Circular and gavitational force

1

Gravity and Circular Motion

• Batte

Circular motion

• When an object undergoes circular motion it must experience a

centripetal force

• This produces an acceleration

towards the centre of the circle

BWH 10/04 AQA 13.3.1-6 2

CALCULATING PLANETARY ORBITS — 1680

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.

BWH 10/04 AQA 13.3.1-6 5

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

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

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)

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.

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

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

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

Acceleration

• The acceleration towards the centre of the circle is

• a = v2/r OR

• a = ω2r

BWH 10/04 AQA 13.3.1-6 13

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

BWH 10/04 AQA 13.3.1-6 14

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!

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

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

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

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

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:

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

?

-1.5 -1 -0.5 0.5 1 1.5

-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

-1.5 -1 -0.5 0.5 1 1.5

-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)

-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'

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

BWH 10/04 AQA 13.3.1-6 25

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

BWH 10/04 AQA 13.3.1-6 26

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

BWH 10/04 AQA 13.3.1-6 27

Graph of Potential against distance

BWH 10/04 AQA 13.3.1-6 28

-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)

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

BWH 10/04 AQA 13.3.1-6 29

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

BWH 10/04 AQA 13.3.1-6 30

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

BWH 10/04 AQA 13.3.1-6 31

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

BWH 10/04 AQA 13.3.1-6 32

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

-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,

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!

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

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

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

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!

Universality of Gravitation: Dark Matter

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

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?

10 0

sin

1sin 0

sin

10

sin0r

rr

r f rr

F

r

Fr r

Central Forces Are Conservative!

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

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

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

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

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.

-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

NEPTUNE AS AN EXAMPLE OF “PHYSICAL SIGNIFICANCE”

seconds of arc

THE “GREAT INEQUALITY” AS A MORE TYPICAL EXAMPLE

minutes of arc

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) .

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

2

2

22

0

mv MmG E

rGM

v Er

Maximum Velocity for An elliptical Orbit

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

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

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.

Scattering and Bound States are All There Are!

E > 0 E < 0

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

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

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

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

Mathematics on r(θ):

Substitute:

Characteristic:

(h = length ang mom vector)

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

Planet pPlanet q

barycenter

Lagrange point

See also: http://janus.astro.umd.edu/javadir/orbits/ssv.html

Balance rotation and gravity

For mp and mq we have:

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

p q

mp = 10mq = 1

Horseshoe and Tadpole orbits

Saturn

Epimetheus

Janus

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

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

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.

12/07/2009 12.215 Lec 21 73

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

12/07/2009 12.215 Lec 21 74

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

12/07/2009 12.215 Lec 21 75

r acosE e

1 e2 sin E

r

cossin

r a(1 ecosE) a(1 e2 )1ecos

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.

12/07/2009 12.215 Lec 21 76

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

12/07/2009 12.215 Lec 21 77

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

12/07/2009 12.215 Lec 21 78

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

12/07/2009 12.215 Lec 21 79

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

12/07/2009 12.215 Lec 21 80

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

Qq

Solution to the Scattering Problem

Solution to the Bound State Problem

Solve Both Problems for r

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

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

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

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

The End!

• With these equations we can reach the stars!

BWH 10/04 AQA 13.3.1-6 86