Mechanics REVIEW: What are Newton’s three laws of motion?Newton s three laws of motion?

Newton’s first law of motion: An object moves jat constant velocity unless a net force acts to change it d di tiits speed or direction.

Aristotle said you need a force to stay in motion..

F ll f B A i G lil h i t dFull of Beans Again says Galileo. .he experiments andDiscovers a law of motion.

G.G. explores idea = Newton’s First Law!

Newton’s second law of motionNewton s second law of motion

Force (net) on body of mass = mass ( ) yacceleration = ma=dp/dt(as originally stated by Newton) With p=mv

p= momentumIf Fnet=ma then a=Fnet/m

Newton’s third law of motion:Newton s third law of motion:

For every force there is always an equal andFor every force, there is always an equal and opposite reaction force.

She pushesp

FAB = -FBA The third law

Reaction on each

The Universal Law of Gravitation:1. Every mass attracts every other mass.2 Att ti i di tl ti l t th d t2. Attraction is directly proportional to the product

of their masses.3 Attraction is inversely proportional to the3. Attraction is inversely proportional to the

square of the distance between their centers.

A few review words on UnitsForce: In English system known as Pounds M t i t d i d f N t 2nd L f M tiMetric system derived from Newtons 2nd Law of MotionF =ma CGS:

1 Dyne = 1gm x 1 cm/sec21 Dyne 1gm x 1 cm/sec2MKS (Systeme International):

1 Newton = 1 kg x 1 m/sec2EEnergy

Derived from changing the state of energies, potential, kineticWe say we do WORK on a system to change it’ energy.y y g gyMechanical work is the Force you apply over some distanceW=E =Fd =Fdr (if F is not a constant-> integral)

U it f E W Fd Other useful unitsUnits of Energy: W =FdCGS: 1erg = 1 dyne x 1cm

MKS: 1 Joule =1Newton x 1 meter

Other useful units Power = E/t

1 watt= 1 joule/secSmall units of energy 1 eV =1.6 x 10-12 erg

Force is mutual each pulls on the other with equal B i f i f iBut opposite force - sign means force attractive

rr

FG = - G m1 m2

r

r 2W= U =S r1

r2 Fdr = -Gm1 m2 S r1 r2

1/r2 drG ( 1/ 1/ )= Gm1 m2 ( 1/r2 -1/r1 )

Or U = - Gm1 m2 /r Equation 9.5 Potential Energy

U of a solid spherical mass,M&R = -(3/5) GM2/R =USee derivation page 158…don’t worry

Escape VelocityVelocity needed

to escape Earth’s gravity from the

vesc

gravity from the surface: vesc ≈

11.6 km/s.

Now, gravitational force decreases with distance (~

vesc(

1/d2) => Starting out high above the surface => lower escape velocity.

vesc

If you could compress Earth to a smaller radiusIf you could compress Earth to a smaller radius => higher escape velocity from the surface.

Non-relativistic!Or falls from infinity

to object at R!to object at R!

= (2GM/R)1/2 (2GM/R)

Examples Vescape

• Earth Fg =GM1M2/R2 units 1 dyne=G g2/cm2

• But F=ma >1 dyne =g cm/s2 or• But F=ma ->1 dyne =g cm/s2 or • units of G=kgm/s2 /kg2 /m2 = cm3/s2 g or• G=6 67 x 10-11 cm3/s2 g• G=6.67 x 10 11 cm3/s2 g • ves=(2GM/R)1/2 • =(2 x 6 67 x 10-7 cm3/s2 g x5 98 x 1027 g/6 37 x 108 cm)1/2• =(2 x 6.67 x 10 cm /s g x5.98 x 10 g/6.37 x 10 cm)• =1.12 x 106 cm/s =11.2 km/s ~ 7mi/s• A NEUTRON STAR Ves = 2 x 108 cm/s ~ 2/3 c• If a star hits the Schwartzchild Radius we solved before,namely• Rs =2GM/c2 WHAT IS Ves = for this case? Solve now!

