13.6 Gravitational Potential Energy

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# 13.6 Gravitational Potential Energy - PowerPoint PPT Presentation

13.6 Gravitational Potential Energy. Gravitational Potential Energy. In Chapter 8: U = mgy (Particle-Earth). Valid only when particle is near the Earth’s surface.

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Presentation Transcript
Gravitational Potential Energy
• In Chapter 8: U = mgy (Particle-Earth). Valid only when particle is near the Earth’s surface.
• Now with the introduction of the Inverse Square Law(Fg 1/r2)we must look for a general potential energy U function without restrictions.
Gravitational Potential Energy, cont
• We will accept as a fact that:
• The gravitational force is conservative (Sec 8.3)
• The gravitational force is a central force
• It is directed along a radial line toward the center
• Its magnitude depends only on r
• A central force can be represented by
Grav. Potential Energy–Work
• A particle moves from A to B 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v), the work done is independent of the path and depends only on rfand ri
Grav. Potential Energy–Work, 2
• The work done by F along any radial segment is
• The work done by a force that is perpendicular to the displacement is 0
• Total work is
• Therefore, the work is independent of the path
Grav. Potential Energy–Work, final
• As a particle moves from A to B, its gravitational potential energy changes by (Recall Eqn 8.15)

(13.11)

• This is the general form, we need to look at gravitational force specifically
Grav. Potential Energy for the Earth
• Since: F(r)= - G(MEm)/r2 Eqn. (13.11) becomes

(13.12)

• Taking Ui= 0 at ri → ∞ we obtain the important result:

(13.13)

Gravitational Potential Energy for the Earth, final
• Eqn. 13.13 applies to Earth-Particle system, with r ≥ RE

(above the Earth’s surface).

• Eqn. 13.13 is not valid for particle inside the Earth r < RE
• U is always negative (the force is attractive and we took Ui= 0 at ri → ∞
• An external agent (you lifting a book) must do + Work to increase the height (separation between the Earth’s surface and the book). This work increases U U becomes less negative as r increases.
Gravitational Potential Energy, General
• For any two particles, the gravitational potential energy function becomes

(13.14)

• The gravitational potential energy between any two particles varies as 1/r
• Remember the force varies as 1/r 2
• The potential energy is negative because the force is attractive and we chose the potential energy to be zero at infinite separation
Systems with Three or More Particles
• The total gravitational potential energy of the system is the sum over all pairs of particles
• Gravitational potential energy obeys the superposition principle
• Each pair of particles contributes a term of U
Systems with Three or More Particles, final
• Assuming three particles:

(13.15)

• The absolute value of Utotal represents the work needed to separate the particles by an infinite distance
Binding Energy
• The absolute value of the potential energy can be thought of as the binding energy
• If an external agent applies a force larger than the binding energy, the excess energy will be in the form of kinetic energy of the particles when they are at infinite separation
• A particle of mass m is displaced through a small vertical displacement y 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:

U = mgy

• From Equation 13.12
Example 13.8The Change in Potential Energy, cont
• Since riand rfare to close to the Earth’s surface, then:
• rf– ri≈y andrfri≈ RE2
• Remember r is measured from center of the Earth

U = mgy

13.7 Energy Considerations in Planetary and Satellite Motion
• An object of mass m is moving with speed v near a massive object of mass M (M >> m).
• The Total Mechanical Energy E, when the objects are separated a distance r, will be:

E = K +U= ½m v2–GMm/r(13.16)

• Depending on the value of v,

E could be +, - , 0

m

v

r

M

Energy and Satellite Motion, cont
• Newton’s 2nd Law for mass m:

GMm/r2= m aR=mv2/r

• Multiplying by r/2 both sides:

GMm/2r = mv2/2 

½mv2 = ½(GMm/r) 

K = ½(GMm/r) (13.17)

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

m

v

r

M

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

E = ½(GMm/r)–GMm/r 

(13.18)

• The total mechanical energy is negative in the case of a circular orbit
• The kinetic energy is positive and is equal to half the absolute value of the potential energy
• The absolute value of E is equal to the binding energy of the system
Energy in an Elliptical Orbit
• For an elliptical orbit, the radius is replaced by the semimajor axis a

(13.19)

• The total mechanical energy is negative
• The total energy is constant if the system is isolated
Summary of Two Particle Bound System
• Conservation of Total Energywill be written as:

(13.20)

• Both thetotal energyand the total angular momentum of agravitationally bound,two-object systemare constants of the motion
Escape Speed from Earth
• An object of mass m is projected vertically upward from Earth’s surface, with initial speed vi.
• Let’s find the minimum value of vineeded by the object to move infinitely far away from Earth.
• The energy of the system at

any point is:

E = ½mv2–GMEm/r

Escape Speed from Earth, 2
• At Earth’s surface:

v = vi & r = ri= RE

• At Maximum height:

v = vf= 0 & r = rf= rmax

• Using conservation of the Energy (Eqn.13.20):

½ mvi2 – GMEm/ri = ½ mvf2 – GMEm/rf 

½ mvi2 – GMEm/RE= 0 – GMEm/rmax

• Solving for vi2:

vi2 = 2GME(1/RE – 1/rmax)(13.21)

