Chapter 6

Gravitation

2

Objectives 6.1 Newton's Law of Universal Gravitation

6.2 Gravitational Field 6.3 Gravitational Potential

6.4 Satellite Motion in Circular Orbits

6.5 Escape Velocity

3

Learning Outcome (a)

and use the formula F = GMm/r2

(b) explain the meaning of gravitational field

(c) define gravitational field strength as force of gravity per unit mass

(d) use the equation g = GM/r2 for a gravitational field

(e) define the potential at a point in a gravitational field;

(f) derive and use the formula V = - GM/r 4

g) use the formula for potential energy U = -GMm/r

h) show that U = mg r = mgh is a special case of U = -GMm/r for situations near to the surface of the Earth

i) use the relationship g = - dV/dr

j) explain, with graphical illustrations, the variations of gravitational field strength and gravitational potential with distance from the surface of the Earth

5

Reflection:

For every action there is an equal and opposite reaction.

or

Whenever on object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. action = opposite reaction

F1 = -F2 or m1a1 = -m2a2

Section 3.4 Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt 6

Reflection:

F1 = - F2 or m1a1 = - m2a2 Jet propulsion exhaust gases in one direction and the rocket in the other direction

Gravity jump from a table and you will accelerate to earth. In reality BOTH you and the earth are accelerating towards each other

You small mass, huge acceleration (m1a1)

Earth huge mass, very small acceleration (-m2a2)

BUT m1a1 = -m2a2 Section 3.4 Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt

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Newton's Laws in Action

Friction on the tires provides necessary centripetal acceleration. Passengers continue straight ahead in original direction and as car turns the door comes toward passenger 1st Law As car turns you push against door and the door equally pushes against you 3rd Law

Section 3.4

Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt 8

Gravity is a fundamental force of nature

We do not know what causes it We can only describe it

Law of Universal Gravitation Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them

Section 3.5 Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt

Gm1m2

r2 Equation form: F =

9

G is the universal gravitational constant

G = 6.67 x 10-11 N.m2/kg2

G:

is a very small quantity

thought to be valid throughout the universe was measured by Cavendish 70 years after

Section 3.5

Gm1m2

r2 Equation form: F =

Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt 10

The forces that attract particles together are equal and opposite F1 = - F2 or m1a1 = - m2a2

Section 3.5 Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt

r

F = Gm1m2 / r2

m1 m2 F2 F1

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6.1 Newton's Law of Gravitation

For a homogeneous sphere the gravitational force acts as if all the mass of the sphere were at its center

Section 3.5

Gm1m2

r2 F =

Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt 12

Two objects with masses of 1.0 kg and 2.0 kg are 1.0 m apart. What is the magnitude of the gravitational force between the masses?

Section 3.5 Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt

1.0 m

1.0 kg 2.0 kg

Force of 9.8 N

Force of 19.6 N

Earth

Negligible force

13

Two objects with masses of 1.0 kg and 2.0 kg are 1.0 m apart. What is the magnitude of the gravitational force between the masses?

Section 3.5

Gm1m2 r2 F =

F = (6.67 x 10-11 N-m2/kg2)(1.0 kg)(2.0 kg)

(1.0 m)2

F = 1.3 x 10-10 N

Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt 14

6.2 Gravitational Field

Section 3.5 Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt

Gravitational field: a region where gravitational force acts on a body.

Gravitational field strength, g at a point in a gravitational field is the gravitational pull per unit mass on a body at that point, thus g = F/m.

Where F = -GMm/r2 ; then, g = GM/r2

15

6.2 Gravitational Field

ME and RE are the mass and radius of Earth

weight (w = mg)

Section 3.5

GME m R2E

F = [force of gravity on object of mass m]

w = mg = GME m RE2

GME R2E

g =

m cancels out g is independent of mass Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt

