is Geometry of common power-law potentials...Many-body 1D collisions Elastic examples: Western...

Many-body 1D collisionsElastic examples: Western buckboard, Bouncing column, Newton’s cradleInelastic examples: “Zig-zag geometry” of freeway crashesSuper-elastic examples: This really is “Rocket-Science”

Geometry of common power-law potentials Geometric (Power) series

“Zig-Zag” exponential geometryProjective or perspective geometry

Parabolic geometry of harmonic oscillator kr2/2 potential and -kr1 force fields Coulomb geometry of -1/r-potential and -1/r2-force fields

Compare mks units of Coulomb Electrostatic vs. GravityGeometry of idealized “Sophomore-physics Earth”

Coulomb field outside Isotropic Harmonic Oscillator (IHO) field insideContact-geometry of potential curve(s)“Crushed-Earth” models: 3 key energy “steps” and 4 key energy “levels”

Earth matter vs nuclear matter: Introducing the “neutron starlet” and “Black-Hole-Earth”

Lecture 6 Thur. 9.12.2013

Topics for Lecture 7

1Thursday, September 12, 2013

Western buckboard = ?????


Western buckboard = 3-ball analogy


Western buckboard = 3-ball analogy Disaster!


(a)mk/mk+1=3

(b)mk/mk+1=7

(c) Bouncingcolumn

(d) Singlepop-up

(+1,-1)

(-1,+1)

(1,0)

(0,1)

mk/mk+1=1

Bang!

Bang!

Bang!

Bang!

Bang!

Bang!

Bang!

Bang!

Bang!

Bang!

Bang!

Bang!

Bang!

-1 0 1 2 3

-1

1

2

3

4

5

6

-1 0 1 2 3

-1

1

2

3

4

5

6

4

Bang(2)12

Bang(3)23

Bang(4)34

Bang(2)12

Bang(3)23

Bang(4)34

Bang(1)01 Bang(1)01

Unit 1Fig. 8.2a-b

4-Body IBM GeometryFig. 8.2c-d

4-Equal-Body Geometry

4-Equal-Body “Shockwave” or pulse wave

Dynamics

Opposite of continuous wave dynamicsintroduced in Unit 2

http://www.uark.edu/ua/modphys/testing/markup/BounceItWeb.html




M1:m2=1:1

0 10 20 30 40 50 6015

10

20

30

““KKaa--rruunncchh!!””

“Ka-runch!”“Ka-runch!”

velocity VM(n...3210) of pileup mass(es)

0 1 2 3 4 5VM(01)=30

0 1 2 3 4 5(VM(0)=60, Vm(1)=0)

0 1 2VM(0123)=15

3

VM(012)=2030 1 2 4 5

4 5

0 1 2VM(01234)=12

3 4 5

0 1 2VM(01235)=10

3 4 5

VM(n...3210)

2:13:1

4:1

40

50

60

Speeding car and five stationary cars

Unit 1Fig. 8.5Pile-up:

One 60mph car hits

five standing cars


M1:m2=1:1

0 10 20 30 40 50 6015

10

20

30




0 1 2 3 4 5VM(01)=30

0 1 2 3 4 5(VM(0)=60, Vm(1)=0)

0 1 2VM(0123)=15

3

VM(012)=2030 1 2 4 5

4 5

0 1 2VM(01234)=12

3 4 5

0 1 2VM(01235)=10

3 4 5

VM(n...3210)

2:13:1

4:1

40

50

60


Unit 1

Fig. 8.5Pile-up:

One 60mph car hits

five standing cars

01(VM(1)=60, Vm(0)=0)

2345

01VM(10)=302345

01VM(210)=40

2345

01VM(3210)=45

2345

01VM(43210)=48

2345

01VM(543210)=50

2345

M1:m2=1:1

0 10 20 30 40 50 60

45

10

20

30




velocity VM(n...3210) of pileup

2:13:14:1

40

50

60

40

30

4850Five speeding cars and a stationary car

Fig. 8.6Pile-up:

