13.6 Gravitational Potential Energy

Gravitational Potential Energy U = mgy Valid only In Chapter 8: U = mgy (Particle-Earth). Valid only when particle is near the Earth’s surface. Inverse Square Law(F g  1/r 2 ) U Now with the introduction of the Inverse Square Law (F g  1/r 2 ) we must look for a general potential energy U function without restrictions.

Gravitational Potential Energy, cont fact We will accept as a fact that: gravitational force is conservative The gravitational force is conservative (Sec 8.3) gravitational force is a central force The gravitational force is a central force It is directed along a radial line toward the center r Its magnitude depends only on r A central force can be represented by

Grav. Potential Energy–Work AB 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 (F  v) r f r i Because the work done along the arcs is zero (F  v), the work done is independent of the path and depends only on r f and r i

Grav. Potential Energy–Work, 2 F The work done by F along any radial segment is 0 The work done by a force that is perpendicular to the displacement is 0 Total work Total work is Therefore, the work is independent of the path

Grav. Potential Energy–Work, final A B As a particle moves from A to B, its gravitational potential energy changes by (Recall Eqn 8.15)(13.11) general form This is the general form, we need to look at gravitational force specifically

Grav. Potential Energy for the Earth F(r) = - G(M E m)/r 2 Since: F(r) = - G(M E m)/r 2  Eqn. (13.11) becomes (13.12) (13.12) U i = 0r i → ∞ Taking U i = 0 at r i → ∞ we obtain the important result:(13.13)

Gravitational Potential Energy for the Earth, final Eqn r ≥ R E Eqn applies to Earth-Particle system, with r ≥ R E (above the Earth’s surface). Eqn r < R E Eqn is not valid for particle inside the Earth r < R E U U i = 0 r i → ∞ U is always negative (the force is attractive and we took U i = 0 at r i → ∞ An external agentyou lifting a book+ Work height U U r 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 two particlesgravitational potential energy For any two particles, the gravitational potential energy function becomes(13.14) 1/r The gravitational potential energy between any two particles varies as 1/r force1/r 2 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 total gravitational potential energy the sum The total gravitational potential energy of the system is the sum over all pairs of particles superposition principle Gravitational potential energy obeys the superposition principle U Each pair of particles contributes a term of U

Systems with Three or More Particles, final three particles Assuming three particles:(13.15) The absolute value of U total work neededseparate the particles The absolute value of U total represents the work needed to separate the particles by an infinite distance

Binding Energy binding energy The absolute value of the potential energy can be thought of as the binding energy larger excess kinetic 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 13.8 The Change in Potential Energy (Example 13.6 Text Book) m  y 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 13.8 The Change in Potential Energy, cont r i r f Since r i and r f are to close to the Earth’s surface, then: r f – r i ≈  y r f r i ≈ R E 2 r f – r i ≈  y and r f r i ≈ R E 2 r Remember r is measured from center of the Earth  U = mg  y

m v MMm An object of mass m is moving with speed v near a massive object of mass M (M >> m). Total Mechanical Energy E r The Total Mechanical Energy E, when the objects are separated a distance r, will be: E = K +U= ½m v 2 –GMmr (13.16) E = K +U = ½ m v 2 – GMm/r (13.16) v Depending on the value of v, E+, -, 0 E could be +, -, 0 M m v r 13.7 Energy Considerations in Planetary and Satellite Motion

m Newton’s 2 nd Law for mass m: GMmr 2 = m a R = mv 2 /r GMm/r 2 = m a R = mv 2 /r r/2 Multiplying by r/2 both sides: GMm2r = mv 2 /2 GMm/2r = mv 2 /2  ½mv 2 = ½(GMmr) ½mv 2 = ½(GMm/r)  K = ½(GMm/r) (13.17) K = ½(GMm/r) (13.17) Kinetic Energy (K) is positive and equal to half the absolute value of the potential Energy (U)!!!! Kinetic Energy (K) is positive and equal to half the absolute value of the potential Energy (U)!!!! M m v r Energy and Satellite Motion, cont