Gravitational field strength = Force of Gravity on Earth

16

Variation of g with Altitude For a mass, m on the surface of the Earth, mg0 = GMm/R2 g0 = GM/R2 GM = g0R2 If r > R surface), mg = GMm/r2 g = GM/r2

g = gR2/r2 (1) g = (R2/r2) g0

r

M

m

h

R

r = R + h, h = altitude

17

Variation of g with Altitude g = (R2/r2) g0 g = g0R2/(R + h)2

g 1/(R + h)2

If r < R surface), the gravitational pull is due to the sphere of radius r1 ; M = (4/3) R3

from go = GM/R2 ; g0 = (4/3) GR ; thus g/g0 = r1/R ; or g r

r

M

m

h

R

m r1

r = R + h, h = altitude 18

For r > R: g 1/(R + h)2 and for r < R: g r

g-r Graph

g 1/r2

0

g0

Acceleration due to gravity, g

r R

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6.3 Gravitational Potential Energy As before, the gravitational potential energy decreases when the separation decreases. We assume that the gravitational potential energy Ep is zero for r = , where r is the separation distance. The potential energy is negative for any finite separation and becomes progressively more negative as the particles move closer together. We take the gravitational potential energy of the two-particle system to be Ep = - GMm / r

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6.3 G. potential Energy -Proof

Let a baseball, starting from rest at a great (infinite) distance from Earth, fall toward point P. The potential energy of the baseball-Earth system is initially zero.

When the baseball reaches P, the potential energy is the negative of the work W done by the gravitational force as the baseball moves to P from its distant position.

21

A baseball of mass m falls towards Earth from infinity, along a radial line (an x axis) passing through point P at a distance R from th ecenter of Earth.

M P m

F x

dx

R

22

Thus dxFWE Rp

dxx

GMm

Fdx

R

R

2

dxx

GMmR2 R

GMm

x

GMmR

23

Acceleration due to Gravity for a Spherical Uniform Object

g = acceleration due to gravity M = mass of any spherical uniform object

Section 3.5 Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt

GM r2

g =

24

Earth Orbit - Centripetal Force

1) Proper Tangential Velocity

2) Centripetal Force

Fc = mac = mv2/r (since ac = v2/r)

The proper combination will keep the moon or an artificial satellite in stable orbit

Section 3.5 Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt

25

Earth as the same rate

Section 3.5 Source:http://campus.kcu.edu/faculty/bhaynie/ips/ch03.ppt 26

Johannes Kepler (1571-1630)

Tycho Brahe/Tyge Ottesen Brahe de Knudstrup

(1546-1601)

27

The law of orbits: All planets move in elliptical orbits, with the Sun at one focus

The law of areas: A line that connects the planet to the Sun sweeps out equal areas

equal time intervals The law of periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit

28

Elliptical orbits of planets are described by a semimajor axis a and an eccentricity e

For most planets, the eccentricities are very small (Earth's e is 0.00167)

29

The law of orbits The orbit in the figure is described by given its semimajor axis a and its eccentricity e, the latter defined so that ea is the distance from the center of

The sum of the perihelion (nearest the Sun) distance Rp and the aphelion (farthest from the Sun) distance Ra is 2a. The sum of the distance from any position in the orbit to two foci is 2a. The equation of any position (x, y) in the orbit is

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The law of orbits An eccentricity of zero corresponds to a circle, in which the two foci merge to a single central point.

The eccentricities of the planetary orbits are not large, so the orbits look circular.

The eccentricity of Earth's orbit is only 0.0167.

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rpL

))((2

1rdrdA

dt

dr

dt

dA

2

2

m

L

dt

dA

2

))(( mvr ))(( rmr 2mr const

2

2rr

For a star-planet system, the total angular momentum is constant (no external torques)

For the elementary area swept by vector

32

For a circular From the definition of a period

For elliptic orbits

))(( 22

rmr

GMmmaF

2

22 42 TT

32

r

GM

32

2 4 rGM

T

32

2 4 aGM

T

Second law

33

For a circular Kinetic energy of a satellite Total mechanical energy of a satellite

r

vm

r

GMm 2

2)(maF

2

U

2

2mvK

UKEr

GMm

r

GMm

2 r

GMm

2K

r

GMm

2

6.4 Satellite motion in circular orbits

34

2

UK

UK

UKEtotal

2

1

6.4 Satellite motion in circular orbits

35

6.4 Satellite motion in circular orbits For an elliptic orbit it can be shown

Orbits with different e but the same a have the same total mechanical energy

a

GMmE

2

36

The Law of Areas The planet will move most slowly when it is farthest from the Sun and most rapidly when it is nearest to the Sun. The law of areas is a direct consequence of the idea that all of the forces are directed exactly toward the sun.

