Chap. 13: Gravitation - Physics and Astronomy at TAMUpeople.physics.tamu.edu/kamon/teaching/phys218/slide/2013A/lec13... · Chap. 13: Gravitation Taken from Fig. 2.16, “the Universe

Chap. 13: Gravitation

1 Taken from Fig. 2.16, “the Universe in a Nutshell” by Stephen Hawking

Newton’s Law of Universal Gravity – Attractive Force (Vector)

– Inverse Square Law

– Universality

Kepler’s Laws

Goals for Chapter 13 • To calculate the gravitational forces that bodies exert

on each other

• To relate weight to the gravitational force

• To use the generalized expression for gravitational potential energy

• To study the characteristics of circular orbits

• To investigate the laws governing planetary motion

• To look at the characteristics of black holes

Introduction • What can we say about the

motion of the particles that make up Saturn’s rings?

• Why doesn’t the moon fall to earth, or the earth into the sun?

• By studying gravitation and celestial mechanics, we will be able to answer these and other questions.

Newton’s Law of Gravitation • Law of universal gravitation:

Every particle of matter 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.

• The gravitational force can be expressed mathematically as: where G is the gravitational constant.

r̂r

mmGFg 2

21

Newoton's Law of Universal Gravitation

Newton’s Law of Universal Gravitation

5

r̂r

mmGF

221


6.67x1011 Nm2/kg2 m2 = ME

m1 = MM

r Attractive Force

Magnitude of the force

between any 2 masses

6



Parameters

• G = 6.67 x 1011 Nm2/kg2

• ME = 5.98 x 1024 kg

• RE = 6.38 x 103 km

• MS = 1.99 x 1030 kg

• RS = 6.96 x 105 km

• rE-S = 150 x 106 km

• Mmoon = 7.35 x 1022 kg

• Rmoon = 1.74 x 103 km

7

Spherically Symmetric Bodies

• The gravitational force between bodies with spherically symmetric mass distributions is the same as if all their mass were concentrated at their centers.

• See Figure 13.2 at the right.


r = 3.80 x 108 m

FME

r

Moon

Earth

FME = 2.03 x 1020 N Answer:

1. Represent the Moon as a

particle and Place the tail

of the force vector on the

Moon’s center.

2. Draw the vector force

as an arrow pointing in the

Earth’s center.

3. Give the vector an

appropriate label.

10

Problem 1: Find FME

(Force on Moon due to Earth)

r̂r

mmGF

221

• G = 6.67 x 1011 Nm2/kg2

• ME = 5.98 x 1024 kg • RE = 6.38 x 103 km • MS = 1.99 x 1030 kg • Mmoon = 7.35 x 1022 kg • Rmoon = 1.74 x 103 km


Gravitational Forces: Vector

MM ME

FME = G

rME2

ME

MS

MM

MM MS

FMS = G

rMS2

Univeral Gravitational Constant

vector

FMnet = FME + FMS

11


Assume that the masses

and the coordinates are

known.

NOTE

The gravitational force is:

• …

(x2, 0)

(x3, y3)

m1

m2

m3

Problem 2: Find the Net Force on Sun.

Sun Earth

Mars

y

x

12


M1

M2

X

Y

x

y

Starship Enterprise

(not to scale)

The starship Enterprise (mass m) is investigating a new star of mass M1, and is

holding position a distance Y from the star. There is another star of mass M2 a

distance Y from the first along a line perpendicular to that connecting the

Enterprise with the first star. Suppose the Enterprise experiences total engine

failure. Express the x and y components of the net gravitational force on the

Enterprise. Use the coordinate system provided. Don’t forget that force is a

vector.

Mo

re E

xa

mp

le

14


Satellite Orbital Period : T = (2pr)/v

v

FG

16

Try to relate between F and v.

Try to relate between F and v.

