circular and gavitational force
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Circular and gavitational forceTRANSCRIPT
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1
Gravity and Circular Motion
• Batte
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Circular motion
• When an object undergoes circular motion it must experience a
centripetal force
• This produces an acceleration
towards the centre of the circle
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CALCULATING PLANETARY ORBITS — 1680
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INFERRING LAWS OF FORCE FROMPHENOMENA OF MOTION
Phenomena: Descriptions of regularities of motion that hold at least quam proxime over a finite body of observations from a limited period of timeThe planets swept out equal areas in equal times quam proxime with respect to the Sun over the period from the 1580s to the 1680s.
Propositions, deduced from the laws of motion, of the form:
“If _ _ _ quam proxime, then …… quam proxime.”
If a body sweeps out equal areas in equal times quam proxime with respect to some point, then the force governing its motion is directed quam proxime toward this point.
Conclusions: Specifications of forces (central accelerations) that hold at least quam proxime over the given finite body of observationsTherefore, the force governing the orbital motion of the planets, at least from the 1580s to the 1680s, was directed quam proxime toward the Sun.
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r
F
For a central force the position and the force are anti-parallel, so rF=0.
0N r F dL
Ndt
0dL
dt
So, angular momentum, L, is constant
N is torqueNewton II, angular
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Since the Angular Momentum, L, is constant:
Its magnitude is fixedIts direction is fixed.
L r p
• By the definition of the cross product, both the position and the momentum are perpendicular to the angular momentum.
• The angular momentum is constant.• Therefore, the position and momentum are
restricted to a plane. The motion is restricted to a plane.
L
rp
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rdr
r+dr
Show that the conservation of angular momentum implies that equal areas are swept out in equal times:
r dr
The green shaded area is dA
The parallelogram formed by r and dr is twice the area of dA. But, by definition, the magnitude of rdr is the area of the parallelogram.
1 1sin
2 2dA r dr r dr
dr sin (height)
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Proof of Kepler’s Second Law
1
2dA r dr
1
2dA r vdt
1
2
dAr p
dt m
2
dA L
dt m
The area swept out in time dt
But dr is just v dt
Divide by dt, and change v to (1/m)p
The angular momentum is constant, so the rate at which area is swept out is also constant.
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Central Forces in Polar Coordinates
Let’s use the Lagrangian:
L T V Don’t confuse the Lagrangian, “L”,with the angular momentum, “L”.
21
ˆ ˆ2 rL m r e re V r Motion is restricted to
a plane, and the potential is that for a central force. 2 2 21
2L m r r V r
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Newton’s Gravitation Equation
• Newton’s Gravitation equation is
• F = -Gm1m2/r2
• MUST BE LEARNED!!
• Negative sign is
• a vector sign
• G is
• Universal Gravitational ConstantBWH 10/04 AQA 13.3.1-6 11
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2 2 21 1
2 2L mr mr V r
Lmr
r
d L
mrdt r
2Lmr f r
r
2mr mr f r
d L L
dt r r
The Radial Equation
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Acceleration
• The acceleration towards the centre of the circle is
• a = v2/r OR
• a = ω2r
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Angular speed
• Angular speed can be measured in ms-1 or
• Rads-1 (radians per second) or
• Revs-1 (revolutions per second)
• The symbol for angular speed in radians per second is
• ω
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2 2 21 1
2 2L mr mr V r
2Lmr
0
L
2 0d
mrdt
d L L
dt
The Angular Equation
But the term in parenthesis is just the angular momentum of a point particle, so conservation of angular momentum falls out of the constraints on the Lagrangian!
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From the Lagrangian, we found Newton’s Laws in Polar Coordinates:
2 Vmr mr
r
Radial eqn.:
2 0mr mr Theta eqn.:
( )V
f rr
2 0r r
2 0d
rdt
2r l L
l r vm
2 ( )f rr r
m
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Obtain solutions for SHAPE, don’t need to solve for tPut it in terms of u and theta:
1r
uLet
1u
r
22 2
1 1 du d du dur u r l
u u d dt d d
Solve for r and its derivatives in terms of u and theta.
2 2 22 2
2 2 2
d du d u d u l d ur l l l l u
dt d d d r d
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To reiterate:
22 2
1 1 du d du dur u r l
u u d dt d d
2 2 22 2
2 2 2
d du d u d u l d ur l l l l u
dt d d d r d
1r
u
2 ( )f rr r
m
2
22 2 21 1/
d ul u lu f m
d u u
22
1r l
u
2
2 2
1/f ud uu
d l u m
Now we have a differential eqn. of u and theta
Replace r’s and thetas
Define the Differential Equation for the Orbit Shape
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Now, we plug our force law into the differential equation we have derived.
