AE-351 Orbital Mechanics

Recap, equations of motion & Newton Categories of OM problems Chapter 1.5-1.6, 2.2

Professor Antunes, office: MAC 202Caantunes@capitol-college.edu

Class: Thursdays 5pm-7:40pm in MAC M104

Homework, review

Chapter 1: 1.3, 1.5 (parts a, b, c, j, k, l, and do a sketch of the path), 1.7 (note answer in book is wrong, should be 275x10-6W), 1.9, 1.10

HW #2: due Thurs Sep 8

1) Chapter 2 #2.1 (similar to 1.9 earlier)

2) Ignoring orbital movement, where and at what distance should you place a point mass so the force of gravity from the Sun and the Earth are balanced?

3) Such a stable point in a properly orbiting frame is a Lagrangian Point. Which “L” point does the above 2-body solution correspond to?

4) Chapter 2 #2.16. Also, how long does it take for that satellite to coast from perigee to apogee?

5) What velocity would you need to achieve to escape Earth's gravity entirely? Assume no air resistance. Also, assuming you instantly launch with that velocity, in what direction, if any, should your velocity vector have (relative to a line from the Earth's surface to your launch point.)

1.5 Newton’s Law of Motion

Figure 1.13 The absolute acceleration a of a particle is in the direction of the net force.

Newton’s second law of motion:where a is the absolute acceleration of the center of mass.a is measured in a frame of reference which itself has neither translational nor rotational acceleration relative to the fixed stars.

This reference is called an absolute or inertial frame of reference.

1.5 Newton’s Law of Motion

Impulse of a force: (a vector quantity)

Angular momentum (moment of momentum):

Angular impulse-momentum principle:

Gravitational Forces

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Figure 2.1 (p. 18)Forces acting on a satellite in a stable orbit around the earth (from Fig. 3.4 of reference 1). Gravitational force is inversely proportional to the square of the distance between the centers of gravity of the satellite and the planet the satellite is orbiting, in this case the earth

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Kinematics & Newton’s Law

• s = ut + (1/2)at2

• v2 = u2 + 2at

• v = u + at

• F = ma

s = Distance traveled in time, t

u = Initial Velocity at t = 0

v = Final Velocity at time = t

a = Acceleration

F = Force acting on the object

Newton’s Second

Law

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FORCE ON A SATELLITE : 1Force = Mass × AccelerationUnit of Force is a NewtonA Newton is the force required to accelerate 1 kg by 1 m/s2

Underlying units of a Newton are therefore (kg) × (m/s2)In Imperial Units 1 Newton = 0.2248 ft lb.

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ACCELERATION FORMULAa = acceleration due to gravity = µ / r2 km/s2

r = radius from center of earthµ = universal gravitational constant G

multiplied by the mass of the earth ME

µ is Kepler’s constant and = 3.9861352 × 105 km3/s2

G = 6.672 × 10-11 Nm2/kg2 or 6.672 × 10-20 km3/kg s2 in the older units

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FORCE ON A SATELLITE : 2

Inward (i.e. centripetal force)

Since Force = Mass × Acceleration

If the Force inwards due to gravity = FIN then

FIN = m × (µ / r2)

= m × (GME / r2)

2.2 Equations of Motion in an Inertial Frame

Figure 2.1: (a) Two masses located in an inertial frame. (b) Free-body diagrams.

Position vector RG of the center of mass G:

The absolute velocity and the absolute acceleration of G:

2.2 Equations of Motion in an Inertial Frame

The equations of motion of the two bodies in inertial space in terms of the components of the position and acceleration vectors in the inertial XYZ frame:

where

The position vector R and velocity vector V are referred to collectivelyas the state vector.

2.2 Equations of Motion in an Inertial Frame

Figure 2.4: The motion of three identical masses as seen from the inertial frame in which m1 and m3 are initially at rest, while m2 has an initial velocity v0 directed upwards and to the right, as shown.

2.2 Equations of Motion in an Inertial Frame

Figure 2.5: The same motion as Figure 2.4, as viewed from the inertial frame attached to the center of mass G.

2.3 Equations of Relative Motion

The equation of motion of m2 relative to m1:

The fundamental equation ofrelative two-body motion.

and µ is the gravitational parameter:

(km2/s3)

where

Figure 2.6: Non-rotating frame xyzattached to point mass of m1.

2.3 Equations of Relative Motion

The equation of motion of m1 relative to the center of mass G is

The equation of motion of m2 relativeto the center of mass G is

where

whereNon-rotating frame xyz attached to the center of mass G.

So what problems would you likely need to solve?

So what problems would you likely need to solve?

2-body problems:

1) stable orbit (what would make it unstable?)2) change orbits3) change orbit shape4) get to orbit: establish, intercept or rendezvous (cue Korean missile launch)5) change orientation or spin6) steal energy (cue STEREO vids)

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Reference Coordinate Axes 1:Earth Centric Coordinate System

Fig. 2.2 in text

The earth is at the center of the coordinate system

Reference planes coincide with the equator and the polar axis

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Reference Coordinate Axes 2: Satellite Coordinate System

Fig. 2.3 in text

The earth is at the center of the coordinate system and reference is the plane of the satellite’s orbit

2.10 Perifocal Frame

Figure 2.29: Perifocal frame.

