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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 (Fv), 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

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

- 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 = mgy

- 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 = mgy

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

- Examples to Read!!!
- 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.
- Scale reading (apparent weight)

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.
- Scale reading (apparent weight)

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.
- Scale reading (apparent weight)

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.

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