Transcript 12_LectureOutline

Lecture Outline
Chapter 12
Physics, 4th Edition
James S. Walker
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Chapter 12
Gravity
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Units of Chapter 12
• Newton’s Law of Universal Gravitation
• Gravitational Attraction of Spherical
Bodies
• Kepler’s Laws of Orbital Motion
• Gravitational Potential Energy
• Energy Conservation
• Tides
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12-1 Newton’s Law of Universal Gravitation
Newton’s insight:
The force accelerating an apple downward is
the same force that keeps the Moon in its orbit.
Hence, Universal Gravitation.
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12-1 Newton’s Law of Universal Gravitation
The gravitational force is always attractive, and
points along the line connecting the two
masses:
The two forces shown are an action-reaction
pair.
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12-1 Newton’s Law of Universal Gravitation
G is a very small number; this means that the
force of gravity is negligible unless there is a
very large mass involved (such as the Earth).
If an object is being acted upon by several
different gravitational forces, the net force on it
is the vector sum of the individual forces.
This is called the principle of superposition.
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12-2 Gravitational Attraction of Spherical
Bodies
Gravitational force between a point mass and a
sphere: the force is the same as if all the mass
of the sphere were concentrated at its center.
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12-2 Gravitational Attraction of Spherical
Bodies
What about the gravitational force on objects at
the surface of the Earth? The center of the Earth is
one Earth radius away, so this is the distance we
use:
Therefore,
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12-2 Gravitational Attraction of Spherical
Bodies
The acceleration of gravity decreases slowly with
altitude:
The acceleration due to gravity at a height h above the Earth’s surface
(a) In this plot, the peak of Mt. Everest is at about h=5.50 mi, and the space shuttle orbit is at roughly h = 150 mi. (b) This
shows the decrease in the acceleration of gravity from the surface of the Earth to an altitude of about 25,000 mi. the orbit of
geosynchronous satellites – ones that orbit above a fixed point on the Earth- is at roughly h = 22,300 mi
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12-2 Gravitational Attraction of Spherical
Bodies
The Cavendish experiment allows us to measure
the universal gravitation constant:
The Cavendish experiment
The gravitational attraction between the masses m and M causes the rod and the suspending thread to
twist. Measurement of the twist angle allows for a direct measurement of the gravitational force.
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12-2 Gravitational Attraction of Spherical Bodies
After landing on Mars, an astronaut performs a simple experiment by dropping
a rock. A quick calculation using the drop height and the time of fall yields a
value of 3.73 m/s2 for the rock’s acceleration. (a) Find the mass of Mars, given
that its radius is RM = 3.39 x 106 m, (b) What is the acceleration of gravity due
to Mars at a distance 2RM from the center of the planet?
F = mgM = G m MM / RM2
MM = gMRM2 / G
MM = (3.73 m/s2)(3.39 x 106 m)2
= 6.43 x 1023 kg
6.67 x 10-11 N.m2 / kg2
Apply Newton’s law of gravity with r = 2RM
F = ma = G m MM
(2RM)2
a = G MM
= ¼ (GMM) = ¼(g M)= ¼ (3.73 m/s2) = 0.933 m/s2
RM2
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12-2 Gravitational Attraction of Spherical
Bodies
Even though the gravitational force is very small,
the mirror allows measurement of tiny deflections.
Measuring G also allowed the mass of the Earth to
be calculated, as the local acceleration of gravity
and the radius of the Earth were known.
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12-3 Kepler’s Laws of Orbital Motion
Johannes Kepler made detailed studies of the
apparent motions of the planets over many years,
and was able to formulate three empirical laws:
1. Planets follow elliptical orbits, with the Sun at
one focus of the ellipse.
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12-3 Kepler’s Laws of Orbital Motion
2. As a planet moves in its orbit, it sweeps out an
equal amount of area in an equal amount of time.
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12-3 Kepler’s Laws of Orbital Motion
3. The period, T, of a planet increases as its
mean distance from the Sun, r, raised to the 3/2
power.
This can be shown to be a consequence of the
inverse square form of the gravitational force.
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12-3 Kepler’s Laws of Orbital Motion
The earth revolves around the Sun once a year at an average distance of 1.50 x 1011 m.
(a) Use this information to calculate the mass of the Sun. (b) Find the period of
revolution for the planet Mercury, whose average distance from the Sun is 5.79 x 1010 m.
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12-3 Kepler’s Laws of Orbital Motion
A geosynchronous satellite is one whose orbital
period is equal to one day. If such a satellite is
orbiting above the equator, it will be in a fixed
position with respect to the ground.