STARS ARE A BALANCE OFGRAVITY AND CORE ENERGYGRAVITY AND CORE ENERGY

FOR MOST OF THEIR LIFESTARS FORM FROM COLLAPSING MATERIAL THIS GIVES RISE FOR THE STARS, OF ENERGY COMING FROM THE RELEASE OF GRAVITATIONAL POTENTIAL ENERGY,U..IT CAN BE SHOWN THATGRAVITATIONAL POTENTIAL ENERGY,U..IT CAN BE SHOWN THAT TYPICALLY CONTAINED IN A SPHERE OF RADIUS R

U = -(3/5) GM2/R

SEE DERIVATION SECTION 9.1.1 (DON’T WORRY)

THERMAL ENERGYTHERMAL ENERGY

• THE THERMAL ENERGY WE SAWTHE THERMAL ENERGY WE SAW BEFORE IS (3/2) k T PER PARTICLE

• N=#PARTICLES=M/m M=STAR MASSN #PARTICLES M/m M STAR MASS• M=HYDROGEN ATOM MASS• HENCE TOTAL THERMAL ENERGY IN• HENCE TOTAL THERMAL ENERGY IN

STAR, K, IS• K =(3/2) NKT equation 9 8 or• K =(3/2) NKT equation 9.8 or • K= (3/2)(M/m) kT equation 9.10

SEC 9.1.2 Gravitational lifetime of our sun• LifeTime =total grav energy/energy given offLifeTime total grav energy/energy given off• Or tE =E /dE/dt recall dE/dt =L (Luminosity)• And E is all the stored Gravitational enegy assembled from gy

inifinity or magnitde of energy =-U for the SUN • E=3/5 GM2/R =2 x 1048 erg see eq. 9.12

i L 4 1033• since L=4 x 1033 ergtE =E/L= 2 x 1014 s =2 x 107 yr• In other words: GIVEN THE SUNS LUMINOSITY ALL THIS• In other words: GIVEN THE SUNS LUMINOSITY ALL THIS

ENERGY WOULD BE EXPENDED IN 20 Million years• This is a very small lifetime in the cosmic scheme of things! y g

Even chemical energy (OIL, COAL, GAS, ETC) Is not sufficient to account for the billions of years our Sun

• has existed based on the Earth’s Moon and Meteoric• has existed based on the Earth s , Moon and Meteoric determined ages. So what keeps it ticking! Note see HW!

How Does the Stars Make Energy for their life cycle?

Why a Steady Luminosity on the

main sequence?Why does a StarWhy does a Star

changeDramatically in

the later stages ofit’s life?

ANS=ANS=The NUCLEARFURNACE IN THE CORE!

Planetary Nebulastellar DEATH!

Some stars near theSome stars near theend are unstableAnd eject theirOuter atmosphere

The Nuclei (REVIEW)• Nuclei consist of a number

of protons, Z, and Neutrons, N We define the MASSN. We define the MASS Number as A = Z+N use notations for Atom AT are

Different numbers of neutrons ↔

• A AT Z or AAT or AT-A different isotopes

• Isotopes have same Z butIsotopes have same Z but different A as 12C6 or 12C or C-12 and 13C or C-13

• This is the isotope• 4He2 or 4He

Helium 4

e2 o e

NUCLEAR PHYSICS IINOTE 10-13J = 1 Mev

Energy generation in the sun (and all other stars): Binding energy

due to strong Nuclear Fusion

= fusing together 2 or more

gforce = on short range, strongest of the 4 known

lighter nuclei to produce heavier ones.

Nuclear fusion can

forces:electromagnetic, weak, strong,gravitationalNuclear fusion can

produce energy up to the production of iron;

gravitational

For elements heavier than iron, energy is gained by nuclear fission.

NUCLEAR PHYSICS I• Using Einstein’s E=mc2 if we add the energies of• Using Einstein s E=mc if we add the energies of

protons and neutrons and the the actual energy of the nucleus we find the latter is less. In other words in the assembly of the nucleus energy went into binding itassembly of the nucleus energy went into binding it, due to the Nuclear force,thus, the Binding Energy, BE

• Mnucleus c2 +BE=Zmpc2 +Nmnc2 eq. 9.13 ->

• BE = Zm c2 + Nm c2 - M c2• BE = Zmpc2 + Nmnc2 - Mnucleus c2

• Recall 1eV=1 6 x 10 -12 ergRecall 1eV=1.6 x 10 erg • BE’s are measured in Mev and are easily found!• See Examples 9.1 and 9.2 p

Examples 9.1 and 9.2Ex 9.1 Rest energy of Proton?