Satellite Orbital Period : T = (2pr)/v


Gravity = Center-seeking Force

v

FG

mS , ME , r = 2rE , G

FG

FG = mS v 2 / (2 rE)

Circular Motion

Dynamics of Dynamics of

Uniform Circular MotionUniform Circular Motion

Center-seeking Acceleration and Force

Frad = m arad



m1 m2

FG = G

r 2

m1 m2

FG = G

r 2

Magnitude of the force

between any 2 masses

6.67x1011 Nm2/kg2 m2 = ME

m1 = MM

rAttractive Force

17


Problem 3: Find T

r = 3.80 x 108 m

FME = 2.03 x 1020 N

FME

r

v

Moon

Earth

18


Problem 4: Geosynchronous Satellite

A geosynchronous satellite, which is one that

stays above the same point on the Earth, is

used for such purpose as cable TV

transmission, for weather forecasting, and as

communication relays.

a) Determine the height above the Earth’s

surface.

b) Determine the speed of the satellite.

20


Gravitational Acceleration:

Constant?

R

FG

m

M

r

FG = m g (Newton’s 2nd Law)

∴ g = FG / m

where

FG = G m M / r 2

= m (G M / r 2)

23


Gravity: Near the Earth’s Surface

r

R

FG h

g(h) = G M / r 2

= G M / (R+h) 2

m

M

25



e.g., h = 0 m g = 9.80 m/s2

MEarth

g(h) = G -----------------

(REarth + h) 2

h = altitude above the surface

26



9

9.1

9.2

9.3

9.4

9.5

9.6

9.7

9.8

9.9

New York Mt. Everest h=1000km

g7.33

m/s

2

9.80 9.77

h = 8848 m h = 0 m

27


Gravity: Near the Moon’s Surface

e.g., h = 0 m g = 1.62 m/s2

MMoon

g(h) = G -----------------

(RMoon + h) 2

h = altitude above the surface

28


1. Accuracy: Dg/g ~ 1 part in 109

2. Applications:

Mineral deposits

r > rave g > gave

Salt domes (i.e., petroleum)

r < rave g < gave

Gravimeters

29


g = 9.80 m/s2

RE = 6.37 x 106 m

ME = 5.96 x 1024 kg

rE = 5.50 x 103 kg/m3

= 5.50 g/cm3 ~ 2 x rRock

What can we conclude about the

structure of the Earth?

Average Density of the Earth

30

Interior of the Earth

• The earth is approximately spherically symmetric, but it is not uniform throughout its volume, as shown in Figure 13.9 at the right.

• Follow Example 13.4, which shows how to calculate the weight of a robotic lander on Mars.


Weight on the Earth

4.22 x 106 N

Weight at h = 350 km?

International Space Station (1)

MS = 4.31 x 105 kg

32

www.station.nasa.gov


Weight on the Earth

4.22 x 106 N (MS = 4.31 x 105 kg)

Weight at h = 350 km

F = MS g(h)

= MS G ME / (RE + h)2

www.station.nasa.gov

33

International Space Station (2)

Weight

• The weight of a body decreases with its distance from the

earth’s center, as shown in Figure 13.8 below.

34


Apparent Weight (FN)

Fnet = FN – mg = m a

y F.B.D.

a = (1/2)g (up)

35


Weightlessness

F.B.D. ?

a = g (down)

y

36

Circular Satellite Orbits • For a circular orbit, the speed of a satellite is just right to keep its distance

from the center of the earth constant. (See Figure 13.15 below.)

• A satellite is constantly falling around the earth. Astronauts inside the

satellite in orbit are in a state of apparent weightlessness because they are

falling with the satellite. (See Figure 13.16 below.)

• Follow Example 13.6.

37

Gravitational Potential Energy

• Derivation of

gravitational potential

energy using Fig. 13.10

at the right.

• The gravitational

potential energy of a

system consisting of a

particle of mass m and

the earth is

U = –GmEm/r.

38

Gravitational Potential Energy

• The gravitational potential

energy of the earth-

astronaut system increases

(becomes less negative) as

the astronaut moves away

from the earth, as shown in

Fig. 13.11 at the right.

39

From the Earth to the Moon to To escape from

the earth, an object must have the escape speed.

