

v2

2

+ ddt

c r

= 0

E= v2

2 +

c r

~r+

r3~r=~0

~r ~r+ r3 ~r ~r= 0

~r ~r= rr


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9ME 461 / 561 - Chapter 1

Conservation of Mechanical Energy

Kinetic energy perunit mass of satelliteKinetic energy perunit mass of satellite

Potential energy per

unit mass of satellite

Potential energy per

unit mass of satellite

Specific mechanical

energy

Specific mechanical

energy

If:If:

c= 0 limrV(r) = 0c= rM V(rM) = 0

E= v2

2 r

If:If:

E= v2

2 +

c r

| {z }

Vis-viva (living force) Eq.:Vis-viva (living force) Eq.:


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10ME 461 / 561 - Chapter 1

Matlab Function: rv2E.m

functionE=rv2E(R,V,mu)

%RV2E FunctiontocalculatethespecificmechanicalenergyofaKeplerian orbit.

% Theinputsofrv2Earethepositionvectorofthesatellite,R,thevelocity

% vectorofthesatellite,V,andthegravitationalparameter ofthecentral

% body,mu. Theoutputisthespecificmechanicalenergy,E.

%

%

Theposition

and

velocity

vectors

must

be

either

3x1

or

1x3 matrices,

and

% thefunctionwillresizebothofthemsothattheyare3x1. Ifavalue

% formuisnotinputed,thedefaultvalueisthatforEarth(3.986e5km^3/s^2)

%Determineifinputvectorsarethecorrectsizeandresizethem

iflength(R)~=3||length(V)~=3

error('Inputvectorsarenotthecorrectsize. Pleaseinputnewvectors.')

end

R=[R

(1);

R

(2);

R

(3)];

V=[V(1);V(2);V(3)];

%Ifnovalueofmuisgiven,usethedefaultvalueforEarth

ifnargin


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11ME 461 / 561 - Chapter 1

Matlab Program using rv2E.m

%Thisprogram(anmfile)calculatestheenergywhilevaryingthemagnitudeofthevelocityvector%foragivenradiusvector,thenmakesaplotofEvs |V|

%Setparametersmu=3.986e5;Npoints =100;

%InitializevariablesR=[7000;0;0];Vrange =linspace(1,15,Npoints);%eachofthesevelocitymagnitudeswillbeusedtocomputeEErange =zeros(1,Npoints);

fori=1:Npoints

V=[0;

Vrange(i);

0];

%

set

V

perpendicular

to

R,

so

either

periapsis or

apoapsis

Erange(i)=rv2E(R,V,mu);end

figurehg=plot(Vrange,Erange); %usehandlegraphics tomakeprofessionalgraphset(hg,'linewidth',2)

xlabel('Velocity magnitude(km/s)','fontsize',12)

ylabel('Energy (km^2/s^2)','fontsize',12)

%Thisprogram(anmfile)calculatestheenergywhilevaryingthemagnitudeofthevelocityvector%foragivenradiusvector,thenmakesaplotofEvs |V|

%Setparametersmu=3.986e5;Npoints =100;

%InitializevariablesR=[7000;0;0];Vrange =linspace(1,15,Npoints);%eachofthesevelocitymagnitudeswillbeusedtocomputeEErange =zeros(1,Npoints);

fori=1:Npoints

V=[0;

Vrange(i);

0];

%

set

V

perpendicular

to

R,

so

either

periapsis or

apoapsis

Erange(i)=rv2E(R,V,mu);end

figurehg=plot(Vrange,Erange); %usehandlegraphics tomakeprofessionalgraphset(hg,'linewidth',2)

xlabel('Velocity magnitude(km/s)','fontsize',12)ylabel('Energy (km^2/s^2)','fontsize',12)


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12ME 461 / 561 - Chapter 1

Matlab Program using rv2E.m

Thisplotistheresultofrunningthe

Matlab mfile that

callstheMatlabfunctionrv2E.m

Notethatforsmallvelocitiestheenergyisnegative,andforlargevelocitiestheenergyispositive

Thinkaboutthat

pointwhere

the

energyiszeroandaskwhatitmightmean

Thisplotistheresultofrunningthe

Matlab mfile that

callstheMatlabfunctionrv2E.m

Notethatforsmallvelocitiestheenergy

isnegative,

and

for

largevelocitiestheenergyispositive

Thinkaboutthat

pointwhere

the

energyiszeroandaskwhatitmightmean

0 5 10 15-60

-40

-20

0

20

40

60

Velocity magnitude (km/s)

Energy

(km

2/s2)


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13ME 461 / 561 - Chapter 1

Conservation of Angular Momentum

We start with the EOM:We start with the EOM:

Cross product withCross product with ~r

Using identity:Using identity:

Then:Then:

~r ~v = ~h=const.