Five 60mph cars hit

one standing cars


M1:m2=1:1

0 10 20 30 40 50 6015

10

20

30




0 1 2 3 4 5VM(01)=30

0 1 2 3 4 5(VM(0)=60, Vm(1)=0)

0 1 2VM(0123)=15

3

VM(012)=2030 1 2 4 5

4 5

0 1 2VM(01234)=12

3 4 5

0 1 2VM(01235)=10

3 4 5

VM(n...3210)

2:13:1

4:1

40

50

60


Unit 1

Fig. 8.5Pile-up:

One 60mph car hits

five standing cars

01(VM(1)=60, Vm(0)=0)

2345

01VM(10)=302345

01VM(210)=40

2345

01VM(3210)=45

2345

01VM(43210)=48

2345

01VM(543210)=50

2345

M1:m2=1:1

0 10 20 30 40 50 60

45

10

20

30




velocity VM(n...3210) of pileup

2:13:14:1

40

50

60

40

30

4850Five speeding cars and a stationary car

Fig. 8.6Pile-up:

Five 60mph cars hit

one standing cars

12345Five speeding cars and five stationary cars

1 2 3 4 5

Fig. 8.7Pile-up:

Five 60mph cars hit

five standing cars(Fug-gedda-aboud-dit!!)


1/2

1/3

1/4

1/51/61/71/81/91/10

1

2

3

4

5

6

7

8

slope

1:1

slope

1:2

slope

1:3

1

110

16

1/5

1/4

1/4

1/4

18

1/8

1/8

19

1/10

1/9

17

1/7

1/7

1/6

1/5

1/9

1/10

1/6

1/5

00ΔV(0)=+1/10Δv

e=-1

11 22 33 44 55pow(0)

66 77 88 99 1100

11 22 33 44 55 66 77 88 99 110000

00 11 22 33 44 55ΔV(1)=+1/9Δv

e=-1

66 77 88 99 1100

00 11 22 33 44 55Δve=-1

66 77 88 99 1100ΔV(2)=+1/8

(a) Harmonic progression

1,1/2,1/3,1/4,1/5,1/6,1/7,...

(b) Harmonic series

1+1/2+1/3+1/4+1/5+1/6+1/7+...

00 11 22 33 44Δve=-1

55ΔV(5)=+1/566 77 88 99 1100

00 33 44 55ΔV(4)=+1/6Δv

e=-1

11 22 66 77 88 99 1100

00 33 44 55Δve=-1

11 22 66 77 88 99 1100ΔV(3)=+1/7

00 11 22 33 44Δve=-1

55ΔV(6)=+1/4

77 88 99 110066

ΔV(7)=+1/388 99 1100

ve(exhaustparticle)

0.5

VM(rocket)

0.5

-1.0

0 1101

213

14

15

16

17

1819

110

pow(0)

slope

1:10

pow(1)

slope

1:9

pow(2)

slope

1:8

pow(3)

slope

1:7

pow(4)

slope

1:6

pow(5)

slope

1:5

pow(6)

slope

1:4

pow(1)

pow(2)

pow(3)

pow(4)

pow(5)

pow(6)

pow(7)

+__=0.336=0.1

ΔV(6)=+1/4

ΔV(5)=+1/5

ΔV(4)=+1/6

ΔV(3)=+1/7

ΔV(2)=+1/8

ΔV(1)=+1/9

ΔV(0)=+1/10

110+__

=1.096

19

18

17

16

14

15

110+__

=0.846

19

18

17

16

15

16

110+__=0.646

19

18

17

17

110+__=0.478

19

18

18

110

19

19

110+__=0.211

1.0

1.0 Unit 1Fig. 8.8a-b

Rocket Science!

