Chapter 14 Chapter 14 GravitationGravitation

1.1. Origin of the Law of Origin of the Law of GravitationGravitation

2.2. Newton’s Law of Newton’s Law of Universal GravitationUniversal Gravitation

3.3. The Gravitational The Gravitational Constant G Constant G

4.4. Gravitation Near the Gravitation Near the Earth’s surfaceEarth’s surface

5.5. The Two Shell The Two Shell TheoremsTheorems

6.6. Gravitational Gravitational Potential EnergyPotential Energy

14-1 Origin of the law of 14-1 Origin of the law of gravitationgravitation

1. In 16th century Copernicus ( 1473~1543 ) proposed a heliocentric ( sun-centered ) scheme, in which the Earth and other planets move about sun.2. Kepler ( 1571~1630 ) proposed three law ( which we discuss in Section 14-7) that describe Planet’s motions. However, Kepler’s Laws were only empirical without any basis in terms of forces.

14-2 Newton’s law of 14-2 Newton’s law of universal gravitationuniversal gravitation

1. Guided by Kepler’s laws, Newton proposed a force law for gravitation:

Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The direction of the force is along the line joining the particles.

Mathematically, the law of gravitation has the following form (14-1)Here G, called the gravitational constant.

We can represent Eq(14-1) in vector’s form.

where and are unit vectors.

221

r

mmGF

2211 )/(1067.6 kgmNG

21221

2121 r

r

mmGF

12212

2112 r

r

mmGF and

21r

12r

Eq(14-3)

The negative sign in Eq(14-3) shows that points in a direction opposite to , which indicates that the gravitational force is attractive.

211212

1212 , rrr

r

rr

1m1m

1m

2m2m2m

12F

12F

12F

21F

21F

12r

12r

21r

122121

2121 , rrr

r

rr

Fig(14-2)

Sample Problem 14-2A properly suited astronaut (ma=105Kg) is drifting through the asteroid belt( 小行星带 ) on a mining expedition （矿业探险） . At a particular instant he is located near two asteroids of masses m1=346Kg (r1=215m) and m2=184Kg (r2=142m) two asteroids form an angle of 120 degree. At that instant, what is the magnitude and direction of the gravitational force on the astronaut due to these two asteroids?

14-3 The gravitational constant G

1.The first laboratory determination of G was done by Cavendish in 1798.

2. It is difficult to improve substantially on the precision of the measured value of G because of its small magnitude.

3. This difficulty of measuring G is unfortunate, because gravitation has such an essential role in theories of the origin and structure of the universe.

This experiment was selected as one of the top 10 beautiful experiments in Phys..

English chemist and physicist who was shy and absent-minded. He was terrified of women, and communicated with his female servants by notes. He performed numerous scientific investigations, but published only twenty articles and no books.

Cavendish, Henry (1731-1810)

The Cavendish Laboratory was founded in 1871, along with the appointment of James Clerk Maxwell as the first Cavendish Professor. It has a distinguished intellectual history, with 29 Nobel prizewinners who worked for considerable periods within its facilities, and is associated with many notable discoveries, including the electron and the structure of DNA.

Cavendish Laboratoryat Cambridge Univ.

(1871~ present)

14-4 Gravitation near the 14-4 Gravitation near the Earth’s surfaceEarth’s surface

1. If we combine the law of gravitation and Newton’s second law, we can obtain the acceleration of free-fall body near Earth’s surface. The distance of the body from the Earth’s center is r.

2r

mMGF E

0mgF ｝ 20 r

MGg E

EM is the mass of the Earth;is the free-fall acceleration due only to the gravitational pull of the Earth

0g

(14-5)

Look at the spacecraft: The acceleration from the gravitational force of Earth at that place is not zero!!!

高度 /Km08.836.640035700

As we will prove in Sec14-5, for spherical mass distributions we can regard the object as a point mass concentrated at its center. It is an exact relationship.

Any assumption in above deduction?

The Earth is regarded as a mass point at its center.

So Eq. (14-5) requires the Earth is spherical and that its density depends only on the radial distance from its center

2. The real Earth differs from our model in three way

(a) The Earth is not uniform ；(b) The Earth is approximately an ellipsoid( 椭球 ) ； flattened at the poles and bulging at the equator

(c) The earth is rotating.

3. How does spin of the Earth affect the measure value g of gravitational acceleration g0?

Fig 14-7 shows the rotating Earth from an inertial

frame positioned in space above the north pole. A

crate of mass m rests on a platform scale( 台秤 ) at the equator.

Because of the Earth’s spin,

the crate is in uniform

circular motion with radius

and period of rotation T (=24

hours ).

From Fig14-7

(14-16)EEc RmRmvmaNmg 22

0 /

)( 20 ERgmN

(a) (b)

N

0mg

North pole

crate

scale

R

ER

Fig 14-7

where N is the normal force on the crate due to the platform scale ( equal to the scale reading mg, and

g is the measure value of gravitational acceleration g0),

The g is smaller than by only , or .

220 /034.0 smRgg E

0g 2/034.0 sm

0mgmg

ERgm

Ng 2

0 If N=mg,

14-5 The two shell theorems14-5 The two shell theorems1.Shell theorem

A uniformly dense spherical shell attracts an external particle as if the mass of the shell were concentrated at its center.The shell theorem indicates : A spherically symmetric body attracts particle outside as if its mass were concentrated at its center.

2.Shell theorem

A uniformly dense spherical shell exerts no gravitational force on a particle located anywhere inside it.

1#

1#

2#

The important of shell theorem can be appreciated by imagining a tunnel drilled along a diameter of the Earth. As we descend into the tunnel, the portions of the Earth outside our radius exert no gravitational force on us. We feel only the effect of the portion of the Earth’s mass inside a sphere whose radius is our distance from the center of the Earth.