~r+ r3~r=~0

~r ~r+

r3~r ~r

| {z }=0= ~0

ddt

~r ~r

=~r ~r= ~0


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14ME 461 / 561 - Chapter 1

Conservation of Angular Momentum

Local

vertic

al

Local

vertic

al

local

loca

l

horizontal

horizo

ntal

~v

M

m~r

is the flight-path angleis the flight-path angle

is the zenith angleis the zenith angle

The motion takes place in a plane that isfixed in inertial space.

This plane is called the orbital plane

The motion takes place in a plane that isfixed in inertial space.

This plane is called the orbital plane

andandto the plane ofto the plane of

oror h=r v cos

~h ~r ~v

~vr

~v

tan = vrv

h=rv sin


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15ME 461 / 561 - Chapter 1

The Trajectory Equation

Using the identity:Using the identity:

LHS of Eq. (a)LHS of Eq. (a)

RHS of Eq. (a) :RHS of Eq. (a) :

= r ~v rr2 ~r= ddt~rr

(c)

( ~A ~B) ~C= ~B( ~A ~C) ~C( ~A ~B)

r3~h ~r) = r3 (~r ~v) ~r= r3 ~v(~r ~r) ~r(~r ~v)

~r= r3~r~r ~h= r3 (~h ~r) (a)

~r ~h= ddt(~r ~h) (b)


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16ME 461 / 561 - Chapter 1

The Trajectory EquationCombining Eqs. (b) & (c)Combining Eqs. (b) & (c)

After integrationAfter integration

Constant vectorConstant vector

Using the identity:Using the identity:

h2 = r+rBcos r= h2/

1+(B/) cos < ~r,~B >

~A ( ~B ~C) = ( ~A ~B) ~C

d

dt(

~r ~

h) =

d

dt~r

r ~r ~h= ~rr +

~B

~r ~r ~h= ~r~rr + ~r ~B


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17ME 461 / 561 - Chapter 1

The Trajectory Equation

Conic Section Eq.Conic Section Eq. Trajectory Eq. for TBPTrajectory Eq. for TBP

r= p1+e cos r= h

2

/1+(B/) cos

The eccentricity:The eccentricity:

The parameter orsemi-latus rectum:

The parameter orsemi-latus rectum:

p= h2

e= B

p

F

~r

CircleCircle

EllipseEllipse

ParabolaParabola

HyperbolaHyperbola

EccentricityEccentricity Type of orbit or TrajectoryType of orbit or Trajectory

e= 0

0< e 1


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18ME 461 / 561 - Chapter 1

Some terminology[2] Satellite: any object which travels in an elliptical orbit around a

planet is called a satellite of that planet.

Space probe: any object which travels in an open trajectory; i.e.,a hyperbolic trajectory in a vicinity of a planet and in ellipticalorbit around the sun is called a space probe.

Interstellar probe: any object with a velocity greater thanheliocentric escape velocity is called an interstellar probe.

Orbitor Trajectory: path of a spacecraft or natural body in

space. We use orbit for closed and trajectory for open paths.

Barycenter: the location of the center of mass of two bodies.

Satellite: any object which travels in an elliptical orbit around aplanet is called a satellite of that planet.

Space probe: any object which travels in an open trajectory; i.e.,a hyperbolic trajectory in a vicinity of a planet and in ellipticalorbit around the sun is called a space probe.

Interstellar probe: any object with a velocity greater thanheliocentric escape velocity is called an interstellar probe.

Orbitor Trajectory: path of a spacecraft or natural body in

space. We use orbit for closed and trajectory for open paths.

Barycenter: the location of the center of mass of two bodies.


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19ME 461 / 561 - Chapter 1

Some important orbit terminology

Peri means the closest point. Perigee means the closest point to theearth. Perihelion means the closest point to the sun, and so forth.