m·Δv1+9m·ΔVM(1)=0


m·Δv0+10m·ΔVM(0)=0







0th: V(0)=1/10=0.1 1st: V(1)=1/10+1/9=0.211 2nd: V(2)=1/10+1/9+1/8=0.336 3rd: V(3)=V(2)+1/7=0.478 4th: V(4)=V(3)+1/6=0.646 5th: V(5)=V(4)+1/5=0.846 6th: V(6)=V(5)+1/4=1.096 7th: V(7)= V(6)+1/3=1.429 8th: V(8)=V(7)+1/2=1.929

1/2

1/3

1/4

1/51/61/71/81/91/10

1

2

3

4

5

6

7

8

slope

1:1

slope

1:2

slope

1:3

1

110

16

1/5

1/4

1/4

1/4

18

1/8

1/8

19

1/10

1/9

17

1/7

1/7

1/6

1/5

1/9

1/10

1/6

1/5

00ΔV(0)=+1/10Δv

e=-1

11 22 33 44 55pow(0)

66 77 88 99 1100

11 22 33 44 55 66 77 88 99 110000

00 11 22 33 44 55ΔV(1)=+1/9Δv

e=-1

66 77 88 99 1100

00 11 22 33 44 55Δve=-1

66 77 88 99 1100ΔV(2)=+1/8


1,1/2,1/3,1/4,1/5,1/6,1/7,...

(b) Harmonic series

1+1/2+1/3+1/4+1/5+1/6+1/7+...

00 11 22 33 44Δve=-1

55ΔV(5)=+1/566 77 88 99 1100

00 33 44 55ΔV(4)=+1/6Δv

e=-1

11 22 66 77 88 99 1100

00 33 44 55Δve=-1

11 22 66 77 88 99 1100ΔV(3)=+1/7

00 11 22 33 44Δve=-1

55ΔV(6)=+1/4

77 88 99 110066

ΔV(7)=+1/388 99 1100

ve(exhaustparticle)

0.5

VM(rocket)

0.5

-1.0

0 1101

213

14

15

16

17

1819

110

pow(0)

slope

1:10

pow(1)

slope

1:9

pow(2)

slope

1:8

pow(3)

slope

1:7

pow(4)

slope

1:6

pow(5)

slope

1:5

pow(6)

slope

1:4

pow(1)

pow(2)

pow(3)

pow(4)

pow(5)

pow(6)

pow(7)

+__=0.336=0.1

ΔV(6)=+1/4

ΔV(5)=+1/5

ΔV(4)=+1/6

ΔV(3)=+1/7

ΔV(2)=+1/8

ΔV(1)=+1/9

ΔV(0)=+1/10

110+__

=1.096

19

18

17

16

14

15

110+__

=0.846

19

18

17

16

15

16

110+__=0.646

19

18

17

17

110+__=0.478

19

18

18

110

19

19

110+__=0.211

1.0

1.0



1/2

1/3

1/4

1/51/61/71/81/91/10

1

2

3

4

5

6

7

8

slope

1:1

slope

1:2

slope

1:3

1

110

16

1/5

1/4

1/4

1/4

18

1/8

1/8

19

1/10

1/9

17

1/7

1/7

1/6

1/5

1/9

1/10

1/6

1/5

00ΔV(0)=+1/10Δv

e=-1

11 22 33 44 55pow(0)

66 77 88 99 1100

11 22 33 44 55 66 77 88 99 110000

00 11 22 33 44 55ΔV(1)=+1/9Δv

e=-1

66 77 88 99 1100

00 11 22 33 44 55Δve=-1

66 77 88 99 1100ΔV(2)=+1/8


1,1/2,1/3,1/4,1/5,1/6,1/7,...

(b) Harmonic series

1+1/2+1/3+1/4+1/5+1/6+1/7+...