See Example 13.5 using Figure 13.12

002

12

2

E

Eescape

r

f,Gf

Rr

i,Gi

ffii

R

mMmv

UKUK

UKUK

E

40

Chap. 13: Gravitation

43 Taken from Fig. 2.16, “the Universe in a Nutshell” by Stephen Hawking

Newton’s Law of Universal Gravity

– Attractive Force (Vector)

– Inverse Square Law

– Universality

Kepler’s Laws


44

Kepler’s Laws of Planetary Motion

s = semi-major axis

F1Q + F1R = 2 x s


Elliptical and Circular Orbits

A special case of elliptical orbit is circular orbit:

e (eccentricity) = 0

48


Kepler’s 2nd Law

(2) A12 = A34

v12 > v34

A d v Dt

d12 d34

49


Example 6

Use Kepler’s second law to show that the ratio

of the speeds of a planet at its nearest and

farthest points from the Sun is equal to the

inverse ratio of the nearest and farthest

distances: vN/vF = dF/dN

52


T = 2 p r / v

G MP MS / r 2 = MP v 2 / r

Frad = m arad

r 3 = (G MS /4 p 2) T 2

T 2 = (4 p 2 / G MS ) r 3 = KSun r 3

r (circular orbit) s (elliptical orbit)

Kepler’s 3rd Law

e = 0 (circular orbit) s = r

53


T = 2 p r / v

G MS ME / r 2 = MS v 2 / r

Frad = m arad

r 3 = (G ME /4 p 2) T 2

T 2 = (4 p 2 / G ME ) r 3 = KE r 3

Satellite about Earth:

Circular Orbit

58


Example 1

A s p a c e s h i p ( m a s s m = 3 0 0 k g ) i s i n a n

elliptical orbit around the Earth. At perigee (apogee) of its

orbit, it is 620 km (3620 km) above

the Earth’s surface. Calculate the

period of the orbit.

Note: RE = 6380 km.

(1) s = …

(2) T 2 = KE s 3 T = …

P

A


ParametersParameters

• G = 6.67 x 1011 Nm2/kg2

• ME = 5.98 x 1024 kg

• RE = 6.38 x 103 km

• MS = 1.99 x 1030 kg

• RS = 6.96 x 105 km

• rE-S = 150 x 106 km

• Mmoon = 7.35 x 1022 kg

• Rmoon = 1.74 x 103 km

T 2 = (4 p 2 / G ME ) r 3 = KE r 3

59




• G = 6.67 x 1011 Nm2/kg2

• ME = 5.98 x 1024 kg

• RE = 6.38 x 103 km

• MS = 1.99 x 1030 kg

• RS = 6.96 x 105 km

• rE-S = 150 x 106 km

• Mmoon = 7.35 x 1022 kg

• Rmoon = 1.74 x 103 km

Example 2

The National Aggie Space Administration

(NASA) wants to send a satellite (300 kg)

into a circular orbit of radius R from the

center of the Earth at the rotational period

of T = 240 hours. Find R and v.

(1) T = …

(2) T 2 = KE R 3 R = …

(3) v = 2 p R / T = …

T 2 = (4 p 2 / G ME ) r 3 = KE r 3

62




• G = 6.67 x 1011 Nm2/kg2

• ME = 5.98 x 1024 kg

• RE = 6.38 x 103 km

• MS = 1.99 x 1030 kg

• RS = 6.96 x 105 km

• rE-S = 150 x 106 km

• Mmoon = 7.35 x 1022 kg

• Rmoon = 1.74 x 103 km

Example 3

(1) T = 1 day = … s

(2) T 2 = KE R 3 R = … m

(3) H = R – RE = … m (… km)

A geosynchronous satellite (m = 300 kg) is one

that stays above the same point on the Earth,

which is possible only if it is above a point of the

equator. Determine the height H above the

Earth’s surface such a satellite must orbit.

T 2 = (4 p 2 / G ME ) r 3 = KE r 3

64




• G = 6.67 x 1011 Nm2/kg2

• ME = 5.98 x 1024 kg

• RE = 6.38 x 103 km

• MS = 1.99 x 1030 kg

• RS = 6.96 x 105 km

• rE-S = 150 x 106 km

• Mmoon = 7.35 x 1022 kg

• Rmoon = 1.74 x 103 km

Example 4

Halley’s comet orbits the Sun roughly once

every 76 years. It comes very close to the

surface of the Sun on its closest approach

about how far out from the Sun is it at its

farthest ?

