Fri. 6.11, 14-17 Visualizing Electric and Rest Energy RE 6...

Fri. 6.11, 14-17 Visualizing Electric and Rest Energy RE 6.d,e

Mon.

Tues.

Things Engineers and Physicists Do

EP6, HW6: Ch 6 Pr’s 58, 59, 91,

99(a-c), 105(a-c)

A huge asteroid smacks into the leading edge of the Earth – stopping

the Earth’s orbit. Subsequently, the Earth falls straight into the sun!

With what speed would the Earth hit the Sun’s surface?

System= Earth + Sun

Active environment = none

Em

kgmS

301099.1

0. iEv

?. fEvmr iSE

11

. 105.1

SEfSE RRr .

m81002.7 extsystemWE 0

0,.., SESESrestErestSE UKKEEE

Not changing

0,, SEESE UKE

iEfEE KKK .. 2

.21

fEEvm

iES

sE

fES

sE

ES

sEES

r

mmG

r

mmG

r

mmGU

..

02

.21

, ESi

sE

ESf

sEfEESE

r

mmG

r

mmGvmE

ESiESf

sfErr

Gmv11

2.

smmm

kgkgNm /1014.6105.1

1

1002.7

11099.1/1067.62 5

118

302211

System: comet + star

Surroundings: negligible

As a comet travels away from a star, how do the

kinetic energy and potential energy of the system

change?

K U

a) increase decrease

b) increase increase

c) decrease increase

d) decrease decrease

e) no change no change

Force as negative gradient (3-D slope) of

Potential Energy

dzFdyFdxFrdFdU zyx .21.21.2121212,1 small change in potential

Say only moves in the x direction, then

dxFdU x.212,1 so xFdx

dU.21

2,1

x̂

2,1U

x̂

xFdx

dU.21

2,1

repuls

ive

attra

ctive

repuls

ive

Similarly, if only moves in the y direction, then

dyFdU y.212,1 so yFdy

dU.21

2,1

or, if only moves in the z direction, then

dzFdU z.212,1 so zFdz

dU.21

2,1

Moving in all directions,

21

2,1

21

2,1

21

2,1

.21.21.2121 ,,,,

dz

dU

dy

dU

x

UFFFF zyx

|r|

K+U = Const K=0 K=0

Energ

y

Limits / Turning

Points

E = Wext=0

(K + U + Erest) = 0

Kf+Uf = Ki + Ui = Constant

y

Conceptual Understanding from Energy Diagrams

Roller Coaster

Ugra

v=

mg

04_potential_energy_well.py

|r|

K+U = Const

r1

K=0 K=0

Energ

y

Limits / Turning

Points

U(r1)

K(r1)+U(r1)=Const

K(r1)=Const - U(r1)


Ex. Nucleus - Proton Potential

E = Wext=0

(K + U + Erest) = 0


|r| r1

K(r1)

K=0 K=0

Energ

y

Limits / Turning

Points

U(r1)

K+U = Const

K(r1)+U(r1)=Const

K(r1)=Const - U(r1)



E = Wext=0

(K + U + Erest) = 0


|r|

K+U = Const

r1

K(r1)

K=0 K=0

Energ

y

Limits / Turning

Points

U(r1)

K(r1)+U(r1)=Const = K(r2)+U(r2)

K(r1)=Const - U(r1)

K(r1)

r2

U(r2)



E = Wext=0

(K + U + Erest) = 0


Which of the following graphs of U vs r represents the

gravitational potential energy, U = –GMm/r?

vi = 0

r

mMGU s

r

(Separation)

A)

Energy Diagrams

Ex. Gravitational, Bound

(K + U)A

Energ

y

vi = 0

r

mMGU s

r

(Separation)

A)

Rstar

B) rmax rmin

(K + U)A

(K + U)B

Energ

y

Energy Diagrams


vi = 0

r

mMGU s

r

(Separation)

Energ

y

A)

Rstar

B) rmax rmin

r C)

(K + U)A

(K + U)B

(K + U)C

Energy Diagrams


Energy Diagrams

Ex. Gravitational, Un-Bound / Escape

r

mMGU

p

r

(Separation)

K+

U

rplanet

A

B

C

D

Consider an rocket launching from a planet’s surface, which of the following

represent un-bound systems (so the rocket could get away and never fall back to

the planet)?

1. A

2. B

3. A & B

4. C

5. D

6. C & D

7. A,B,C, & D

Special Case: K + U = 0 and Escape Speed

In which graph does the cyan line correctly represent

the sum of kinetic energy plus potential energy?

1) A 2) B 3) C 4) A,B 5) B,C 6) A,B,C

The system is a comet and a star. In which case(s) will the

comet escape from the star and never return?

Gravitational Potential Energy

Newton’s Universal Law of

Gravitation

12r

1m

2m

12F

21F

122

12

21

rr

mmG ˆ12F

22

1110676kg

mNG

.

