13.6 Gravitational 13.6 Gravitational Potential EnergyPotential Energy

Gravitational Potential Gravitational Potential EnergyEnergy

In Chapter 8: UU = mgy = mgy (Particle-Earth). Valid onlyValid only when particle is near the Earth’s surface.

Now with the introduction of the Inverse Square LawInverse Square Law ((FFgg 1/r 1/r22)) we must look for a general potential energy UU function without restrictions.

Gravitational Potential Energy, Gravitational Potential Energy, contcont

We will accept as a fact fact that: The gravitational force is conservativegravitational force is conservative

(Sec 8.3) The gravitational force is a central forcegravitational force is a central force

It is directed along a radial line toward the center

Its magnitude depends only on rr A central force can be represented by ˆF r r

Grav. Potential Energy–Grav. Potential Energy–WorkWork

A particle moves from AA to BB while acted on by a central force F

The path is broken into a series of radial segments and arcs

Because the work done along the arcs is zero (F(Fv)v), the work done is independent of the path and depends only on rrff and rrii

Grav. Potential Energy–Work, Grav. Potential Energy–Work, 22

The work done by FF along any radial segment is

The work done by a force that is perpendicular to the displacement is 00

Total workTotal work is

Therefore, the work is independent of the path

( )dW d F r dr F r

( ) f

i

r

rW F r dr

Grav. Potential Energy–Work, Grav. Potential Energy–Work, finalfinal

As a particle moves from AA to BB, its gravitational potential energy changes by (Recall Eqn 8.15)

(13.11)(13.11) This is the general formgeneral form,

we need to look at gravitational force specifically

( ) f

i

r

f i rU U U F r dr

Grav. Potential Energy for Grav. Potential Energy for the Earththe Earth

Since: F(r)F(r) = - = - GG(M(MEEm)/rm)/r22 Eqn. (13.11) becomes

(13.12)(13.12)

Taking UUii = 0= 0 at rrii → ∞ → ∞ we obtain the important result:

(13.13)(13.13)

( ) EGM mU r

r

ifE

r

rE

r

rEif rrmGM

rmGMdr

rmGMUU

f

i

f

i

11112

Gravitational Potential Gravitational Potential Energy for the Earth, finalEnergy for the Earth, final

Eqn. 13.13Eqn. 13.13 applies to Earth-Particle system, with r r ≥ R≥ REE (above the Earth’s surface).

Eqn. 13.13 Eqn. 13.13 is not valid for particle inside the Earth r r < R< REE

UU is always negative (the force is attractive and we took UUii = 0= 0 at rrii → ∞ → ∞

An external agentAn external agent (you lifting a you lifting a bookbook) must do + Work+ Work to increase the heightheight (separation between the Earth’s surface and the book). This work increases U U UU becomes less negative as rr increases.

Gravitational Potential Gravitational Potential Energy, GeneralEnergy, General

For any two particlestwo particles, the gravitational gravitational potential energypotential energy function becomes

(13.14)(13.14)

The gravitational potential energy between any two particles varies as 1/1/rr Remember the forceforce varies as 1/1/r r 22

The potential energy is negative because the force is attractive and we chose the potential energy to be zero at infinite separation

1 2GmmU

r

Systems with Three or Systems with Three or More ParticlesMore Particles

The total gravitational total gravitational potential energypotential energy of the system is the sumthe sum over all pairs of particles

Gravitational potential energy obeys the superposition superposition principleprinciple

Each pair of particles contributes a term of UU

Systems with Three or Systems with Three or More Particles, finalMore Particles, final

Assuming three particlesthree particles:

(13.15)(13.15)

The absolute value of The absolute value of UUtotaltotal represents the work neededwork needed to separate the separate the particlesparticles by an infinite distance

total 12 13 23

1 2 1 3 2 3

12 13 23

U U U U

mm mm m mG

r r r

Binding EnergyBinding Energy The absolute value of the potential

energy can be thought of as the binding energybinding energy

If an external agent applies a force largerlarger than the binding energy, the excessexcess energy will be in the form of kinetic energykinetic energy of the particles when they are at infinite separation

Example 13.8Example 13.8The Change in Potential EnergyThe Change in Potential Energy(Example 13.6 Text Book)(Example 13.6 Text Book)

A particle of mass mm is displaced through a small vertical displacement yy near the Earth’s surface. Show that the general expression for the change in gravitational potential energy (Equation 13.12) reduces to the familiar relationship:

UU = = mmggyy From Equation 13.12

fi

ifE

fi

fiE

ifEif

rr

rrmGMU

rr

rrmGM

rrmGMUUU

11

Example 13.8Example 13.8The Change in Potential Energy, The Change in Potential Energy, contcont

Since rrii and rrff are to close to the Earth’s surface, then:

rrff – – rrii ≈≈ y y and rrffrrii ≈ ≈ RREE2 2

Remember rr is measured from center of the Earth

UU = = mmggyy

ymgy

R

GMm

R

ymGM

rr

rrmGMU

E

E

EE

fi

ifE 22

An object of mass mm is moving with speed v v near a massive object of mass MM (MM >> mm).