Apo means the farthest point. Apogee means the farthest pointfrom the earth. Aphelion means the farthest point from the sun.

Keplerian orbit:is an orbit in which

1. The only force is gravity2. The central body is spherically symmetric

3. The primary mass is much greater than secondary mass

4. There are no other masses in the system

Also called two-body problem orbit or unperturbed orbit

Peri means the closest point. Perigee means the closest point to theearth. Perihelion means the closest point to the sun, and so forth.

Apo means the farthest point. Apogee means the farthest pointfrom the earth. Aphelion means the farthest point from the sun.

Keplerian orbit:is an orbit in which

1. The only force is gravity2. The central body is spherically symmetric

3. The primary mass is much greater than secondary mass

4. There are no other masses in the system

Also called two-body problem orbit or unperturbed orbit

A memory aid:a b c d e f g h I j k l m n o p q r s t u v w x y zmemory aid: a b c d e f g h I j k l m n o p q r s t u v w x y z


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20ME 461 / 561 - Chapter 1

Planet Symbols


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21ME 461 / 561 - Chapter 1

Conic Sections

A conic section is the locus of points such that the ratio of the absolutedistance from a given point (a focus) to the absolute distance from agiven line (a directrix) is a positive constant e.

A conic section is the locus of points such that the ratio of the absolutedistance from a given point (a focus) to the absolute distance from agiven line (a directrix) is a positive constant e.

Hyperbola

branches

Hyperbola

branches

CircleCircle

EllipseEllipse

ParabolaParabola

FFFF

AA

DirectrixDirectrix

rr

r/er/e

HH


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22ME 461 / 561 - Chapter 1

Conic Sections

CircleCircle

EllipseEllipse

ParabolaParabola

a=

c=

2c

2pF

2c

FF0

2p

2p

2a

F0F,

2a

2a

F F0 2p

HyperbolaHyperbola


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23ME 461 / 561 - Chapter 1

Conic Sections

= p1+e =a(1 e)

=

p

1e =a(1 + e)

e= caEccentricity:Eccentricity:

The parameter:The parameter:

Periapsis radius:Periapsis radius:

Apoapsis radius:Apoapsis radius:

The eccentricity vector:The eccentricity vector: ~e= ~v~h

~rr

(Except for parabola)(Except for parabola)

(Except for parabola)(Except for parabola)

p= h2

p=a(1 e2

)

ra=

p

1+e cos 180o

rp = p

1+e cos0o

~e= (v2 r )~r (~r ~v)~v


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24ME 461 / 561 - Chapter 1

Conic Sections Relating Relating E andand

Circle and Ellipse:Circle and Ellipse:

Parabola:Parabola:

Hyperbola:Hyperbola: E>0

E= 0

E


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25ME 461 / 561 - Chapter 1

The Circular Orbit

For Circular orbit we have:For Circular orbit we have: e= 0

Circular velocity:Circular velocity:

The period is:The period is:

v2cs2

a =

2a vcs =

p

a

T Pcs = 2

= 2 avcsTPcs=

2

a(3/2)

r= p1+0cos = h2

=a


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26ME 461 / 561 - Chapter 1

Pop Quiz (doesnt count)

An earth satellite is in a circular orbit with

altitude 400 km. What are the satellitesspeed and orbital period?

Useful information: = 3.986 10

5 km3/s2

R = 6378 km

Answer:


altitude 400 km. What are the satellitesspeed and orbital period?

Useful information: = 3.986 10

5 km3/s2

R = 6378 km

Answer:


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27ME 461 / 561 - Chapter 1

The Elliptic Orbit

bSemi-minorAxisSemi-minor

Axis

FocusFocus

Empty

Focus

Empty

Focus

c

a

aSemi-majorAxis

Semi-major

Axis

Center of mass\

Barycenter

Center of mass\

Barycenter

a2 =b2 + c2

rp+ ra = 2a

e= rarp

ra+rp

The eccentricity: The eccentricity:

Periapsis

(Perifocal)

Periapsis

(Perifocal)

Apoapsis

(Apofocal)

Apoapsis

(Apofocal)

The period is: The period is:

from calculus:from calculus:

h= r2ddt dt= r2

hd (a)

dA= 1

2r2

d (b)Eliminating between (a) & (b):Eliminating between (a) & (b):r

2d dt= 2hdA

For one period:For one period: T P= 2abh T P= 2

a3/2

&&

Line of

apses

Line of

apses


28/34

28ME 461 / 561 - Chapter 1

Pop Quiz (doesnt count) An earth satellite is in an elliptical orbit with

perigee altitude 400 km and apogee altitude

900 km. What are the satellites speed at

perigee and apogee, and orbital period?

