Chapter 13 - United States Naval Academy...¥ Gravitation is a field force that always exists...
Transcript of Chapter 13 - United States Naval Academy...¥ Gravitation is a field force that always exists...
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Chapter 13
Universal Gravitation
Prof. Raymond Lee,revised 11-10-2010
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• Newton’s law of universal gravitation
• Every particle in universe attracts every otherparticle with force !: (1) product of their masses m (2) 1/(separation distance r)2: Fg = Gm1m2/r
2 (Eq. 13-1, p. 331)
• G is universal gravitational constant;G = 6.673 x 10-11 N•m2/kg2
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• Law of gravitation, 2
• Eq. 13-1 an example of an inversesquare law:
• magnitude of force varies as inversesquare of particles’ separation distance
• In vector form, this is:
(Eq. 13-3, p. 332)
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• Notation
• F12 = force exerted by particle 1 on 2
• – sign in Eq. 13-3 shows particle 2 isattracted toward particle 1
• F21 = force exerted by particle 2 on 1
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• More about forces
• F12 = -F21
• forces form a Newton’s 3rd law action-reaction pair
• Gravitation is a field force thatalways exists between 2 particles,regardless of intervening medium
• Force decreases rapidly as distanceincreases, a consequence of inversesquare law
(compare Fig. 13-2, p. 331)
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• G vs. g
• Always distinguish between G & g
• G = universal gravitational constantwhich, by definition, is same everywhere
• g = acceleration due to earth’s gravity
• g = 9.80 m/s2 at earth’s surface
• g varies with latitude & altitude
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• Moon’s centripetal acceleration
• Newton’s comparison of accelerations ofmoon (aM) & objects near earth’s surface:
• aM/g = rM-2/RE
-2 = (RE/rM)2 = 2.75 x 10-4
• So aM = 9.8 m/s2 * 2.75 x 10-4 ~ 0.00270 m/s2
• Remarkably, this aM differs from ac calculatedfrom moon’s orbital period T by < 1%
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• Moon’s centripetal acceleration, 2
• Compare this aM withcentripetal acceleration ofmoon as it orbits earth:
• ac = v 2/rM = (2"rM/T)2/rM =
4"2rM/T 2 = 0.00272 m/s2;
differs by < 1% from inverse-square law value for aM
(SJ 2004 Fig. 13.2, p. 392)
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• Measuring G
• G first measured by HenryCavendish in 1798
• This apparatus shows howattractive force between 2spheres makes rod rotate
• Mirror just amplifies this rotation
• Experiment was repeated forvarious masses
(SJ 2004 Fig. 13.4, p. 393)
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• Finding g from G
• Force acting on object of mass m infreefall near earth’s surface = mg
• Set this force = force of universalgravitation acting on object
• Thus mg = GMEm/RE2 becomes
g = GME/RE2 (Eq. 13-11, p. 334)
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• g as f (h) & f (#)
• If object is a distance h above earth’s surface,radius r = RE + h and sog(h) = GME/(RE+h)2 (SJ 2008 Eq. 13.6, p. 394)
• So g $ as h %, & object’s weight & 0 as r & !
• At h = 0, centripetal acceleration of earth’ssurface at latitude # reduces g(#) as:
g(#) ~ gpole– 'E2*R(*cos(#)
= gpole– 'E2*RE*cos2(#) {see Eq. 13-14, p. 336}
where R( = ( distance to earth’s axis at #
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• Variation of g with h
(SJ 2008 Table 13.1,
p. 365)
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• Newton’s assumption
• Newton treated earth as if its masswere all at its center
• Newton’s calculus shows that thisinitially troubling assumption followsnaturally from his law of universalgravitation
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• Gravitational force due tomass distribution m(r)
• Gravitational force Fg exerted by a finite-size, spherically symmetric m(r) on particleoutside distribution is same as ifdistribution’s entire mass were at its center
• For earth, Fg = GMEm/RE2 (Eq. 13-9, p. 334)
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• Kepler’s laws
• German astronomer Johannes Kepler wasassistant to Tycho Brahe, last of the “nakedeye” astronomers
• Kepler analyzed Brahe’s data & formulated 3descriptive laws of planetary motion
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• Kepler’s laws, 2 (pp. 343-344)
• Kepler’s 1st law• All planets move in elliptical orbits with sun at 1 focus
• Kepler’s 2nd law• Radius vector drawn from sun to any planet sweeps
out = areas in = time intervals
• Kepler’s 3rd law• Any planet’s (orbital period)2 ! (semimajor axis)3 of its
elliptical orbit
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• Notes about ellipses
• F1 & F2 are each a focus ofellipse; foci are at distancec from its center