E=mpc2 =(1.67 x 10-24g)(3.0 x 1010cm/s)2 =1.5 x 10-3 ergRecall 1eV=1.6 x 10 -12 erg

So E= =1.5 x 10-3 erg/ 1.6 x 10 -12 erg /eV=9.4 x 108 eV= 940 MeV

EX 9.2 Deuteron binding energy ? givenmd=3.3436x 10 -24 g mp = 1.6726x 10 -24 g mn = 1.6749x 10 -24 g

9.13 -> BE = mpc2 + mnc2 - md c2 = (mp+ mn - md )c2 =(0.0039x 10 -24 erg)(2.99779x 10 10 cm/s)2

(3 9 10 27 )(2 99 9 10 10 / )2=(3.9x 10 -27 erg)(2.99779x 10 10 cm/s)2 =3.6 x10 -6 erg

= 3.6x 10 -6 erg / 1.6 x 10 -12 erg /eVg g=2.2 x 10 6 eV = 2.2 MeV

Nuclear ReactionsThe liberation of Binding EnergyThe liberation of Binding Energy

Reaction types that happen in the Nuclear world3 Decays that release Energy!3 Decays that release Energy!: Alpha: Large Nucleus emits 4He nuclei hence Z and A go down by ? Energy goes out as Kinetic in g y gy g: Beta: np +e- +anti- Z, N and A ??

: Gamma: Nucleus goes from higher energy state tolower emits this E&M Radiation( a way of seeing into the nucleus!FISSION: BREAK APART OF NUCLEUS as in atomic

reactors and bombs.FUSION: low mass nuclei build larger ones, as in starsAlso H-bomb and on a small scale in the laboratory

Star Power Nuclear Fusion ReactionsThis reaction by astrophysics defines a Star!

Coulomb OR FUSION barrierIS OVERCOME with?

y p y

T=20,000,000 KIS OVERCOME with?

Note: leaving!

Don’t worryYou’ll meet

A nice electronAnd

AnnihilateN t AnnihilateEach other

2H

Note: pn +e+ +Needs energy HNeeds energyie: mn >mp

Positron Star Power by Proton-Proton chain conversion of H to He

Deuterium 1H + 1H -> 2H + e+ + neutrinoFUSION

N t i ?2H + 1H -> 3He +GammaFUSION

Proton Neutrino?Helium 3 Hey Protons are

leavingGamma ray

FUSION

g

Helium 4!

3He + 3He -> 4He + 1H + 1H

Helium 4!

Proton-proton chain

1H + 1H -> 2H + e+ + neutrino2H + 1H -> 3He +Gamma ray

3He + 3He > 4He + 1H + 1H

H + H > H + e + neutrino

3He + 3He -> 4He + 1H + 1HNet Fusion Reactions in the Sun

NOTE 6 H -> 1 He and 2 H’s or…..4 1H -> 4He +ENERGY

4 hydrogen nuclei =4mp = 4 x1.6726x10 -24 g=6.6904 x 10-24g=4mp1 h li 4 l 6 6466 10 24 (4 )1 helium 4 nucleus = 6.6466 x 10-24g= m(4

He)Missing mass ? = Energy released via E = m c 2

4mp - m(4He) = 0.0438 x 10-24g= .0065 x6.6904 x 10-27 kg

Or ~4mp - m(4He) = 0.007 (4mp ) equation 9.18

4mp - m(4He) = 0.007 (4mp )

Implies that every proton gives up ABOUT 007 OF ITS MASSABOUT .007 OF ITS MASS

Ex 9.3 LIFE TIME OF THE SUN FROM CONVERTING PROTONS TO ENERGYPROTONS TO ENERGY

Assume the Sun’s mass is all protons and each one will lose over the lifetime of the sun .007 of its mass then how long

ld th l t if thi th l ?would the sun last if this was the only energy source?Total Energy reserve is

E=0.007Msun c2 =(0.007)x(2.0 x 1033g)x(3.0 x 1010 cm/s)2sun ( ) ( g) ( )

= 1.3 x 1053 ergs (that’s a lot!) Lifetime

t = E/L =1 3 x 1053 ergs/4x 1033 erg/s =3 2 x 1018s =1 x 10 11 yrOur Sun requires 1038 reactions per second To Balance Gravity or

5 million tons of mass per second

t E = E/L =1.3 x 1053 ergs/4x 1033 erg/s =3.2 x 1018s =1 x 10 11 yr

5 million tons of mass per secondIt takes 10 Billion years to consume core Hydrogen this way!