The shell theorem are true only for the inverse-square force.

2#

3. Proof of the shell theorems

Fig 14-8 shows a thin shell and the ring we will consider. The shell has total mass M, thickness t, and uniform density ( mass per unit volume ). A point mass m is located at point P, a distance r from the center of the shell ( point o ). Our goal is to find the force exerted on m first by the ring and then by the entire shell.

o

d A

Q

cosR

r

x

mP

Fig 14-8t

R

The ring gives the total force dF exerted on m:

(14-7)

where the factor gives the axial component of the force.

(14-8)

cos2

dMx

GmdF

cos

dRtdM sin2 2

From the three variables ( , and ), we choose

to eliminate and , leave as the single

variable.

so

(14-9)

Using the law of cosines on triangle Aop we obtain

or (14-10)

xx

coscos RrxPQ

x

Rr coscos

cos2222 rRRrx

r

xRrR

2cos

222

Differentiate Eq(14-10) to find (14-11)dx

rR

xd sin

Put Eq(14-10) into Eq(14-9) and then substitute

the result for into Eq(14-7).

cos

dxx

Rr

r

mRGtdF )1(

2

22

2

,)4(

)1(

2

2

22

2

2

r

MmGR

r

mRGt

dxx

Rr

r

mRGtdFF

Rr

Rr

tRM 24

cos2

dMx

GmdF

dRtdM sin2 2

At this point we notice that this equation is beautifully

simplified if we assume that space-time has 92

dimensions.

The proof of the second shell theorem is exactly same up to the final step, but the lower limit of the integral is now R-r rather than r-R. This small change causes the value of the integral to be zero.

o

d A

r

xm

t

R

★

14-6 Gravitational potential 14-6 Gravitational potential energyenergy

In chapter 12, we obtained the potential energy change due to gravity for a body :

(Eq(12-9))

where we regard the gravitational force as approximately constant.

Our goal now is to find a general expression of potential energy for universal gravitation.

mgyU

1. The potential energy difference can be found

from Eq(12-4).

However this equation applies only if the force is

conservative.

abab WUUU

Is the gravitational force conservative?

Fig14-11 shows a particle of

mass m moving in a region

where a gravitational force

is exerted on it by a particle

M. Particle m moves from a

to b along several different

paths: path 1 (a A b ),

path 2 (a B b) and

path 3 ( a c D E F b).

mPath 1

Path 3

Path 2B

CD

EF

A

a

b

M

Fig 14-11

WAb is zero, because is perpendicular to . So

F

sd

sdFsdFWWWWb

a

A

a

r

r

r

raAAbaA1

bA rr

sdFsdFWWWWb

a

b

B

r

r

r

rBbBbaB2

For Path2:

aB rr

For Path1:

sdF

sdFsdFsdFsdF

WWWWWWWW

b

a

b

D

D

a

F

E

D

C

r

r

r

r

r

r

r

r

r

r

EFCDFbEFDEcDac3

For Path3:

It is clear from this calculation that , the work is independent of the path, and the

gravitational force is conservative.

321 WWW

mPath 1

Path 3

Path 2B

CD

EF

A

a

b

M

Fig 14-11

2. Calculating the potential energy.

In Fig 14-12. M is at rest. As m moves from a to b,

the work done on m by the gravitational force is

)11

(2

ab

r

r

b

a

b

aab

rrGMm

r

drGMm

drFrdFW

b

a

(14-14)

ma

b

M

F^

r

The negative sign in the first line of this equation arises because the force is in opposite direction of .

Fig 14-12

^

r

Applying Eq(12-4) (14-15)

We can obtain the value of the potential energy at a single point if we define a reference point.

)11

(ba

abab rrGMmWUUU

br 0bU

)01

()()( r

GMmrUU

r

GMmrU a )(

3 From to mgy

From Eq(14-15), the difference in potential energy between the location at a height y (y is a small value) above the surface and the surface is

mgyRmyGMyRR

ymGM

yRRmGMRUyRUU

EEEE

E

EEEEE

2/)(

)11

()()(

Ea Rr

ERy

r

GMmrU )(

)11

(ba rr

GMmU

4 The three universal speeds A projectile fired upward from the Earth’s surface will usually slow down, come momentarily to rest, and return to Earth.

(a) The first universal speed, km/s9.71 vFor a certain initial speed, however, it will move around the Earth, not return back to the Earth immediately.

vh

``````

)(2

1

E

E20 R

mmGmE v

)(2

1

E

E2

hR

mmGm

v

2E

E

E

2

)( hR

mmG

hRm

v

Newton’s second law:

hR

Gm

R

Gm

E

E

E

E0

2v

If hR E

E0 gRv

vh

``````

0Em/s109.7 30

)(2 E

E

hR

GmmE

For a certain initial speed, the object can move upward forever, with its speed decreasing gradually to zero just as its distance from Earth approaches to infinitely. The initial speed for this case is called the “escape speed”.

0 ffii KUKU 0)(2

1 20

R

GMmmv

skmR

GMv /2.11

20

(b) The second universal speed, km/s2.112 v

E=0 ``````

v

h

中国“嫦娥一号”绕月探测卫星在轨飞行

(c) The third universal speed, sm.4k16 3 v

v

h

E>0

vh

抛 体 的 轨 迹 与 能 量 的 关 系

0E

0E 椭 圆 (包括圆 )

km/s9.71 v

0E

0E 抛物线

km/s2.112 v

0E

0E 双曲线

sm.4k16 3 v

牛顿的《自然哲学的数学原理》插图，抛体的运动轨迹取决于抛体的初速度

0E