• Longest distance throughcenter is major axis• a is semimajor axis
(compare Fig. 13-12, p. 343)
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• Notes about ellipses, 2
• Shortest distance throughcenter is minor axis• b is semiminor axis
• Ellipse’s eccentricity e ise = c/a• For a circle, e = 0
• Ellipses’ e range from 0 < e < 1
• N.B.: e = 1 for parabola,e > 1 for hyperbola compare Fig. 13-12, p. 343)
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• Notes about ellipses, planetary orbits
• Sun is at 1 focus & other focus is empty
• Aphelion is point farthest from sun• Aphelion distance = a + c (apogee for earth orbits)
• Perihelion is point nearest sun• Perihelion distance = a – c (perigee for earth orbits)
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• Kepler’s 1st law
• Circular orbit is special case of an elliptical orbit
• 1st law a direct result of inverse-square nature ofFg (Eq. 13-3, p. 332)
• Elliptical (& circular) orbits allowed for boundobjects, one that repeatedly orbits center
• Unbound objects pass by & don’t return {pathscan be parabolas (e = 1) or hyperbolas (e > 1)}
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• Orbit examples
• Pluto has highesteccentricity of any (former)planet: ePluto = 0.25
• Halley’s comet has an orbitwith high eccentricity:eHalley’s comet = 0.97
(compare Fig. 13-15, p. 345)
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• Kepler’s 2nd law
• A consequence ofconservation of L
• Fg || r yields no ), so L
is conserved
• L = r x p = MP r x v
= constant
(compare Fig. 13-13, p. 343)
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• Kepler’s 2nd law, 2
(compare Fig. 13-13, p. 343)• In time dt, radius vector r
sweeps out area dA, which= 1/2 area of parallelogram|r x dr|
• r ’s displacement is dr = vdt
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• Kepler’s 2nd law, 3
• Since vector L = constant, can show thatdA/dt = L/(2Mp) = constant (Eq. 13-32, p. 343)
• So radius vector from sun to any planetsweeps out = areas in = times
• Law applies to any central force, whetherinverse-square or not
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• Kepler’s 3rd law
• Predict from inverse-square law (p. 332)
• Start by assuming acircular orbit
• Fg & centripetal force
• Ks is a ! constant
(inverse-square) = centripetal
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• Kepler’s 3rd law, 2
• Now extend to elliptical orbits
• Replace r with a, the semimajor axis
• Ks is independent of planet’s mass, & sois valid for any planet
(Eq. 13-34, p. 344)
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• Kepler’s 3rd law, 3
• If 1 object orbits another, K ’s valuedepends on object being orbited
• E.g., for moon orbiting earth, replaceKsun with Kearth
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• Calculating the sun’s mass
• Using sun-earth distance & Tearth, use Kepler’s3rd law to find Msun
• Similarly, mass of any orbited object followsfrom data on object(s) orbiting it
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• Geosynchronous satellite(SJ 2008 Ex. 13.5, pp. 371-372)
• Geosynchronous (strictly,geostationary) satelliteremains above same pointon earth
• Fg & centripetal force
• From this, find h or v
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• Gravitational field
• Gravitational field exists everywhere inspace
• When particle of mass m is placed at pointwhere gravitational field = g, particleexperiences a force Fg = m g
• Field exerts a force on particle
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• Gravitational field, 2
• Define gravitational field g as:
• g = gravitational force experienced by testparticle at a point, normalized by particle’s mass
• Test particle isn’t necessary for field to exist;merely a diagnostic tool
(SJ 2008 Eq. 13.9, p. 372)
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• Gravitational field, 3
• g vectors point in directionthat particle wouldaccelerate if in field
• |g| = freefall a at givenlocation
(SJ 2008 Fig. 13.9, p. 373)
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• Gravitational field, 4
• g-field describes how object affects spacearound it in terms of force generated ifanother object were present:
(compare Eq. 13-3,p. 332)
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• Gravitational PE {* U(r)}
• Gravitational force is both conservative &a central force (i.e., directed along radialline toward CM)
• Magnitude depends only on r
• Represent a central force by F (r)r (compare
Fig. 13-9, p. 339)
^
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• Gravitational PE & work
• Particle moves A&B while actedon by central force F
• Now break path into myriadradial segments & arcs
• Because W done along arcs = 0,W done is independent of path &depends only on rf & ri
