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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