2( )
kf r
r
k GMmThe force is:
2
2 2
2
2 2 2
1/f u ku
l u
d u ku
d mlm l u m
This equation is similar to a simple harmonic oscillator with a constant force offset, except the independent variable is theta not time.
2cos
ku A
ml 2
1
cosr
kA
ml
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2
1
cosr
kA
ml
0o Choose theta nought so that theta equal to 0 yields the distance of closest approach
2
2
/
1 cos(
ml kr
mlA
k
“Look Ma, an ellipse!”“But Johnny, It doesn’t look like an ellipse???”“Oh Ma… don’t you know nothin’?”
Manipulate the Shape Equation a Bit More:
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2
2
/
1 cos(
ml kr
mlA
k
-1.5 -1 -0.5 0.5 1 1.5
-1
-0.5
0.5
1
We want to show that this equation is an ellipse
?
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-1
-0.5
0.5
1
The ellipse below has the equation: 2 22 3x y
a
b
2 2
2
2 3
3
3
x y
x
x
Semimajor axis:
2 2
2
2 3
2 3
3
2
x y
y
y
Semiminor Axis
Review of Some Basic Ellipse Properties
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-1
-0.5
0.5
1
To solve for the eccentricity of an ellipse, use the defining relationship for the ellipse and solve equations for two special cases:
2 2 2
22 2
2
constant
when , constant 2
( ) constant /4
constant
constant 2
1
r r
r r r
b a
a a a a
a
b a a
b
a
Pythagorean Triangle
When touching the right edge
a
r rb
a
Plug (2) into (1)
(1)
(2)
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-1.5 -1 -0.5 0.5 1 1.5
-1
-0.5
0.5
1
r
22 2 2
2 2 2 2 2 2
2 2 2 2
2 2
2
sin 2 cos
sin cos 4 cos
4 4 cos
4 cos
r r a
r r a r
r r a r ar
r r a ar
r r a a r
Defining equation for an ellipse
a a
r
Put our Equation for the Ellipse in the Form of Our Shape Equation
r'
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Gravitational field
• A gravitational field is an area of space subject to the force of gravity. Due to the inverse square law relationship, the strength of the field fades quickly with distance.
• The field strength is defined as
• The force per unit mass OR• g = F/m in Nkg-1
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Radial Field
• Planets and other spherical objects exhibit radial fields, that is the field fades along the radius extending into space from the centre of the planet according to the equation
• g = -GM/r2
• Where M is
• the mass of the planet
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Gravitational Potential
• Potential is a measure of the energy in the field at a point compared to an infinite distance away.
• The zero of potential is defined at• Infinity• Potential at a point is• the work done to move unit mass from infinity
to that point. It has a negative value.• The equation for potential in a radial field is• V = -GM/r
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Graph of Potential against distance
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-1.2
-1
-0.8
-0.6
-0.4
-0.2
0
0 1 2 3 4 5 6 7 8 9 10
Distance
Potential
Series1
Pow er (Series1)
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Orbits
• Circular orbits follow the simple rules of gravitation and circular motion. We can put the force equations equal to each other.
• F = mv2/r = -Gm1m2/r2
• So we can calculate v
• v2 = -Gm1/r
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More orbital mechanics
• Period T is the time for a complete orbit, a year. It is given by the formula.
• T = 2π / ω
• and should be calculated in
• seconds
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Example
• The Moon orbits the Earth once every 28 days approximately. What is the approximate radius of its orbit? G = 6.67 x 10-11 Nm2kg-2, mass of Earth M = 6.00 x 1024 kg
• ω = 2π / T = 2π/(28x24x60x60) =• 2.6 x 10-6 rads-1
• F = mω2r = -GMm/r2 so• r3 = -GM/ω2
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Example continued
• r3 = 6.67 x 10-11 x 6.00 x 1024/(2.6x10-6)2
• = 5.92 x 1025
• r = 3.90 x 108 m
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2
2
2 2 2 2 2
2 2 2
2
2
2
2
2
4 cos
4 cos
4 4 4 4 cos
4 4 4 4 cos
cos
1 cos 1
2
1
1 os
2
c
r a a r
r a a r
a r ar r a ar
a ar a ar
a r a r
r a
a
r
r
r
r a
a r
Put our Equation for the Ellipse in the Form of Our Shape Equation
From the last slide
From the defining relationship
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-1.5 -1 -0.5 0.5 1 1.5
-1
-0.5
0.5
1The latus rectum is the distance to a focus from a point on the ellipse perpendicular to the major axis
21a
2 2 2 2
2 2 2
2 2
2
2
2
2
2
4
1
4
4
2
4
4
4
4
r a
a
a
a a
r
r
a a
a
a
a
Defining relationship for the ellipse
Solve it for r
Define a Pythagorean relationship
Solve the resulting relationship for alpha
r
2a
Define the Latus Rectum,
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21
1 cos 1 cos
ar
2
2
/
1 cos(
ml kr
mlA
k
So, our general equation for an ellipse is
And the solution to our shape equation was
They have the same form! We’ve derived Kepler’s Second Law. The trajectories of planets for a central force are described by ellipses.