T 2 = (4 p 2 / G MS ) r 3 = KS r 3

66


x

s s

67




• G = 6.67 x 1011 Nm2/kg2

• ME = 5.98 x 1024 kg

• RE = 6.38 x 103 km

• MS = 1.99 x 1030 kg

• RS = 6.96 x 105 km

• rE-S = 150 x 106 km

• Mmoon = 7.35 x 1022 kg

• Rmoon = 1.74 x 103 km

Example 4

Halley’s comet orbits the Sun roughly once

every 76 years. It comes very close to the

surface of the Sun on its closest approach

about how far out from the Sun is it at its

farthest ?

(1) T = 76 yrs = … s

(2) T 2 = KS s 3 s = … m

(3) s >> RS x = … m

T 2 = (4 p 2 / G MS ) r 3 = KS r 3

68


Example 5

When it orbited the Moon, the Apollo 11

spacecraft’s mass was 9.979 x 103 kg, its

period was 7.140 x 103 s, and its mean

distance from the Moon’s center was

1.849 x 106 m. Find v and MMoon.

(1) v = 2 p R / T = … m/s

(2) arad = v2 / R = … m/s2

(3) G m MMoon / R2 = m aradMMoon= … kg

70


73

Appendix

Gravitation

Determining the Value of G • In 1798 Henry Cavendish made the first measurement of the

value of G. Figure 13.4 below illustrates his method.

Some gravitational calculations

• Example 13.1 shows how to calculate the gravitational force between two masses.

• Example 13.2 shows the acceleration due to gravitational force.

• Example 13.3 illustrates the superposition of forces, meaning that gravitational forces combine vectorially. (See Figure 13.5 below.)

Weight

• The weight of a body is the total gravitational force exerted

on it by all other bodies in the universe.

• At the surface of the earth, we can neglect all other

gravitational forces, so a body’s weight is w = GmEm/RE2.

• The acceleration due to gravity at the earth’s surface is

g = GmE/RE2.

The motion of satellites • The trajectory of a projectile fired from A toward B depends on

its initial speed. If it is fired fast enough, it goes into a closed

elliptical orbit (trajectories 3, 4, and 5 in Figure 13.14 below).

Kepler’s laws and planetary motion • Each planet moves in an

elliptical orbit with the sun at one focus.

• A line from the sun to a given planet sweeps out equal areas in equal times (see Figure 13.19 at the right).

• The periods of the planets are proportional to the 3/2 powers of the major axis lengths of their orbits.

Some orbital examples • Follow Conceptual Example 13.7 on orbital speeds.

• Follow Example 13.8 involving Kepler’s third law.

• Example 13.9 examines the orbit of Comet Halley. See Figure

13.20 below.

Spherical mass distributions • Follow the proof that the gravitational interaction between two spherically

symmetric mass distributions is the same as if each one were concentrated at its

center. Use Figure 13.22 below.

A point mass inside a spherical shell • If a point mass is inside a spherically symmetric shell, the

potential energy of the system is constant. This means that

the shell exerts no force on a point mass inside of it.

• Follow Example 13.10 using Figure 13.24 below.

Apparent weight and the earth’s rotation

• The true weight of an

object is equal to the

earth’s gravitational

attraction on it.

• The apparent weight of

an object, as measured

by the spring scale in

Figure 13.25 at the

right, is less than the

true weight due to the

earth’s rotation.

Black holes • If a spherical nonrotating body has radius less than the Schwarzschild radius,

nothing can escape from it. Such a body is a black hole. (See Figure 13.26

below.)

• The Schwarzschild radius is RS = 2GM/c2.

• The event horizon is the surface of the sphere of radius RS surrounding a black

hole.

• Follow Example 13.11.

Detecting black holes • We can detect black holes by looking for x rays emitted

from their accretion disks. (See Figure 13.27 below.)

Documents

Chap. 13: Gravitation - Physics and Astronomy at TAMUpeople.physics.tamu.edu/kamon/teaching/phys218/slide/2013A/lec13... · Chap. 13: Gravitation Taken from Fig. 2.16, “the Universe