00 11 22 33 44Δve=-1

55ΔV(5)=+1/566 77 88 99 1100

00 33 44 55ΔV(4)=+1/6Δv

e=-1

11 22 66 77 88 99 1100

00 33 44 55Δve=-1

11 22 66 77 88 99 1100ΔV(3)=+1/7

00 11 22 33 44Δve=-1

55ΔV(6)=+1/4

77 88 99 110066

ΔV(7)=+1/388 99 1100

ve(exhaustparticle)

0.5

VM(rocket)

0.5

-1.0

0 1101

213

14

15

16

17

1819

110

pow(0)

slope

1:10

pow(1)

slope

1:9

pow(2)

slope

1:8

pow(3)

slope

1:7

pow(4)

slope

1:6

pow(5)

slope

1:5

pow(6)

slope

1:4

pow(1)

pow(2)

pow(3)

pow(4)

pow(5)

pow(6)

pow(7)

+__=0.336=0.1

ΔV(6)=+1/4

ΔV(5)=+1/5

ΔV(4)=+1/6

ΔV(3)=+1/7

ΔV(2)=+1/8

ΔV(1)=+1/9

ΔV(0)=+1/10

110+__

=1.096

19

18

17

16

14

15

110+__

=0.846

19

18

17

16

15

16

110+__=0.646

19

18

17

17

110+__=0.478

19

18

18

110

19

19

110+__=0.211

1.0

1.0

By calculus: M·ΔV=-ve·ΔM or: Integrate:dV = −vedMM

dVVINVFIN∫ = −ve M

dMMIN

MFIN∫

ve known as“Specific Impulse”



1/2

1/3

1/4

1/51/61/71/81/91/10

1

2

3

4

5

6

7

8

slope

1:1

slope

1:2

slope

1:3

1

110

16

1/5

1/4

1/4

1/4

18

1/8

1/8

19

1/10

1/9

17

1/7

1/7

1/6

1/5

1/9

1/10

1/6

1/5

00ΔV(0)=+1/10Δv

e=-1

11 22 33 44 55pow(0)

66 77 88 99 1100

11 22 33 44 55 66 77 88 99 110000

00 11 22 33 44 55ΔV(1)=+1/9Δv

e=-1

66 77 88 99 1100

00 11 22 33 44 55Δve=-1

66 77 88 99 1100ΔV(2)=+1/8


1,1/2,1/3,1/4,1/5,1/6,1/7,...

(b) Harmonic series

1+1/2+1/3+1/4+1/5+1/6+1/7+...

00 11 22 33 44Δve=-1

55ΔV(5)=+1/566 77 88 99 1100

00 33 44 55ΔV(4)=+1/6Δv

e=-1

11 22 66 77 88 99 1100

00 33 44 55Δve=-1

11 22 66 77 88 99 1100ΔV(3)=+1/7

00 11 22 33 44Δve=-1

55ΔV(6)=+1/4

77 88 99 110066

ΔV(7)=+1/388 99 1100

ve(exhaustparticle)

0.5

VM(rocket)

0.5

-1.0

0 1101

213

14

15

16

17

1819

110

pow(0)

slope

1:10

pow(1)

slope

1:9

pow(2)

slope

1:8

pow(3)

slope

1:7

pow(4)

slope

1:6

pow(5)

slope

1:5

pow(6)

slope

1:4

pow(1)

pow(2)

pow(3)

pow(4)

pow(5)

pow(6)

pow(7)

+__=0.336=0.1

ΔV(6)=+1/4

ΔV(5)=+1/5

ΔV(4)=+1/6

ΔV(3)=+1/7

ΔV(2)=+1/8

ΔV(1)=+1/9

ΔV(0)=+1/10

110+__

=1.096

19

18

17

16

14

15

110+__

=0.846

19

18

17

16

15

16

110+__=0.646

19

18

17

17

110+__=0.478

19

18

18

110

19

19

110+__=0.211

1.0

1.0

By calculus: M·ΔV=-ve·ΔM or: Integrate:dV = −vedMM

dVVINVFIN∫ = −ve M

dMMIN

MFIN∫

VFIN −VIN = −ve lnMFIN − lnMIN[ ]=ve lnMFIN

MIN⎡⎣

⎤⎦The Rocket Equation:

ve known as“Specific Impulse”





Compare mks units of Coulomb Electrostatic vs. Gravity


1

1

2 3 4

2

3

4

s1=1.5

s2=2.25

s3=3.375

s4=5.0625

s0=1.0

s-1=0.667s-2=0.444

1-1-2

y = s·x (s=1.5)y = s·x“Zig-Zag” geometry of a power sequence

Example: s=1.5

The y=

x line

at 45

°


1

1

2 3 4

2

3

4

s1=1.5

s2=2.25

s3=3.375

s4=5.0625

s0=1.0

s-1=0.667s-2=0.444

1-1-2


...and exponential function...