The Total Mechanical Energy Total Mechanical Energy EE, when the objects are separated a distance rr, will be:

E = K +UE = K +U = ½= ½ m m vv22 –– GGMMmm/rr (13.16)(13.16)

Depending on the value of vv,

EE could be +, - , 0+, - , 0

MM

m

vv rr

13.7 Energy Considerations 13.7 Energy Considerations in Planetary and Satellite in Planetary and Satellite MotionMotion

Newton’s 2nd Law for mass mm:

GGMMmm/rr22 = = m m aaRR == mmvv22//rr Multiplying by rr/2/2 both sides:

GGMMmm/22r r = = mmvv22//2 2 ½½mmvv2 2 == ½ ½((GGMMmm/rr) )

K = ½(GMK = ½(GMmm//rr) ) (13.17)(13.17)

Kinetic Energy (K) is Kinetic Energy (K) is positive and equal to half positive and equal to half the absolute value of the the absolute value of the potential Energy (U)!!!!potential Energy (U)!!!!

MM

m

vv rr

Energy and Satellite Motion, Energy and Satellite Motion, contcont

Energy in a Circular OrbitEnergy in a Circular Orbit Substituting (13.17) into (13.16):

E = E = ½½ (G(GMMmm/rr)) –– GGMMmm/r r

(13.18)(13.18)

The total mechanical energytotal mechanical energy is negativenegative in the case of a circular orbit

The kinetic energykinetic energy is positivepositive and is equal to half the absolute value of the potential energy

The absolute value of absolute value of EE is equal to the binding energybinding energy of the system

2

GMmE

r

Energy in an Elliptical Energy in an Elliptical OrbitOrbit

For an elliptical orbit, the radius is replaced by the semimajor axis aa

(13.19)(13.19)

The total mechanical energytotal mechanical energy is negativenegative

The total energytotal energy is constantconstant if the system is isolatedsystem is isolated

2

GMmE

a

Summary of Two Particle Summary of Two Particle Bound SystemBound System

Conservation of Total EnergyConservation of Total Energy will be written as:

(13.20)(13.20)

Both the total energytotal energy and the total total angular momentum angular momentum of a gravitationally bound,gravitationally bound, two-object system are constants of the motionare constants of the motion

2 21 1

2 2i fi f

GMm GMmE mv mv

r r

An object of mass mm is projected vertically upward from Earth’s surface, with initial speed vvii.

Let’s find the minimum value of vvii needed by the object to move infinitely far away from Earth.

The energy of the system at any point is:

E = E = ½½ mmvv22 –– GMGMEEmm/rr

Escape Speed from EarthEscape Speed from Earth

At Earth’s surface:

v v = vvii & r r = rrii = RREE

At Maximum height:

vv = vvff = 0= 0 & r r = rrff = rrmaxmax

Using conservation of the Energy (Eqn.13.20):

½ ½ mmvvii22 – GM – GMEEmm//rrii = ½ = ½ mmvvff

22 – GM – GMEEmm//rrf f

½ ½ mmvvii22 – GM – GMEEmm//RREE = 0 – GM= 0 – GMEEmm//rrmaxmax

Solving for vi2:

vvii2 2 = 2GM= 2GMEE(1/R(1/REE – 1/r – 1/rmaxmax)) (13.21)(13.21)

Knowing vvii we can calculate the maximum altitude h:h: h = rrmaxmax – RREE

Escape Speed from Earth, Escape Speed from Earth, 22

Escape Speed from Earth, Escape Speed from Earth, 33

Let’s calculate Escape SpeedEscape Speed (v(vescesc ))

Minimum speedMinimum speed for the object must have at the Earth’s surfaceEarth’s surface in order to approach an infiniteinfinite separation distance from Earth.