GO OVER EXAMPLE 9.3!

HWHW

• Read Chap 9 -9 3Read Chap 9 9.3• Study examples 9.1,9.2,9.3 remember

E=mc2E=mc2.• Problems 9.1(HW#1), look up size of a

t ll bl k h lstellar black hole• 9.4(HW#2),,9.6(HW#3) , 9.8 (HW#3B)• HAND IN ONE WEEK FROM TODAY!

Energy Generation in the Sun: The Proton-Proton Chain --A REVIEW

Basic reaction:

4 1H 4He + energyNeed large proton speed ( high

temperature) to overcome

4 protons have

temperature) to overcome Coulomb barrier (electrostatic repulsion between protons).

T ≥ 107 0K =4 protons have 0.048*10-27 kg (= 0.7 %) more mass than 4He.

T ≥ 107 0K = 10 million 0K

Energy gain = m*c2

= 0.43*10-11 J

per reaction.

Sun needs 1038 reactions, transforming 5 million tons of mass into energy every second, to resist its own gravity.

Triple- Process Massive stars or in evolution to a Red Giant Evolution as below

H-burning shell keeps dumping He

t th

4 H → He

onto the core.

He-core gets denser and hotter until the

He next stage of nuclear burning can begin in

the core: T=108 KHe fusion through the

“Triple-Alpha Process”Process

4He + 4He 8Be +

Some Be breaks up but

AM I STARDUST?PROFESSOR

8Be + 4He 12C +

In soup of 4He

The CNO Cycle In stars slightly more massive than the sun, a more powerful

energy generationenergy generation mechanism than

the PP chain t ktakes over:

The CNO Cycle.

Net ResultNet Result4 H4He+2 e+

+2+3

Fusion Into Heavier Elements

Slow n capture (s-process) decay(Z A)+n(Z A+1)(Z+1 A+1)+e-+anti (CORRECTED)

With lots of neutrons around Fusion into heavier elements than C, O: is

(Z,A)+n(Z,A+1)(Z+1,A+1)+e-+anti (CORRECTED)

Or (Z,A)(Z-1,A)+e+

possible.

Neutron capture: (Z,A)+n(Z,A+1) 9.29Neutron capture: (Z,A) n(Z,A 1) 9.29Fast n capture =r-process another n captu

we get (Z,A)+n(Z,A+2) 9.31

QUIZ: What’s up?

Other possible NUCLEAR FUSION reactions IN STARS Releases Energy

All these reaction take place in the last stages of a Stars life

Main Sequence Band

ZAMS LINE FIRST APPEARENCE OF STAR

NUCLEAR REACTIONS CHANGE COMPOSITION WHICH CHANGES

B0 K0

LUMINOSITY AND SPECTRAL TYPE..STAR EVOLVES ALONG LINES INDICATED

Hence We See a Band of Stars

Nuclear Fission LET’S MAKE THE DESERTS BLOOM

Weapons or Electric Power It’s up to YOU!

Weapons

Electric Power for all

B t P f N l P Pl t S ME!But Professor – Nuclear Power Plants Scare ME!

BIG PROBLEM: only 1/3 Solar Neutrino’s detected Either wedo not understand Energy Generation or our Neutrino Model

It look’s today like our “Snu” model needs

refinement!refinement!

Super-Kamiokande -50,000 ton neutrino experiment located 1 km underground in an unused zinc mine near Kamioka, Japan.

The Solar Neutrino ProblemThe solar interior can not be observed directly because it is highly opaque to radiation.g y p qBut neutrinos can penetrate huge amounts of material without being absorbed.Early solar neutrino experiments detected a muchexperiments detected a much lower flux of neutrinos than expected ( the “solar neutrino problem”)neutrino problem ).Recent results have proven that neutrinos change (“oscillate”) b t diff t t (“fl ”) Davis solar neutrino

experimentbetween different types (“flavors”), thus solving the solar neutrino problem. ,MAYBE????