Finally, Show that our Central Force Yielded an Elliptical Orbit!
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2 2
22
/ /
/
ml k l GM
AmlAl GM
k
21
1 cos 1 cos
ar
2
2
os(
/
1 cr
m
ml
lk
k
A
Relations Between Orbit Parameters
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2
LA
m
0
Adt A
0 0 0 02 2 2 2
L l l lAdt dt dt dt
m
2
2
l AA
l
Integration of the area dA/dt over one period gives A (1)
Kepler’s Second Law
Use Kepler II to relate integral and l. (2)
Set (1) and (2) equal
Kepler’s Third Law
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2 22 2 12
A ab a
l l l
2 4 2 2 3 2 2 32
2 2 2
4 1 4 1 4a a a a
l l l
2 2 2ml ml l
k GMm GM 2
2 34a
GM
Kepler’s Third Law Derivation Continued
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22 34
aGM
Earth presumably fits this rule… Using Earth years for time, and Astronomical Units (1 AU = 1 Earth-Sun distance) for distance renders the constants equal to 1.
Look How Well The Solar System Fits Kepler’s Third Law!
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Universality of Gravitation: Dark Matter
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3
3
4
34
/3
M r
M R
Only the mass interior to the star in question acts.
Density, assuming a uniform density sphere.
Keplerian Motion: All the mass of the galaxy would be assumed to be focused very close to
the origin.2
2nucleus
nucleus
M m vG m
r r
GMv
r
Velocity curve, Assuming a Constant Density Sphere All the Way Out
32
23
4
3
4r mmv
Gr r
v r G
Keplerian Motion vs. Constant Density Sphere Motion
r
v
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1 1 1w v u w v uF hF gF u fF hF v gF fF w
gh v w fh w u fg u v
, ,
1, , sin
u r v w
f g r h r
1sin
sin
1sin
sin
1sin
sin
r
r
F
F
F F F rr
r Fr r
r Fr r
Are Central Forces Conservative?
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10 0
sin
1sin 0
sin
10
sin0r
rr
r f rr
F
r
Fr r
Central Forces Are Conservative!
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Energy Equation of an Orbit in a Central Field
2 2 2 2
2 2
22
2
2 1
1constant
2
1
2
duml u V u E
d
v r r
m r r V r E
2
22 2
2
1r
u
lu
dur l
d
d ur l u
d
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Orbital Energies in an inverse Square Law
kV r ku
r
22 21
2
duml u ku E
d
2
22
2
22
2 2
2 2
(1/ 2)
2
1
2 2
2
du E ku
d duE ku
uml ml
ud ml
du E kuu
d ml ml
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Integrating Orbital Energy Equation
2
2 2
2
21
2
2 4
2
2
1
1
2 21, ,
1
221
cos1
1 2cos
4
4 8
o
o
d duc bu au
k Ea b c
ml ml
duc bu
b au
a b acau
ku
ml
k Em l ml
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2
21
2
2 4 2
2
2 2
2 2 2
2 2
2
2
2
/
221
cos1 4 8
cos2
1 2 / cos
1 2 / co
/ cos
s
1 1 2
o
o
o
o
o
ku
ml
k Em l ml
k uml
k Eml
k Eml k k uml
uml k k Eml k
ml kr
Eml k
Integrating Orbital Energy Equation Continued
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2
2
2
2
2 2
2
/
/
1 1 2 / cos
2
21
1
21
2
o
ml kE
k
a
E
kE
ml kr
Eml k
ka
E
l
a
m
k
k
Total Energy of an Orbit
To fit earlier form for an ellipse
Earlier Relations
Total energy of the orbit.