Example: s=1.5

The y=

x line

at 45

°


1

1

2 3 4

2

3

4

s1=1.5

s2=2.25

s3=3.375

s4=5.0625

s0=1.0

s-1=0.667s-2=0.444

1-1-2

AApppprrooxxiimmaattiinnggyy ==ssxx


...and exponential function...

Example: s=1.5

The y=

x line

at 45

°


1

1

2 3 4

2

3

4

s1=1.5

s2=2.25

s3=3.375

s4=5.0625

s0=1.0

s-1=0.667s-2=0.444

1-1-2

AApppprrooxxiimmaattiinngg::xx ==ssyy oorr yy==llooggss xx

AApppprrooxxiimmaattiinnggyy ==ssxx


...and exponential function... ...and

logarithm function...

Example: s=1.5The

y=x l

ine at

45°


y=2x

y=− x12

y=−2x

y= x12

y=x

y=-x

y=x

y=-x

s3

s2

s

1/s1/s21/s3-1/s3-1/s2

-1/s

-s

1/s1/s2 s s2

-1/s2

-1/s

-1/s3

-s

ss2s3

1/s

-s3-s2

1/s 1/s2s

1

1

1

-1

-1

0

0 1

(a) Slope factor s=2 (b) s=1/2

90°90°

90°90°

Unit 1Fig. 9.2

“Zig-Zags” give perspective geometry(1D-vanishing point )


y=2x

y=− x12

y=−2x

y= x12

y=x

y=-x

y=x

y=-x

s3

s2

s

1/s1/s21/s3-1/s3-1/s2

-1/s

-s

1/s1/s2 s s2

-1/s2

-1/s

-1/s3

-s

ss2s3

1/s

-s3-s2

1/s 1/s2s

1

1

1

-1

-1

0

0 1

(a) Slope factor s=2 (b) s=1/2

90°90°

90°90°

Unit 1Fig. 9.2

“Zig-Zags” give perspective geometry(1D-vanishing point )

1st-day-of-schoolperspective of 1st-grader

1st-day-of-schoolperspective of 12th-grader


Each y=x2 parabola point found by just one“Zig-Zag”1. Pick an (x=?)-line

x=?x=1

2. “Zig” from its y=xintersection to x=1 line

x=?

x=1

y=x y=x

3. “Zag” from originback to (x=?)-line

x=?

x=1

y=x“Zag” line is y=(?)·xand hits (x=?)-line aty=(?)·(?)=(?)2



x=?x=1


x=?

x=1

y=x y=x


x=?

x=1


Unit 1Fig. 9.1

21x=0-1-227Thursday, September 12, 2013


x=?x=1


x=?

x=1

y=x y=x


x=?

x=1


Unit 1Fig. 9.1


1

y = x

(a) Oscillator potential U(x)=x2

F(-1.5)

(b)Hooke-Law FFoorrccee FF((xx)) ==--22xx

x=0 1 2

FF((--00..55)) ==11

FF((--11)) ==22

FF((00)) ==00

-2 -1

FF((00..55)) ==--11

F(-1.25)

F(-1.0)

FF((11)) ==--22

F(+1.25)

F(+1.0)U=1

U=2

U=3

x=0 1 2-2 -1

F(2.0)F(-2.0)

F(2.4)F(-2.4)

F(1.5)

