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
r̂r
mmGF
221
Newoton's Law of Universal Gravitation
6.67x1011 Nm2/kg2 m2 = ME
m1 = MM
r Attractive Force
Magnitude of the force
between any 2 masses
6
Newton’s Law of Universal Gravitation
Newoton's Law of Universal Gravitation
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.
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
Gravitational Forces: Vector
MM ME
FME = G
rME2
ME
MS
MM
MM MS
FMS = G
rMS2
Univeral Gravitational Constant
vector
FMnet = FME + FMS
11
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
Newton’s Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
Problem 3: Find T
r = 3.80 x 108 m
FME = 2.03 x 1020 N
FME
r
v
Moon
Earth
18
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
Gravity: Near the Earth’s Surface
r
R
FG h
g(h) = G M / r 2
= G M / (R+h) 2
m
M
25
Newoton's Law of Universal Gravitation
Gravity: Near the Earth’s Surface
e.g., h = 0 m g = 9.80 m/s2
MEarth
g(h) = G -----------------
(REarth + h) 2
h = altitude above the surface
26
Newoton's Law of Universal Gravitation
Gravity: Near the Earth’s Surface
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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.
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
Apparent Weight (FN)
Fnet = FN – mg = m a
y F.B.D.
a = (1/2)g (up)
35
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
Newoton's Law of Universal Gravitation
44
Kepler’s Laws of Planetary Motion
s = semi-major axis
F1Q + F1R = 2 x s
Newoton's Law of Universal Gravitation
Elliptical and Circular Orbits
A special case of elliptical orbit is circular orbit:
e (eccentricity) = 0
48
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
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
Newoton's Law of Universal Gravitation
Newoton's Law of Universal Gravitation
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
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
Newoton's Law of Universal Gravitation
Newoton's Law of Universal Gravitation
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
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
Newoton's Law of Universal Gravitation
Newoton's Law of Universal Gravitation
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
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
Newoton's Law of Universal Gravitation
Newoton's Law of Universal Gravitation
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
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
Newoton's Law of Universal Gravitation
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
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.