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-4 -2 2 4
-4
-2
2
4
E < 0 Ellipse or Circle e<1E=0 Parabola e=1E>0 Hyperbolic e>1
2
2 2
2
2
/
1 1 2 / cos
1 2 /
o
ml kr
E
E
m k
ml k
l
Ellipses, Parabolas, Hyperbolas
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NEPTUNE AS AN EXAMPLE OF “PHYSICAL SIGNIFICANCE”
seconds of arc
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THE “GREAT INEQUALITY” AS A MORE TYPICAL EXAMPLE
minutes of arc
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Second-Order Phenomena Often Underdetermine Their Physical Source
Example ExampleDeviation of surface gravity from Newton’s ideal variation implies the value of (C-A)/Ma2 and hence a correction to the difference (C-A) in the Earth’s moments of inertia, and the lunar-solar precession implies the value of (C-A)/C and hence a correction to the polar moment C; these two corrected values constrain the variation (r) of density inside the Earth, but they do not suffice to determine (r) .
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TAKING THE THEORY TO BE EXACTTHE PRIMARY IMPLICATION
Every systematic discrepancy between observation and any theoretically deduced result ought to stem from a physical source not taken into account in the theoretical deduction
– a further density variation
– a further celestial force
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2
2
22
0
mv MmG E
rGM
v Er
Maximum Velocity for An elliptical Orbit
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5 10 15 20
-1.5
-1
-0.5
0.5
1
1.5
Limits of Radial Motion
2
2
2
2
2
2
2
2
2
mr E
m
lV r
r
U r
mlU r
r E
V rr
E2
22
ml
r
( )V r( )U r
U is the effective potential
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2
2
2 2
2 2
2 2
1,0
2 2
1,0
0
02
2 2 0
2 2 0
2 4 8
4
2
2
U r E
ml kE
r r
ml kr Er
Er kr ml
k k Emlr
E
k k Emlr
E
Limits Continued
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2 2
1,0
2
min 2
min
2
2
2
2o
k k Emlr
E
kE
mlk
rE
Minimum Orbital Energy
Radical ought to remain real for elliptical orbits.
Value of E for which the radical is 0
Extremes of motion merge to one value.
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Scattering and Bound States are All There Are!
E > 0 E < 0
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2
2 2 11
2
duml u V u E
d
Energy Equation for a Central Force Again
kV ku
r
QqV Qqu
r
For attractive inverse square For repulsive inverse square
k Qq
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2
2 2 11
2
duml u V u E
d
The Scattering Calculation Proceeds Exactly Like the Bound
State Calculation
22 21
2
duml u ku E
d
22 2
1
2 2d du
E kuu
ml ml
But, we’re going to let k go to –Qq after we integrate
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2
2
2 2
2
1
2 2
1
1,
2Ac
1o
4
,
so
d duc bu au
k Ea b c
ml ml
duc bu a
b a
au
u
a b c
Integrate Both the Bound State And the Scattering Problem
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2
2 2
22
2 2
2
2
2
2
22 22 2
2
2
2
2
2
Acos f u1
2
41
2
4
1 1
22 1144
4 4
4 4 44 4 4
Acos f u
1 ( )
4
2
2 4
1 4 2
b auf u
b ac
b aub aub acb ac
b ac
f ud
du f u
af u
b ac
f u
b ac
a aub b
b ac
a u abu acac a u abu
f ud a b ac
b acdu a auf u
bu c
2
2 2
2 2
2
1 1 2Acos
A4
4
cos
bu c
d b au a
du b ac au
b audu
aau
b
bu c b a
u c
c
Verify Integral in 6.10.4
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Mathematics on r(θ):
Substitute:
Characteristic:
(h = length ang mom vector)
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Xi
Yi
ZiPerigee
Right ascension
Nodal line
I
Kepler’s solution in an inertial coordinate system
vrH r
v Satellite
XYZ: inertial cs
: right ascension
: argument van perigee
: true anomaly
I: Inclination orbit plane
H: angular momentum vector
r: position vector satellite
v: velocity satellite
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Planet pPlanet q
barycenter
Lagrange point
See also: http://janus.astro.umd.edu/javadir/orbits/ssv.html
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Balance rotation and gravity
For mp and mq we have:
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Equations of motion after rotationAfter straightforward differentiation we get
If we ignore the Coriolis term, then we obtain
So that we can plot the length of the acceleration vector on the left hand side to demonstrate the existence of the Lagrangian points L1 till L5
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p q
mp = 10mq = 1
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Horseshoe and Tadpole orbits
Saturn
Epimetheus
Janus
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Equations of motion
Tzyxx
zyxr
rr
GMxU
xUx
),,(
)(
)(
222
xr
x3
These equations hold in the inertial coordinate frame and they are only valid for the Kepler problem
r
U
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2
2
2
2
2
2 2 4 2
2
1
2Acos
4
2 4 cos
2 4 cos
2cos
1 2Acos
4o
o
o
o
o
duc bu au
b u
b c
u b b c
u b b c
k k Eu
ml m
b au
a c
l
b a
ml
Solve Both Bound State and Scattering Problem for u
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Summary of last class– Ionospheric delay effects in GPS
• Look at theoretical development from Maxwell’s equations• Refractive index of a low-density plasma such as the Earth’s
ionosphere.• Most important part of today’s class: Dual frequency ionospheric
delay correction formula using measurements at two different frequencies
– Effects of ionospheric delay are large on GPS (10’s of meters in point positioning); 1-10ppm for differential positioning
– Largely eliminated with a dual frequency correction (most important thing to remember from this class) at the expense of additional noise (and multipath)
– Residual errors due to neglected terms are small but can reach a few centimeters when ionospheric delay is large.