U=4

U=5

FF((xx))−ΔUU

UU((xx))Δxx

−ΔUUFF((xx))11 ==______

Δxx

12

12

12

12

12

12


x=?x=1


x=?

x=1

y=x y=x


x=?

x=1


Unit 1Fig. 9.1


A more conventional parabolic geometry…(uses focal point)

(a) Parabolic Reflector y=x2

y=1

y=2

y=3

y=4

y=5

(b)Parabolic geometry

λ

λ

Directrix

Latusrectum

reflectsintofocus

Verticalincomingray

DistancetoFocus

Distance =to

directrix

Mλ/2λ/2

Unit 1Fig. 9.3


A more conventional parabolic geometry...

(a) Parabolic Reflector y=x2

y=1

y=2

y=3

y=4

y=5

(b)Parabolic geometry

λ

λ

Directrix

Latusrectum

reflectsintofocus

Verticalincomingray

DistancetoFocus

Distance =to

directrix

Mλ/2λ/2

Unit 1Fig. 9.3

Better name† for λ : latus radius†Old term latus rectum is exclusive copyright of

X-Treme Roidrage GymsVenice Beach, CA 90017

radius


...conventional parabolic geometry...carried to extremes...

p=λ/2

y =-p

0

p

2p

3p

4p

x =0 p 2p 3p 4p

Parabola4p·y =x2=2λ·y

pdirectrix

Latus rectumλ =2p

slope=1tangent slope=-2

Circleofcurvatureradius=λ

(a)

p

p

slope=1/2y =-p

0

p

2p

3p

4p

x =0 p 2p 3p 4p


tangent slope=-5/2

(b)

Unit 1Fig. 9.4

radius


x=2.0x=0.5 x=1(0,0)

-0.5

-1

-2

-3

-4

--11//xx

--11//xx22

--11//xx

--11//xx22

2.00.5 x=1(0,0)

-1

-2

-3

-4

UU((xx))==--11//xx

--11//xx

--11//xx22

Step1 : Follow line from origin (0,0)

through (x,-1) intercept to (+1,-1/x) intercept.

Transfer laterally to draw (x,-1/x) point.


through (x,-1/x) point to (+1,-1/x2) intercept.

Transfer laterally to draw (x,-1/x2) point.

FF((xx))==--11//xx22

FF((11))

FF((00..55))

FF((00..77))

FF((xx))

11

Δxx

−ΔUU

UU((xx))

Δxx−ΔUUFF((xx))

11==

______

Step3 :(Optional) Display Force vector

using similar triangle constuction based

on the slope of potential curve.

Unit 1Fig. 9.4

Coulomb geometryForce and Potential F(x)=-1/r2 U(x)=-1/r

x1

1/x

1/x2

x:1=1:(1/x)

1x:(1/x)=1:(1/x2)


x=2.0x=0.5 x=1(0,0)

-0.5

-1

-2

-3

-4

--11//xx

--11//xx22

--11//xx

--11//xx22

2.00.5 x=1(0,0)

-1

-2

-3

-4

UU((xx))==--11//xx

--11//xx

--11//xx22


through (x,-1) intercept to (+1,-1/x) intercept.

Transfer laterally to draw (x,-1/x) point.


through (x,-1/x) point to (+1,-1/x2) intercept.

Transfer laterally to draw (x,-1/x2) point.

FF((xx))==--11//xx22

FF((11))

FF((00..55))

FF((00..77))

FF((xx))

11

Δxx

−ΔUU

UU((xx))

Δxx−ΔUUFF((xx))

11==

______

Step3 :(Optional) Display Force vector

using similar triangle constuction based

on the slope of potential curve.