Energy in a Circular Orbit Substituting (13.17) into (13.16): E = ½ (GMmr) –GMmr E = ½ (GMm/r) – GMm/r (13.18) total mechanical energy negative The total mechanical energy is negative in the case of a circular orbit kinetic energypositive The kinetic energy is positive and is equal to half the absolute value of the potential energy absolute value of E binding energy The absolute value of E is equal to the binding energy of the system

Energy in an Elliptical Orbit a For an elliptical orbit, the radius is replaced by the semimajor axis a(13.19) total mechanical energynegative The total mechanical energy is negative total energyconstantsystem is isolated The total energy is constant if the system is isolated

Summary of Two Particle Bound System Conservation of Total Energy Conservation of Total Energy will be written as: (13.20) (13.20) total energy total angular momentum gravitationally bound, are constants of the motion Both the total energy and the total angular momentum of a gravitationally bound, two-object system are constants of the motion

m v i An object of mass m is projected vertically upward from Earth’s surface, with initial speed v i. v i Let’s find the minimum value of v i needed by the object to move infinitely far away from Earth. The energy of the system at any point is: E = ½mv 2 –GM E mr E = ½ mv 2 – GM E m/r Escape Speed from Earth

At Earth’s surface: v v i r r i R E v = v i & r = r i = R E At Maximum height: vv f = 0 rr f r max v = v f = 0 & r = r f = r max Using conservation of the Energy (Eqn.13.20): ½ mv i 2 – GM E m/r i = ½ mv f 2 – GM E m/r f  ½ mv i 2 – GM E m/R E = 0 – GM E m/r max  Solving for v i 2 : v i 2 = 2GM E (1/R E – 1/r max ) (13.21) v i h: r max R E Knowing v i we can calculate the maximum altitude h: h = r max – R E Escape Speed from Earth, 2

Escape Speed from Earth, 3 Escape Speed (v esc ) Let’s calculate Escape Speed (v esc ) Minimum speed Earth’s surface infinite Minimum speed for the object must have at the Earth’s surface in order to approach an infinite separation distance from Earth. minimum speed Traveling at this minimum speed the object continues to move farther away, such that we take: r max → ∞ or 1/r max → 0 r max → ∞ or 1/r max → 0  Equation (13.21) becomes: v i 2 = v esc 2 = 2GM E (1/R E – 0) v i 2 = v esc 2 = 2GM E (1/R E – 0)  (13.22) (13.22)

Escape Speed From Earth, final v esc independentmass of the object (Question 8 Homework) v esc is independent of the mass of the object (Question 8 Homework) v esc v esc 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, General M 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 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 This explains why some planets have atmospheres and others do not Lighter moleculeshigher average speeds 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) v esc Calculate v esc from the Earth for a 5,000 kg spacecraft. v esc = (2GM E /R E ) ½ = 1.12 x 10 4 m/s v esc = (2GM E /R E ) ½ = 1.12 x 10 4 m/s v esc = 11.2 km/s ≈ 40,000 km/h v esc = 11.2 km/s ≈ 40,000 km/h K Find the kinetic energy K it must have at Earth’s surface in order to move infinitely far away from the Earth K = ½mv esc 2 K = ½ mv esc 2 K = ½(5,000 kg)(1.12 x 10 4 m/s) 2 K = ½(5,000 kg)(1.12 x 10 4 m/s) 2  K = 3.14 x J ≈ 2,300 gal of Gasoline

Examples to Read!!! Examples to Read!!! Example 13.7 Example 13.7 (page 407) Homework to be solved in Class!!! Homework to be solved in Class!!! Question: 16 Question: 16 Problem: 27 Problem: 27 Problem: 44 Problem: 44 Material for the Midterm

What keeps a satellite in orbit? Centripetal Acceleration Centripetal Acceleration Due to: Gravitational Force! F g = ma c F g = ma c  F g = (mv 2 )/r = G(mM E )/r 2 F g = (mv 2 )/r = G(mM E )/r 2 Its high speed or centripetal acceleration keeps it in orbit!! Its high speed or centripetal acceleration keeps it in orbit!! Satellites & “Weightlessness”