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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
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Vector to satellite
• At a specific time past perigee; compute Mean anomaly; solve Kepler’s equation to get Eccentric anomaly and then compute true anomaly.
• Vector r in orbit frame is
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r acosE e
1 e2 sin E
r
cossin
r a(1 ecosE) a(1 e2 )1ecos
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Perturbed motions
• The central force is the main force acting on the GPS satellites, but there are other significant perturbations.
• Historically, there was a great deal of work on analytic expressions for these perturbations e.g. Lagrange planetary equations which gave expressions for rates of change of orbital elements as function of disturbing potential
• Today: Orbits are numerically integrated although some analytic work on form of disturbing forces.
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J2 Perturbations
• Notice that only and n are effected and so this perturbation results in a secular perturbation
• The node of the orbit precesses, the argument of perigee rotates around the orbit plane, and the satellite moves with a slightly different mean motion
• For the Earth, J2 = 1.08284x10-3
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Gravitational perturbation styles
Parameter Secular Long period Short period
a No No Yes
e No Yes Yes
i No Yes Yes
Yes Yes Yes
Yes Yes Yes
M Yes Yes Yes
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Other perturbation on orbits and approximate size
Term Acceleration (m/sec2)
Central 0.6
J2 5x10-5
Other gravity 3x10-7
Third body 5x10-6
Earth tides 10-9
Ocean tides 10-10
Drag ~0
Solar radiation 10-7
Albedo radiation 10-9
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GPS Orbits
• Orbit characteristics are– Semimajor axis 26400 km (12 sidereal hour period)– Inclination 55.5 degrees– Eccentricity near 0 (largest 0.02)– 6 orbital planes with 4-5 satellites per plan
• Design lifetime is 6 years, average lifetime 10 years
• Generations: Block II/IIA 9729 kg, Block IIR 11000 kg
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2
2 2 2
22
2 2
2
2
2
2
2
2cos
1 21 1 c
21 1
os
21 1 cos
cos
o
o
o
o
Qq Qq Eu
ml ml ml
Qq E ml
r ml ml Qq
mlkr
Eml
mlQq
rEm
k
l
Solution to the Scattering Problem
Solution to the Bound State Problem
Solve Both Problems for r
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Scattering in an Inverse Square Field
2
2
2
2
d u Qqu
d
d u ku
d ml ml
2 2
1
co ss
1
corr
kA
QqA
mml l
2
2 2 2
/
1 1 2 / cos o
ml qQr
Eml Q q
Solving the energy equation for r
attractive
RepulsiveK goes to -Qq
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b
so
o
rmin
A Drawing of the Scattering Problem
A small range of impact parameter
scattering center
Trajectory
2
2 2 2
/
1 1 2 / cos o
ml qQr
Eml Q q
Impact parameter b
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b
so
o
rmin
A small range of impact parameter
scattering center
Trajectory
Impact parameter b
2
2 2 2
/
1 1 2 / cos o
ml qQr
Eml Q q
0
cos 0
cos 2 cos cos
2 0 asymptotes
o
o o o o
o
r
r
r
or
Asymptotes
min
2o
s o
r
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2
2 2 2
/
1 1 2 / cos o
ml qQr
Eml Q q
(1/ 2)2
2
2 2 2
2 2 2
(1/ 2)2
2
2
2
1 1 2 / cos 2 0
1cos
1 2 /
2tan
2
1
2
1tan tan
2
2cot
2
o o
o
o
s o
o s
o ss
Eml Q q
Eml Q q
Eml
Q q
Eml
Q q
(1/ 2)2
2 2
2Eml
Q q
1
2
2 2
21
Eml
Q q
o
/2-o
Angles
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The End!
• With these equations we can reach the stars!
BWH 10/04 AQA 13.3.1-6 86