12r̂

12r̂12

12

r

r

1q

2q

2

29

41 109

C

mNo

122

12

214

1

rr

qqo

ˆ

Coulomb’s Electric Force Law

21

212,1

r

mmGU

21

21

41

2,1

r

qqU

o

Electric Potential Energy

21r

electricU .2,1

opposite charges

21r

electricU .2,1

like charges

Example: Ionize Hydrogen. In a hydrogen atom the electron averages around

10-10 m from the proton. When a hydrogen atom is ionized, the electron is stripped

away. What is the change in electric potential energy when such an atom is

ionized?

per

electricpeU .,

mri

1010 fr

pe

pe

electper

qqU

o

41

.,

System= electron + proton

Active environment = none

21

2

41

.,

r

eU

oelectpe

if

elctper

e

r

eU

oo

2

41

2

41

,,

fi

e

rro

114

2

∞

0

m

CU

Nm

Celctpe 10

219

1085.84

1,,

10

106.1

2

212

JU elctpe

18

,, 103.2

Or in eV’s (divide by electron charge)

C

eJ 19106.1

118103.2

eV14

Comparison:

Electric vs. Gravitational

21

21

2

41

.,

.,

r

mmG

r

e

U

U

pegravpe

electpeo

pe

G

gravpe

electpe

mm

e

U

U

o

2

41

.,

.,

39

.,

.,106.5

gravpe

electpe

U

U

kgkg

C

kg

Nm

C

Nm

273111

2199

107.1109107.6

106.1109

2

2

2

2

Return to Rest Energy and Mass

Pair (electron and positron) Annihilation

-e

+e

per peU ,

ir

ir

-e

+e

22 cmcmE ee

eeinitial final

Electron and positron Two photons (light pulses)

EcmE e 22 2

EccMeV 22/511.0

EMeV 511.0

Application Note:

Positron Electron Tomography (PET)

Scans

Return to Rest Energy and Mass Neutron Decay

per peU ,

fr

ir

-e p+

2cmE n

initial final

neutron Proton, electron, and neutrino

pepepepen UUUKKKcmcmcmcmE ,,,

2222

eepn 0

n e

KKKcmcmcmE pepen 222

Finally infinitely far apart Nearly massless

222 cmcmcmKKK penpe

MeVMeVMeV 3.938511.06.939 MeV79.0

Mass as Energy and Energy as Mass

Box o’ decaying Neutrons

Viewed from outside

2cmEE boxrest

Peaking inside

particlesall

pepepe UKKKcmmm.

,

2

Box’s mass includes internal kinetic and potential energies

What is Mass

Quantification of…

Gravitational ‘charge’

Inertia

Internal Energy

r

Energ

y

Efree=2mc2

Mfree=Efree/c2=2m

rfree

K+U=0

Return to Rest Energy and Mass O2 bonding

rbound r

Energ

y

U

rfree

K

K+U

Efree=2mc2

Mfree=Efree/c2=2m

Return to Rest Energy and Mass O2 bonding

rbound r

Energ

y

U

rfree

K

K+U

Ebound=2mc2+ (K+ U)

Mbound=Ebound/c2 =2m + (K+U)/c2

K+U


Note: would have shed excess energy

by emitting photon / light pulse

O2 bonding

2/ cEEMMM bfboundfree

Energy / Mass difference

kgceVcUKM bound

3622 109/5/

222 /22 cUKmcmcM bound

Noticeable?

101032

om

M

Efree=2mc2

Mfree=Efree/c2=2m

not really


Nuclear Binding: Iron nucleus

initial final

Iron nucleus Protons and neutrons

pairsallparticlesall

npFe UKcmcmcm..

222 3026

npFe 302626

56

fi EE


rFer UKEE..

.%1009.0

2

2

cm

mc

Fe


npFe UKcmcmcm..

222 3026

2

2

2

.

/3.938

/9.939

/52107

cMeVm

cMeVm

cMeVM

p

n

nucFe

If an iron nucleus were disintegrated, how much K + U energy would be

consumed /produced?

Useful info


UKMeVMeVMeV..

3.938309.9392652107


UKMeV..

482

Noticeable?

yes

Rest and Electric-Potential and Kinetic A U-235 nucleus is struck by a slow-moving neutron, so that the merge and become U-

236, with mass MU-236 This nucleus is unstable to falling apart – fission. One way it

could do so is to first slosh into something of a dumbbell shape, now most of the into two

symmetric nuclei, Pd-118, with mass MPd-118 , each has ½ the original number of protons,

i.e., qPd = 46e. Having fallen apart, the two palladium nuclei no longer experience a Strong

interaction holding them together, just the Electric repulsion of each other’s protons.

Subsequently, they accelerate away.

a) What’s the final speed of one of the Pd atoms, when they have sped far, far apart?

b) What is the distance between the Pd atoms just after fission?

Fri. 6.11, 14-17 Visualizing Electric and Rest Energy RE 6.d,e

Mon.

Tues.

Things Engineers and Physicists Do

EP6, HW6: Ch 6 Pr’s

58, 59, 91, 99(a-c),

105(a-c)

Fri. 6.11, 14-17 Visualizing Electric and Rest Energy RE 6...

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Transcript of Fri. 6.11, 14-17 Visualizing Electric and Rest Energy RE 6...