Traveling at this minimum speedminimum speed the object continues to move farther away, such that we take:

rrmaxmax → ∞ or 1/r → ∞ or 1/rmaxmax → 0 → 0 Equation (13.21) becomes:

vvii2 2 = v= vescesc

2 2 = 2GM= 2GMEE(1/R(1/REE – 0) – 0)

(13.22)(13.22)

2 Eesc

E

GMv

R

Escape Speed From Earth, Escape Speed From Earth, finalfinal

vvescesc is independentindependent of the mass mass of the object of the object (Question 8 (Question 8 Homework)Homework)

vvescesc is the same for a spacecraft or a molecule.

The result is independent of the direction of the velocity and ignores air resistance

Escape Speed, GeneralEscape Speed, General The Earth’s result can be

extended to any planet of mass MM

(13.22a)(13.22a)

The table at right gives some escape speeds from various objects

esc

2GMv

R

Escape Speed, Escape Speed, ImplicationsImplications

Complete escape from an object is Complete escape from an object is not really possiblenot really possible The gravitational field is infinite and so

some gravitational force will always be felt no matter how far away you can get

This explains why some planets This explains why some planets have atmospheres and others do have atmospheres and others do notnot Lighter moleculesLighter molecules have higher average higher average

speedsspeeds and are more likely to reach escape speeds

Example 13.9 Example 13.9 Escape Speed of a RocketEscape Speed of a Rocket(Example 13.8 Text Book)(Example 13.8 Text Book)

Calculate vvesc esc from the Earth for a 5,000 kg spacecraft.

vvescesc = (2GM= (2GMEE/R/REE))½½ = 1.12 x 10 = 1.12 x 1044 m/s m/s

vvescesc = 11.2 km/s ≈ 40,000 km/h= 11.2 km/s ≈ 40,000 km/h

Find the kinetic energy KK it must have at Earth’s surface in order to move infinitely far away from the Earth

K K = ½= ½ mvmvescesc22

K K = ½(5,000 kg)(= ½(5,000 kg)(1.12 x 101.12 x 1044 m/s m/s))22 K K = 3.14 x 10= 3.14 x 101111 J ≈ 2,300 gal of J ≈ 2,300 gal of GasolineGasoline

Examples to Read!!!Examples to Read!!! Example 13.7Example 13.7 (page 407)

Homework to be solved in Homework to be solved in Class!!!Class!!! Question: 16Question: 16 Problem: 27Problem: 27 Problem: 44Problem: 44

Material for the Material for the MidtermMidterm

What keeps a satellite in orbit?

Centripetal AccelerationCentripetal Acceleration

Due to: Due to: Gravitational Gravitational Force!Force!

FFgg = ma = mac c

FFgg = (mv = (mv22)/r = G(mM)/r = G(mMEE)/r)/r22

Its high speed or Its high speed or centripetal acceleration centripetal acceleration keeps it in orbit!!keeps it in orbit!!

Satellites & Satellites & “Weightlessness”“Weightlessness”

Satellites & “Weightlessness”, 2Satellites & “Weightlessness”, 2 If the satellite stops it will fall

directly back to Earth But at the very high speed it has,

it will quick fly out into space In absence of gravitational gravitational

force (Fforce (Fgg)), the satellite would move in a straight line!straight line!

Newton’s 1Newton’s 1stst Law Law Due to FFgg of the Earth the satellite

is pulling into orbit In fact, a satellite is fallingis falling

(accelerating toward Earth), but its high tangential speedtangential speed keeps it from hitting Earth

Objects in a satellite, including people, experience “EFFECTIVEEFFECTIVE weightlessnessweightlessness”. What causes this?

NOTE:NOTE: THIS DOESTHIS DOES NOTNOT MEAN MEAN THAT THE GRAVITATIONAL FORCE THAT THE GRAVITATIONAL FORCE ON THEM IS ZERO!ON THEM IS ZERO!

To understand this, first, look at a simpler problem:simpler problem: A person in an elevator

Satellites & “Weightlessness”, Satellites & “Weightlessness”, finalfinal

Person in elevator. Mass mm attached to scale. Elevator at rest (a = 0)(a = 0)

∑∑F = F = mma = 0a = 0 W W (weight) = upward force on

mass mm By Newton’s 3Newton’s 3rdrd Law:Law: WW is equal & oppositeequal & opposite to the

reading on the scale.