Unit 1Fig. 9.4

Coulomb geometryForce and Potential F(x)=-1/r2 U(x)=-1/r

x1

1/x1/x2

x:1=1:(1/x)

1x:(1/x)=1:(1/x2)


Felec.(r) = ± 14πε0

qQr2 where : 1

4πε0

≅ 9,000,000,000 Newtons ⋅meter ⋅ squareper square Coulomb

Compare mks units for Coulomb fields1. Electrostatic force between q(Coulombs) and Q(C.)



qQr2 where : 1

4πε0



More precise value for electrostatic constant : 1/4πε0=8.987,551·109Nm2/C2 ~9·109~1010

Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+)

quantum of charge: |e|=1.6022·10-19 Coulomb

...but 1 Ampere = 1 Coulomb/sec.



qQr2 where : 1

4πε0




Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+)


...but 1 Ampere = 1 Coulomb/sec.

“Fingertip Physics” of Ch. 9 notes that 1 (cm)3 of water (1/38 Mole) has (1/38) 6·1023 molecules (about 6·1023 electrons) That’s about 6·10231.6022·10-19 Coulombor about 10+5 C or 100,000 Coulomb



qQr2 where : 1

4πε0



Fgrav.(r) = −G mMr2 where :G = 0.000,000,000,067 Newtons ⋅meter ⋅ square

per square Coulomb

2. Gravitational force between m(kilograms) and M(kg.)


More precise value for gravitational constant : G=6.67384(80)·10-11Nm2/C2 ~(2/3)10-10

COMPARE!BIGvssmall

Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+)vsAlways Attractive (so far)




qQr2 where : 1

4πε0



Repulsive (+)(+) or (-)(-)Attractive (+)(-) or (-)(+) Discussion of repulsive force and PE in Ch. 9...

1(a). Electrostatic potential energy between q(Coulombs) and Q(C.)

U(r) = 14πε0

qQr

where : 14πε0

≅ 9,000,000,000 Jouleper square Coulomb




qQr2 where : 1

4πε0





U(r) = 14πε0

qQr

where : 14πε0


Nuclear size ~ 10-15 m = 1 femtometer =1fm Atomic size ~ 1 Angstrom = 10-10 m Big molecule ~ 10 Angstrom = 10-9 m = 1nanometer=1nm


H-atom diameter

1 A=10-1nm



qQr2 where : 1

4πε0





U(r) = 14πε0

qQr

where : 14πε0




H-atom diameter

1 A=10-1nm



qQr2 where : 1

4πε0





U(r) = 14πε0

qQr

where : 14πε0



also:1fm = 10 -13 cm =1Fermi =1Fm

1 Fermi


H-atom diameter

1 A=10-1nm



qQr2 where : 1

4πε0





U(r) = 14πε0

qQr

where : 14πε0



also:1fm = 10 -13 cm =1Fermi =1Fm

1 Fermi

nuclear radii are 100,000 to 1,000,000 times smaller than atomic/chemical radii...so nuclear qQ/r energy 100,000 to 1,000,000 times bigger that of atomic/chemical...


H-atom diameter

1 A=10-1nm


Geometry of idealized “Sophomore-physics Earth” Coulomb field outside Isotropic Harmonic Oscillator (IHO) field insideContact-geometry of potential curve(s)“Crushed-Earth” models: 3 key energy “steps” and 4 key energy “levels”



d

D

You areHere!

Shell mass element

Shell mass elementM =(solid-angle factor A)D2

Gravity at rdue to shell mass elementsG M - G mD2 d2

D2 - d2

D2 d2( )A = 0

B

Θ

Θ

r

m =(solid-angle factor A) d2

M

m

=(...andweightless!)

You areHere!

OdΩsinΘ

A=

Coulomb force vanishes inside-spherical shell (Gauss-law)

Coulomb force inside-spherical body due to stuff below you, only.

M<

Gravitational force at r< isjust that of planet below r<

r<

M<

Unit 1Fig. 9.6

Cancellation of


d

D

You areHere!

Shell mass element



D2 - d2

D2 d2( )A = 0

B

Θ

Θ

r


M

m


You areHere!

OdΩsinΘ

A=



M<


r<

M<

F inside(r< ) = GmM <

r<2 = Gm 4π

3

M <

4π

3r<3r< = Gm

4π

3ρ⊕r< = mg

r<R⊕

≡ mg ⋅ x

Note:Hooke’s (linear) force law for r< inside uniform body

Unit 1Fig. 9.6

Cancellation of


d

D

You areHere!