2.10 Perifocal Frame

Figure 2.30: Position and velocity relative to the perifocal frame.

Position vector:

Velocity:

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KEPLER’S THREE LAWSOrbit is an ellipse with the larger body (earth) at one focusThe satellite sweeps out equal arcs (area) in equal time (NOTE: for an ellipse, this means that the orbital velocity varies around the orbit)The square of the period of revolution equals a CONSTANT × the THIRD POWER of SEMI-MAJOR AXIS of the ellipse

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Ellipse analysis

• Points (-c,0) and (c,0) are the foci.

•Points (-a,0) and (a,0) are the vertices.

• Line between vertices is the major axis.

• a is the length of the semimajor axis.

• Line between (0,b) and (0,-b) is the minor axis.

• b is the length of the semiminor axis.

x2

a2y 2

b2=1

a2=b2+c2

Standard Equation:

y

V(-a,0)

P(x,y)

F(c,0)

F(-c,0)

V(a,0)

(0,b)

x

(0,-b)

A=πab

Area of ellipse:

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KEPLER 1: Elliptical Orbits

Equation 2.17 in text: (describes a conic section, which is an ellipse if e < 1)r 0=

p1+e*cos φ0

e = eccentricity e<1 ⇒ ellipse e = 0 ⇒ circler0 = distance of a point in the orbit to the center of the earthp = geometrical constant (width of the conic section at the focus) p=a(1-e2)ϕ0 = angle between r0 and the perigee

p

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KEPLER 2: Equal Arc-SweepsFigure 2.5

Law 2

If t2 - t1 = t4 - t3

then A12 = A34

Velocity of satellite is SLOWEST at APOGEE; FASTEST at PERIGEE

(Why?) 2.4 Angular Momentum and the Orbit Formulas

Specific relative angular momentum:

Conservation of angular momentum:

Zero angular momentum characterizes rectilinear trajectories on which m2 moves towards or away from m1 in a straight line.

The angular momentum depends only on the transverse component of the relative velocity:

2.4 Angular Momentum and the Orbit Formulas

Figure 2.8: The path of m2 around m1 lies in a plane whose normal is defined by h.

2.4 Angular Momentum and the Orbit Formulas

Figure 2.9: Components of the velocity of m2, viewed above the plane of the orbit.

2.4 Angular Momentum and the Orbit Formulas

Figure 2.10: Differential area dA swept out by the relative position vector r during time interval dt.

dA/dt - areal velocity.

Kepler’s second law - Equal areas are swept out in equal times.

2.4 Angular Momentum and the Orbit Formulas

The first integral of is

e is the dimensionless eccentricity vector, which is a constant and lies in the orbital plane.

The line defined by the vector e is commonly called the apse line.

The vector C = µe is called the Laplace vector.

2.4 Angular Momentum and the Orbit Formulas

Figure 2.11: The true anomaly θ is the angle between the eccentricity vector e and the position vector r.

2.4 Angular Momentum and the Orbit Formulas

From we obtain the orbit formula:

This is the equation of a conic section. Therefore, it is a statement ofKepler’s First Law: The planets follow elliptical paths around the sun.

The point of closest approach (θ = 0) lies on the apse lineand is called periapsis.

Keplerian orbits: Two-body orbits

2.4 Angular Momentum and the Orbit Formulas

Transverse component of v:

Radial component of v:

Figure 2.12: Position and velocity of m2 in polar coordinates centered at m1, with the eccentricity vector being the reference for true anomaly (polar angle) θ. γ is the flight path angle.

2.4 Angular Momentum and the Orbit Formulas

Figure 2.13: Illustration of latus rectum, semi-latus rectum p, and the chord between any two points on an orbit.

Semi-latus rectum (parameter):

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KEPLER 3: Orbital PeriodOrbital period and the Ellipse are related by

T2 = (4 π2 a3) / µ (Equation 2.21)

That is the square of the period of revolution is

equal to a constant × the cube of the semi-major axis.

IMPORTANT: Period of revolution is referenced to inertial space, i.e., to the galactic background, NOT to an observer on the surface of one of the bodies (earth).

µ = Kepler’s Constant = GME

(and... we're back to the realm of the practical)

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Numerical Example 2.1.1 p 29The Geostationary Orbit:

Sidereal Day = 23 hrs 56 min 4.1 secCalculate radius and height of GEO

orbit:T2 = (4 π2 a3) / µ (eq. 2.21)Rearrange to a3 = T2 µ /(4 π2) T = 86,164.1 seca3 = (86,164.1) 2 x 3.986004418 x 105 /(4 π2) a = 42,164.172 km = orbit radiush = orbit radius – earth radius = 42,164.172 –

6378.14 = 35,786.03 km