(compare Fig. 13-10, p. 340)
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• Gravitational PE & work, 2
• W done by F along any radial segment is
• W done by F ( displacement = 0
• So total W is:
& W is independent of path
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• Gravitational PE & work, 3
• As particle moves A&B, its
gravitational U(r) changes by:
• This is general form; nowconsider Fg specifically
(SJ 2008 Eq. 13.11, p. 373)
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• Gravitational PE for earth
• Set gravitational U(r) = 0 where Fg = 0,so that Ui = 0 at ri = !:
U(r) = –GMEm/r (Eq. 13-21, p. 339)
• Valid only for r + RE & not valid for r < RE
• U(r) < 0 because of our Ui choice, or:Uf – Ui = –GMEm(1/rf – 1/ri)(SJ 2008, Eq. 13.12, p. 373)
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• Gravitational PE for earth, 2
• Plot gravitational U(r)for object above earth’ssurface
• Its U(r) & 0 as r & ,(SJ 2008Fig. 13.11,p. 374)
(note U(r) is undefined for r < RE)
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• General gravitational PE
• For any 2 particles m1& m2, gravitational U(r)is: U(r) = –Gm1m2/r (Eq. 13-21, p. 339)
• U(r) between any 2 particles ! 1/r, although|Fg| ! 1/r 2
• U(r < ,) < 0 since Fg is attractive & we choseU(r = ,) = 0 (p. 339)
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• General gravitational PE, 2
• External agent must do positive W asr % between 2 objects
• W done by external agent & increase in
gravitational U(r) as particles are separated(i.e., U(r) becomes less negative as r %)
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• Binding energy
• Think of |U(r)| as a binding energy
• If external agent applies F > |U(r)|,then excess energy appears as KE ofparticles when their separation = ,
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• Systems with ! 3 particles
• Total gravitational PE ofsystem = sum over allparticle pairs
• Gravitational PE obeys thissuperposition principle
(compare Fig. 13-8,p. 339)
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• Systems with ! 3 particles, 2
• Each particle pair contributes a term to Utotal
• For 3 particles:
• Again, |Utotal| = W needed to separateparticles by , distance
(Eq. 13-22,p. 339)
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• Energy & satellite motion
• Object of mass m travels at v(m) near
massive object of mass M, where M » m
• Assume v(M) = 0 in inertial reference frame
• Total system energy = its KE + PE
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• Energy & satellite motion, 2
• Total energy E = K +U
• In bound system such as an object inquasi-circular orbit, E < 0 because wechose U(r = ,) = 0 (p. 339)
(see p. 341)
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• Energy in circular orbits
• Object of mass m moves in circularorbit about M
• Because the centripetal &gravitational forces are equal,KE = (1/2)mv
2 = GMm/(2r)(Eq. 13-38, p. 345)
• Now ME is E = GMm(1/(2r) – 1/r) = –GMm/(2r)(Eq. 13-40, p. 345; note minus sign)
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• Energy in circular orbits, 2
• Thus total ME < 0 for circular-orbit case
• KE > 0 & KE = |PE|/2
• |ME| = system binding energy, energyrequired to separate objects to ,
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• Energy in elliptical orbits
• For elliptical orbits, replace r withsemimajor axis a: E = –GMm/(2a)(Eq. 13-42, p. 345)
• Thus total ME < 0 for any bound,2-object system
• System ME & angular momentum L are
both constant for isolated systems
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• Summary of 2-particle bound system
• Total ME is (see p. 342 Sample Problem):
• Total ME & L for gravitationally bound,
2-object system are both motion-dependent constants
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• Earth escape speed
(SJ 2008 Fig. 13.14, p. 377)
• Project object of mass mupward from earth’s surfacewith initial speed vi
• Use energy to get minimum vi
needed for object to move to ,
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• Earth escape speed, 2
• This minimum speed is escape speed
• Note vesc is independent of object’s mass
• This result also is: (1) independent of velocity direction, (2) ignores air resistance (doesn’t changevesc , just makes it more difficult to achieve)
(Eq. 13-28, p. 341)
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• General escape speed
• Extend earth’s resultto any planet
• Table 13-2 (p. 341)gives vesc from variousobjects
(Eq. 13-28,p. 341)
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• Escape speed implications
• Complete escape from object isn’t reallypossible since g-field is infinite & thus|Fg| > 0 at all r < ,
• Explains why some planets haveatmospheres & others don’t• Lighter molecules have higher average v &
so are likelier to reach vesc