Shell mass element



D2 - d2

D2 d2( )A = 0

B

Θ

Θ

r


M

m


You areHere!

OdΩsinΘ

A=



M<


r<

M<


r<2 = Gm 4π

3

M <

4π

3r<3r< = Gm

4π

3ρ⊕r< = mg

r<R⊕

≡ mg ⋅ x


Earth surface gravity acceleration: g = G M⊕

R⊕2 = G M⊕

R⊕3 R⊕ = G 4π

3

M⊕

4π

3R⊕

3R⊕ = G 4π

3ρ⊕R⊕ = 9.8m / s

G=6.67384(80)·10-11Nm2/C2 ~(2/3)10-10

Unit 1Fig. 9.6

Cancellation of


d

D

You areHere!

Shell mass element



D2 - d2

D2 d2( )A = 0

B

Θ

Θ

r


M

m


You areHere!

OdΩsinΘ

A=



M<


r<

M<


r<2 = Gm 4π

3

M <

4π

3r<3r< = Gm

4π

3ρ⊕r< = mg

r<R⊕

≡ mg ⋅ x


Earth surface gravity acceleration: g = G M⊕

R⊕2 = G M⊕

R⊕3 R⊕ = G 4π

3

M⊕

4π

3R⊕

3R⊕ = G 4π

3ρ⊕R⊕ = 9.8m / s

Earthradius :R⊕ = 6.371⋅106m 6.4 ⋅106m

Solarmass :M = 1.9889 × 1030 kg. 2.0 ⋅1030 kg. Earthmass :M⊕ = 5.9722 × 1024 kg. 6.0 ⋅1024 kg. Solar radius :R = 6.955 × 108m. 7.0 ⋅108m.

G=6.67384(80)·10-11Nm2/C2 ~(2/3)10-10

Unit 1Fig. 9.6

Cancellation of


Example of contacting lineand contact point

directrixdistance

Directrix

Sub-directrix

focal distance =

2.00.5 x=1(0,0)

-1

-2

UU((xx))==--11//xx

FF((11..00))

FF((xx))==--11//xx22((oouuttssiiddee EEaarrtthh))

FF((xx))==-- xx((iinnssiiddee EEaarrtthh))

FF((11..44))FF((22..00))

FF((22..88))

FF((00..88))

FF((00..44))

FF((--11..00))FF((--11..44))

FF((--22..00))

FF((--00..88))

FF((--00..44))

-0.5-1.5

Directrix

Latusrectum

λ

Focus

UU((xx))==((xx22--33))//22

Parabolic potentialinside Earth

Unit 1Fig. 9.7

The ideal “Sophomore-Physics-Earth” model of geo-gravity


...conventional parabolic geometry...carried to extremes…

(Review of Lect. 6 p.29)

p=λ/2

y =-p

0

p

2p

3p

4p

x =0 p 2p 3p 4p


pdirectrix

Latus rectumλ =2p

slope=1tangent slope=-2

Circleofcurvatureradius=λ

(a)

p

p

slope=1/2y =-p

0

p

2p

3p

4p

x =0 p 2p 3p 4p


tangent slope=-5/2

(b)

Unit 1Fig. 9.4

radius


surface gravity: g = −G M⊕

R⊕2

surface potential: PE= −G M⊕

R⊕

surface escape speed: ve= 2G M⊕

R⊕

KE = PE relation: 12mv2

e=GmM⊕

R⊕

-orbit energy: E = −G M⊕

2R⊕

-orbit speed: v = G M⊕

R⊕

The “Three (equal) steps from Hell”

1

2

3

Ground level : PE= −G 3M⊕

2R⊕

Surface level : PE= −G M⊕

R⊕

Orbit level : PE= −G M⊕

2R⊕

Dissociation threshold : PE= 0


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