• Knowing viwe can calculate the maximum altitude h:h = rmax – RE
Escape Speed from Earth, 3
• Let’s calculate Escape Speed(vesc)

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

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

rmax → ∞ or 1/rmax → 0 

• Equation (13.21) becomes:

vi2 = vesc2 = 2GME(1/RE – 0)

(13.22)

Escape Speed From Earth, final
• vescis independent of the mass of the object (Question 8 Homework)
• vescis the same for a spacecraft or a molecule.
• The result is independent of the direction of the velocity and ignores air resistance
Escape Speed, General
• The Earth’s result can be extended to any planet of mass M

(13.22a)

• The table at right gives some escape speeds from various objects
Escape Speed, Implications
• Complete escape from an object is not 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 have atmospheres and others do not
• Lighter molecules have higher average speeds and are more likely to reach escape speeds
Example 13.9 Escape Speed of a Rocket(Example 13.8 Text Book)
• Calculate vesc from the Earth for a 5,000 kg spacecraft.

vesc= (2GME/RE)½ = 1.12 x 104 m/s

vesc= 11.2 km/s ≈ 40,000 km/h

• Find the kinetic energy K it must have at Earth’s surface in order to move infinitely far away from the Earth
• K = ½mvesc2

K = ½(5,000 kg)(1.12 x 104 m/s)2

K = 3.14 x 1011 J ≈ 2,300 gal of Gasoline

Material for the Midterm

• Example 13.7(page 407)
• Homework to be solved in Class!!!
• Question: 16
• Problem: 27
• Problem: 44
Satellites & “Weightlessness”
• What keeps a satellite in orbit?
• Centripetal Acceleration

Due to: Gravitational Force!

Fg = mac 

Fg = (mv2)/r = G(mME)/r2

• Its high speed or centripetal acceleration keeps it in orbit!!
Satellites & “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 force (Fg), the satellite would move in a straight line!
• Newton’s 1st Law
• Due to Fg of the Earth the satellite is pulling into orbit
• In fact, a satellite is falling (accelerating toward Earth), but its high tangential speed keeps it from hitting Earth
Satellites & “Weightlessness”, final
• Objects in a satellite, including people, experience “EFFECTIVEweightlessness”. What causes this?
• NOTE:THIS DOESNOTMEAN THAT THE GRAVITATIONAL FORCE ON THEM IS ZERO!
• To understand this, first, look at a simpler problem: A person in an elevator
“Weightlessness” Case 1
• Person in elevator. Mass mattached to scale. Elevator at rest (a = 0) 

∑F = ma = 0

• W (weight) = upward force on mass m
• By Newton’s 3rd Law:
• Wis equal & opposite to the reading on the scale.

∑F = 0 = W – mg 

Scale reading is W = mg

“Weightlessness” Case 2
• Elevator moves up with acceleration a

∑F = maorW – mg = ma

• W (weight) = upward force on mass m
• Wis equal & opposite to the reading on the scale.

W = mg + ma > mg

• For a = ½gW = 3/2mg

Person is “heavier”

Person experiences: 1.5 g’s!

“Weightlessness” Case 3
• Elevator moves down with acceleration a

∑F = maorW –mg = – ma

• W (weight) = upward force on mass m
• Wis equal & opposite to the reading on the scale.

W = mg – ma< mg

• For a = ½gW = 1/2mg

Person is “lighter”

Person experiences: 0.5 g’s!

½mg

“Weightlessness” Case 4
• Elevator cable breaks! Free falls downwith acceleration a = g

∑F = maorW –mg = – ma

• W (weight) = upward force on mass m
• Wis equal & opposite to the reading on the scale.

W = mg – ma= mg – mg

W = 0!

• Elevator & person are apparently“weightless”.
• Clearly, though: THE FORCE OF GRAVITY STILL ACTS!
“Weightlessness” in a satellite
• Apparent “Weightlessness” experienced by people in a satellite orbit?
• Similar to the elevator case 4
• With gravity, satellite (mass m) free “falls” continually due to

F = G(mME)/r2 = mv2/r

• Consider scale in satellite:

∑F = ma = – mv2/r (towards Earth center!)

• W – mg = – mv2/r

W = mg – mv2/r<<mg

• Objects in the satellite experience anapparent weightlessnessbecause they, and the satellite,are acceleratingas infree fall
Reporters are Wrong
• TV reporters are just plainWRONG when they say things like:

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

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

• Why? Because the gravitational force F = G(mME)/r2

exists (& is NOT zero) even for HUGEdistances raway from Earth

F  0only forr  

Black Holes
• A black hole is the remains of a star that has collapsed under its own gravitational force
• The escape speed for a black hole is very 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, cont
• The critical radius at which the escape speed equals c is called the Schwarzschild radius, RS
• The imaginary surface of a sphere with this radius is called the event horizon
• This is the limit of how close you can approach the black hole and still escape
Black Holes and Accretion Disks
• 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 disk around the black hole
• Friction among the particles in the disk transforms mechanical energy into internal energy
Black Holes and Accretion Disks, 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 Galaxies
• 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.