∑∑F = 0 = W F = 0 = W –– mg mg

Scale readingScale reading is W = mgW = mg

““Weightlessness” Case 1Weightlessness” Case 1

Elevator moves up up with acceleration aa

∑∑F = F = mmaa oror W W – – mmg = g = mmaa W W (weight) = upward force on mass

mm WW is equal & oppositeequal & opposite to the

reading on the scale. Scale readingScale reading (apparent weightapparent weight)

W = mg + maW = mg + ma > > mmgg For a = ½ga = ½g W = 3/2mgW = 3/2mg

Person is “heavier”“heavier” Person experiences: 1.5 g’s!1.5 g’s!

““Weightlessness” Case 2Weightlessness” Case 2

Elevator moves down down with acceleration aa

∑∑F = F = mmaa oror W W –– mmg = g = – – mmaa W W (weight) = upward force on mass mm

WW is equal & oppositeequal & opposite to the

reading on the scale. Scale readingScale reading (apparent weightapparent weight)

W = mg W = mg – – mama < < mmgg For a = ½ga = ½g W = 1/2mgW = 1/2mg

Person is “lighter”“lighter” Person experiences: 0.5 g’s!0.5 g’s!

““Weightlessness” Case 3Weightlessness” Case 3

½mg

Elevator cable breaks! Free falls downdown with acceleration a = ga = g ∑∑F = F = mmaa oror W W –– mmg = g = – – mmaa

W W (weight) = upward force on mass mm

WW is equal & oppositeequal & opposite to the reading on the scale.

Scale readingScale reading (apparent weightapparent weight) W = mg W = mg – – mmaa = = mg mg – – mgmg

W = 0!W = 0! Elevator & person are apparentlyapparently

“weightless”.“weightless”. Clearly, though: THE FORCE OF THE FORCE OF

GRAVITY STILL ACTS!GRAVITY STILL ACTS!

““Weightlessness” Case 4Weightlessness” Case 4

Apparent Apparent ““WeightlessnessWeightlessness”” experienced by people in a satellite orbit?

Similar to the elevator case 4case 4 With gravity, satellite (mass mm) free “falls” “falls”

continually continually due to

F = G(mMF = G(mMEE)/r)/r22 = mv = mv22/r/r Consider scale in satellite:∑∑F = ma = F = ma = – – mvmv22/r/r (towards Earth center!) (towards Earth center!) W W –– mg = mg = – – mvmv22/r/r W = mg W = mg –– mv mv22/r/r <<<< mmgg Objects in the satellite experience an

apparent weightlessnessapparent weightlessness because they, and the satellite, are acceleratingare accelerating as in free fallfree fall

““Weightlessness” in a Weightlessness” in a satellitesatellite

TV TV reporters are just plain WRONG WRONG when they say things like:

““The space shuttle has escaped the Earth’s The space shuttle has escaped the Earth’s gravity”gravity”

““The space shuttle has reached a point The space shuttle has reached a point outside the Earth’s gravitational pull”outside the Earth’s gravitational pull”

Why? Because the gravitational force F = F = G(mMG(mMEE)/r)/r22

exists (& is (& is NOTNOT zero) zero) even for HUGEHUGE distances rr away from Earth

F F 0 0 only foronly for r r

Reporters are WrongReporters are Wrong

Black HolesBlack Holes A black holeblack hole is the remains of a star that

has collapsed under its own gravitational gravitational forceforce

The escape speedescape speed for a black hole is very largevery large due to the concentration of a large mass into a sphere of very small radius If the escape speed exceeds the speed of

light, radiation cannot escape and it appears black

Black Holes, contBlack Holes, cont The critical radius at which

the escape speed equals c is called the Schwarzschild radiusSchwarzschild radius, RS

The imaginary surface of a sphere with this radius is called the event horizonevent horizon This is the limit of how close

you can approach the black hole and still escape

Black Holes and Accretion Black Holes and Accretion DisksDisks

Although light from a black hole cannot escape, light from events taking place near the black hole should be visible

If a binary star system has a black hole and a normal star, the material from the normal star can be pulled into the black hole

This material forms an accretion diskaccretion disk around the black hole

Friction among the particles in the disk transforms mechanical energy into internal energy

Black Holes and Accretion Black Holes and Accretion Disks, contDisks, cont

The orbital height of the material above the event horizon decreases and the temperature rises

The high-temperature material emits radiation, extending well into the x-ray region

These x-rays are characteristics of black holes

Black Holes at Centers of Black Holes at Centers of GalaxiesGalaxies

There is evidence that supermassive black holes exist at the centers of galaxies

Theory predicts jets of materials should be evident along the rotational axis of the black hole

An HST image of the galaxy M87. The jet of material in the right frame is thought to be evidence of a supermassive black hole at the galaxy’s center.