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Ch13 Newton’s Theory of Gravity

講者： 許永昌 老師

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ContentsA Little HistoryNewton’s Law of GravityLittle g and Big GGravitational Potential EnergySatellite Orbits and Energies

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A Little History ( 請預讀 P385~P386) Ancient Greeks:

The earth is at the center of the universe. 2AD, Ptolemy (P is silent) developed an elaborate mathematical model of the solar system.

Copernicus (1543): The sun is at the center of the universe.

Tycho and Kepler (1570~1600): Tycho developed ingenious mechanical sighting device to measure the positions of

stars and planets. Kepler use algebra, geometry and trigonometry to analyze the data.

Kepler’s Law. He suggested that the sun was a center of force that somehow caused the planetary motions.

Hooke: He suggested that the planets might be attracted to the sun with a strengh

proportional to the inverse square of the distance between the sun and the planet. Issac Newton (1642~1727)

Universal gravitation: The force of the sun on the planets was identical to the force of the earth on the apple.

Newton did not publish his results for a long 22 years, because he need to show that all of these forces exerted by a planet add up to give a result that is identical with treating this planet as a single particle. Development of Calculus.

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Newton’s Law of Gravity ( 請預讀P387~P391)

Quiz:

1 21 on 2 2 on 1 12 12 2 12

12

11 2 2

ˆ , where ,

Gravitational constant: 6.67 10 Nm / kg

GmmF F r r r r

r

G

12 ?r

211 on 2F

2 on 1F

21(1)

21(2)

We need the concept of field and Gauss’s Law shown in Ch28.

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Action (the motion of the moon)Purpose:

Realize the motion of the moon.Action:

Please calculate the ratio Fearth on moon/Fsun on moon.How about its motion?

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Little g and Big G ( 請預讀 P391~P393)

Gravity exerted by a planet: FG=mg.g is independent of mass m.

E.g. the acceleration of a free fall is gearth~9.80 m/s2, which is independent of the object’s mass.

g is the function of position. Universal gravitation: FG=GMm/r2

g=GM/r2. How to weight the earth?

Cavendish’s contribution.

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Gravitational Potential Energy ( 請預讀 P394~P395)

The potential contributed by the gravity pair:

Proof:

1 2 , Gmm

U rr

211 on 2F

2 on 1F

r

O

1 2 2 on 1 1 1 on 2 2

1 on 2 2 1

1 22

1 22

1 2

ˆ

r

U W W F dr F dr

F d r r

Gm mr dr

r

Gm mdr

rGm m

r

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Exercise ( 請預讀 P396~P397) Since U=-GMm/r for universal gravitation,

why can we write Ug=mgy in Chapter 10? Hint: RE>>y, (Flat-earth approximation)

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Satellite Orbits and Energies ( 請預讀 P398~P402)Kepler’s Law

1. Planets move in elliptical orbits, with the sun at one focus of the ellipse.

軌道形狀 要證明它需要大二的工數。

2. A line drawn between the sun and a planet sweeps out equal areas during equal intervals of time.

dA/dt=0 (Derived from L=constant.)

3. The square of a planet’s orbital period is proportional to the cube of the semi-major-axis length.

Exercise: 請利用圓形軌道驗算看看。

22 34

.T aGM

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ExerciseHow to determine the mass of the sun from

figure 13.18?It obeys log T = 1.500 log10 r – 9.264.

The height of the geosynchronous orbits.

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Orbital EnergeticsPotential Ug=-GMm/r.

Both for circular and ellipse motion.For a CIRCULAR MOTION

Centripetal acceleration: -v2/r=ar.FG is a central force: FG=GMm/r2=|mar|

(circular motion)=2K/r. K=GMm/(2r) for a circular motion.

E=K+U= for a circular motion.02 2

gUGMm

r

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Orbital Energetics (continue)The condition of a Circular Motion

滿足不代表一定是 circular motion ，不滿足就一定是橢圓軌道。Example:

1: 延切線推進，透過做功增加動能。12: 利用 r1v1=r2v2 與 E1=E2 ，就可

求得 v1 & v2 。 (r1, r2 給定 )

當然，實際上會複雜一點。 2: 增加動能，使之 E=-Ug/2.

2 2gUGMm

Er

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HomeworkStudent Workbook

13.4, 13.6, 13.8

Student Textbook13.2013.2913.54