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 gravitational force (F g ) straight line! In absence of gravitational force (F g ), the satellite would move in a straight line! Newton’s 1 st Law Newton’s 1 st Law F g Due to F g of the Earth the satellite is pulling into orbit is falling tangential speed In fact, a satellite is falling (accelerating toward Earth), but its high tangential speed keeps it from hitting Earth

EFFECTIVE weightlessness Objects in a satellite, including people, experience “EFFECTIVE weightlessness”. What causes this? NOTE: THIS DOES NOT MEAN THAT THE GRAVITATIONAL FORCE ON THEM IS ZERO! NOTE: THIS DOES NOT MEAN THAT THE GRAVITATIONAL FORCE ON THEM IS ZERO! simpler problem: To understand this, first, look at a simpler problem: A person in an elevator Satellites & “Weightlessness”, final

m (a = 0) Person in elevator. Mass m attached to scale. Elevator at rest (a = 0)  ∑F = ma = 0 W m W (weight) = upward force on mass m Newton’s 3 rd Law: By Newton’s 3 rd Law: W equal & opposite W is equal & opposite to the reading on the scale. ∑F = 0 = W – mg ∑F = 0 = W – mg  Scale readingW = mg Scale reading isW = mg “Weightlessness” Case 1

up a Elevator moves up with acceleration a ∑F = ma or W – mg = ma W m W (weight) = upward force on mass m W equal & opposite W is equal & opposite to the reading on the scale. Scale readingapparent weight Scale reading (apparent weight) W = mg + ma > mg a = ½gW = 3/2mg For a = ½g W = 3/2mg “heavier” Person is “heavier” 1.5 g’s! Person experiences: 1.5 g’s! “Weightlessness” Case 2

down a Elevator moves down with acceleration a ∑F = ma or W – mg = – ma W m W (weight) = upward force on mass m W equal & opposite W is equal & opposite to the reading on the scale. Scale readingapparent weight Scale reading (apparent weight) W = mg – ma< mg W = mg – ma < mg a = ½gW = 1/2mg For a = ½g W = 1/2mg “lighter” Person is “lighter” 0.5 g’s! Person experiences: 0.5 g’s! “Weightlessness” Case 3    ½mg

down a = g Elevator cable breaks! Free falls down with acceleration a = g ∑F = ma or W – mg = – ma W m W (weight) = upward force on mass m W equal & opposite W is equal & opposite to the reading on the scale. Scale readingapparent weight Scale reading (apparent weight) W = mg – ma= mg – mg W = mg – ma = mg – mg W = 0! apparently “weightless”. Elevator & person are apparently “weightless”. THE FORCE OF GRAVITY STILL ACTS! Clearly, though: THE FORCE OF GRAVITY STILL ACTS! “Weightlessness” Case 4

Apparent “Weightlessness” Apparent “Weightlessness” experienced by people in a satellite orbit? case 4 Similar to the elevator case 4 m“falls” continually With gravity, satellite (mass m) free “falls” continually due to F = G(mM E )/r 2 = mv 2 /r Consider scale in satellite: ∑F = ma = – mv 2 /r (towards Earth center!) W – mg = – mv 2 /r W – mg = – mv 2 /r  W = mg – mv 2 /r << mg W = mg – mv 2 /r << mg apparent weightlessness are accelerating free fall Objects in the satellite experience an apparent weightlessness because they, and the satellite, are accelerating as in free fall “Weightlessness” in a satellite

TV WRONG TV reporters are just plain WRONG 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” F = G(mM E )/r 2 Why? Because the gravitational force F = G(mM E )/r 2 (& is NOT zero)HUGEr exists (& is NOT zero) even for HUGE distances r away from Earth F  0only forr   F  0 only for r   Reporters are Wrong

Black Holes black hole gravitational force A black hole is the remains of a star that has collapsed under its own gravitational force escape speedvery large 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 Schwarzschild radius The critical radius at which the escape speed equals c is called the Schwarzschild radius, R S event horizon 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 accretion disk 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.