37

The Law of Periods Consider a circular orbit with radius r. See figure. Applying Newton's second law, F = ma, to the orbiting planet yield If is replaced with 2 /T, where T is the period of the motion, yield

))(( 22

rmr

GMm

32

2 4 rGM

T

M

r

m

The law holds also for elliptical orbits, provided we replace r with a, the semimajor axis of the ellipse. 38

Example 1

kg100.6)6089)(1067.6(

)1060.6(44 24211

362

2

32

GT

rM

A satellite in circular orbit at an altitude h of 230 km above Earth's surface has a period T of 89 min. What mass of Earth follows from these data?

Sol: From Kepler's law of periods we have

The radius r of the satellite orbit is

r = R + h = 6.37 X 106 m + 230 X 103 m = 6.60 X 106 m, where R = radius of Earth.

39

Example 2 Comet Halley orbits about the Sun with a period of 76 years and, in 1986, had a distance of closest approach to the Sun, its perihelion distance Rp, of 8.9 1010 m. (a) What is the comet's farthest distance from the Sun, its aphelion distance Ra? (b) What is the eccentricity of the orbit of comet Halley?

Sol:(a) From Kepler's law of period we have

a

m107.2

4

)104.2)(1099.1)(1067.6(

4

12

3/1

2

2930113

1

2

2GMT

40

Ra = 2a Rp = 5.3 X 1012

Since ea = a Rp We have e = (a Rp) / a = 1 Rp / a = 1 (8.9 X 1010) / (2.7 X 1012)

= 0.97

41

6.4 Satellite motion in a circular orbit The mechanical energy KE + PE of the satellite remains constant. We first assume that the orbit of the satellite is circular.

The potential energy is

where r is the radius of the orbit.

By Newton's second law,

r

GMmPE

r

mv

r

GMm 2

2 Where r

v2

is the centripetal acceleration of the satellite.

a = v2/r = v = r 2

42

6.4 Satellite motion in a circular orbit The kinetic energy of a satellite is

The total mechanical energy is

For a satellite in an elliptical orbit of semimajor axis a, we have

r

GMmmvKE

22

1 2

r

GMm

r

GMm

r

GMmPEKE

22

a

GMmPEKE

2

r

mv

r

GMm 2

2

43

For a satellite orbiting Earth, the gravitational pull of Earth upon the satellite,

6.4 Satellite motion in a circular orbit

22

2

2

Tmrmr

r

GMmF

33

23

3

3

21

2

3

2or constant

4

r

T

r

T

GMr

T

44

gRvgR

vac

2

6.5 Escape velocity Accounting for the shape of Earth, projectile motion has to be modified:

45

6.5 Escape Speed There is a certain minimum initial speed that will cause a projectile to move upward forever, theoretically coming to rest only at infinity. This initial speed is called the escape speed. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because this is our zero-potential energy configuration.

46

002

12

1

planet

planet

R

mGmvm

ffii UKUK

planet

planetescape R

Gmv

2

6.5 Escape velocity Escape speed: speed required for a particle to escape from the planet into infinity (and stop there)

47

csmR

Gmv

object

objectescape /103

2 8

6.5 Escape velocity If for some astronomical object

Nothing (even light) can escape from the surface of this object a black hole

48

6.5 Escape velocity From the principle of conservation of energy, we have Where M is the mass of the planet and R is its radius. Thus

02

1 2R

GMmmvEE pk

R

GMv

2

49

Example An asteroid headed directly toward earth, has a speed of 12000 m/sec relative to the planet when it is at a distance of 10 Earth radii from Earth's center. Ignoring the effects of the terrestrial atmosphere on the asteroid, find the asteroid's speed when it reaches Earth's surface.

50

Answer Because the mass of an asteroid is much less than that of Earth, we can assign the gravitational potential energy of the asteroid-Earth system to the asteroid alone, and we can neglect any change in the speed of Earth relative to the asteroid during the asteroid's fall.

51

Thus, Let m represent the mass of the asteroid, M the mass of the earth (=5.98X1024kg), and R the radius of Earth (= 6.37X106 m), Thus

pikipfkf EEEE

R

GMmmv

R

GMmmv if 102

1

2

1 22

52

10

11

222R

GMvv if

228

6

241123

sm 10567.2

9.01037.6

)1098.5)(1067.6(2)1012(

-14 ms 1060.1fv

53

Summary: Gravitation

Gravitational Field

G = GM/r2

Law of Gravitation

F = Gm1m2/r2

Gravitational Potential

V = -GM/r

U = -GMm/r

g = -dV/dr

Satellite

v = (GM/r)1/2

T2 r3

Escape Velocity

v = (2GM/R)1/2