Useful information:

= 3.986 105 km3/s2

R

= 6378 km

Answer:

An earth satellite is in an elliptical orbit with

perigee altitude 400 km and apogee altitude

900 km. What are the satellites speed atperigee and apogee, and orbital period?

Useful information:

= 3.986 105 km3/s2

R = 6378 km

Answer:


29/34

29ME 461 / 561 - Chapter 1

The Parabolic Trajectory

For a parabolic trajectory we have:For a parabolic trajectory we have:

Escape velocity :Escape velocity :

e= 1

rp= p1+cos 0 = p2

vesc=q

2r

Note:Note: vesc=

2vcs

v2esc2

r =

v22|{z}0

r|{z}0

= 0


30/34

30ME 461 / 561 - Chapter 1

Pop Quiz (doesnt count) An earth satellite is in a circular orbit with

altitude 400 km. What is the satellites speed?

What is the satellites escape speed? Suppose

the satellites speed is changed to the escape

speed, then what is the satellites speed when

the radius reaches infinity?

Useful information:

= 3.986 105 km3/s2

R = 6378 km


altitude 400 km. What is the satellites speed?

What is the satellites escape speed? Supposethe satellites speed is changed to the escape

speed, then what is the satellites speed when

the radius reaches infinity?

Useful information:

= 3.986 105 km3/s2

R = 6378 km


31/34

31ME 461 / 561 - Chapter 1

The Hyperbolic Orbit Turning angle: Turning angle:

Hyperbolic excess velocity: Hyperbolic excess velocity:

E= v2bo2

rbo=

v2

2

r

v2 =v2bo 2rbo =v2bo v2esc

F F0

c b

c2 =a2 + b2

a

Note: is denoted byand it is called departure energy.

Note: is denoted byand it is called departure energy.v

2 C3

vbovbo

rborbo

rrv

v

sin 2 = ac =

1e


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32ME 461 / 561 - Chapter 1

Canonical Units and The Reference Orbit In order to simplify the arithmetic, we use the following canonical

units in Astrodynamics.

Mass: We assume the mass of the central body is 1, i.e. 1 mass

unit. Distance: Mean distance from earth to sun is called Astronomical

unit, AU.

Distance: For orbits about a planet, typically use radius of planetas 1 Distance unit, or D.U.

Time: We choose the time unit so that the period of the orbit isTP = 2 T.U.

In order to simplify the arithmetic, we use the following canonicalunits in Astrodynamics.

Mass:We assume the mass of the central body is 1, i.e. 1 mass

unit. Distance: Mean distance from earth to sun is called Astronomical

unit, AU.

Distance: For orbits about a planet, typically use radius of planetas 1 Distance unit, or D.U.

Time: We choose the time unit so that the period of the orbit isTP = 2 T.U.

1AU/TU1AU

SunSunPlanet

1 DU

1 DU/TU1 DU/TUExercise:

What is ?

Exercise:

What is ?


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33ME 461 / 561 - Chapter 1

Extrasolar Planets

More than 700 planets orbitingother stars have been discovered

since the first was detected in 1995

Most of these are large planets

(Jupiter-sized and larger)

Detection is mainly by wobble of

the stars motion

At least one planet has been

detected by observing its transit

across the star

Planet Period (days) Eccentricity,e

Mass, m(Jupiter masses)

Semi-majoraxis, a(AU)

B 4.617 0.02 0.69 0.059

C 241.3 0.24 2.05 0.828

D 1308 0.31 4.29 2.556

Our sun wobbles 12 m/s due to Jupiter

Upsilon Andromedaehas 3 planets!


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34ME 461 / 561 - Chapter 1

References

[1] http://solarsystem.nasa.gov

[2] Mission Geometry; Orbit Constellation

Design and Management, J. Wertz, 2001.

[1] http://solarsystem.nasa.gov

[2] Mission Geometry; Orbit Constellation

Design and Management, J. Wertz, 2001.

orbital mech. chapter 1

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