These satellites are used for communications and
and weather forecasting.
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12-3 Kepler’s Laws of Orbital Motion
GPS satellites are not
in geosynchronous
orbits; their orbit
period is 12 hours.
Triangulation of
signals from several
satellites allows
precise location of
Global Positioning System
objects on Earth. Thesystem
of 24 satellites in orbit about the earth makes it
A
possible to determine a person’s location with great accuracy.
Measuring the distance of a person from satellite 2 places the
person somewhere here on the red circle. Similar measurements
using satellite 11 place the person’s position somewhere on the
green circle, and further measurement can pinpoint the person’s
location.
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12-4 Gravitational Potential Energy
Gravitational potential energy of an object of
mass m a distance r from the Earth’s center:
Gravitational potential energy as a function of the distance r from the center
of the Earth
The lower curve in this plot shows the gravitational potential energy, U = -G m ME/r, for r greater than RE.
Near the Earth’s surface, U is approximately linear, corresponding to the result U = mgh
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12-4 Gravitational Potential Energy
Very close to the Earth’s surface, the
gravitational potential increases linearly with
altitude:
Gravitational potential energy, just like all
other forms of energy, is a scalar. It
therefore has no components; just a sign.
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12-5 Energy Conservation
Total mechanical energy of an object of mass m
a distance r from the center of the Earth:
This confirms what we already know – as an
object approaches the Earth, it moves faster
and faster.
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12-5 Energy Conservation
Potential and kinetic energies of an object falling toward Earth
As an object with zero total energy moves closer to the earth, its gravitational potential energy, U, becomes increasingly
negative. In order for the total energy to remain zero, E = U + K = 0, it is necessary for the kinetic energy to become
increasingly positive.
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12-5 Energy Conservation
Escape speed: the initial upward speed a
projectile must have in order to escape
from the Earth’s gravity
G = 6.67 x 10-11 N.m2 / kg2
ME = 5.87 x 1024 kg
RE = 6.37 x 106 m
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12-5 Energy Conservation
Calculate the escape speed for an object launched from the Moon.
G = 6.67 x 10-11 N.m2 / kg2
Mm = 7.35 x 1022 kg
RE = 1.74 x 106 m
Ve = [(2 * 6.57 x 10-11 N.m2/kg2)(7.35 x 1022 kg) / 1.74 x 106 m)] 1/2
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12-5 Energy Conservation
Speed of a projectile as it leaves the Earth,
for various launch speeds
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12-5 Energy Conservation
Black holes:
If an object is sufficiently massive and
sufficiently small, the escape speed will
equal or exceed the speed of light –
light itself will not be able to escape the
surface.
This is a black hole.
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12-5 Energy Conservation
Light will be bent by any
gravitational field; this can
be seen when we view a
distant galaxy beyond a
closer galaxy cluster. This is
called gravitational lensing,
and many examples have
been found.
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12-6 Tides
Usually we can treat planets, moons, and stars
as though they were point objects, but in fact
they are not.
When two large objects exert gravitational
forces on each other, the force on the near side
is larger than the force on the far side, because
the near side is closer to the other object.
This difference in gravitational force across an
object due to its size is called a tidal force.
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12-6 Tides
This figure illustrates a general tidal force on
the left, and the result of lunar tidal forces on
the Earth on the right.
The reason for two tides a day
(a) Tides are caused by a disparity between the gravitational force exerted at various points on a finite-sized objects (dark
red arrows) and the centripetal force needed for circular motion (light red arrows). Note that the gravitational force
decreases with distance, as expected. On the other hand, the centripetal force required to keep an object moving in a
circular path increase with distance. On the near side, therefore, the gravitational force is stronger that required, and the
object is stretched inward. On the far side, the gravitational force is weaker than required and the object stretches
outward. (b) On the Earth, the water in the oceans responds more to the deforming effects of tides than do the solid rocks
of the land. The result is two high tides and two low tides daily on opposite sides of the Earth.
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12-6 Tides
Tidal forces can result in orbital locking,
where the moon always has the same face
towards the planet – as does Earth’s Moon.
If a moon gets too close to a large planet, the
tidal forces can be strong enough to tear the
moon apart. This occurs inside the Roche
limit; closer to the planet we have rings, not
moons.
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Summary of Chapter 12
• Force of gravity between two point masses:
• G is the universal gravitational constant:
• In calculating gravitational forces,
spherically symmetric bodies can be replaced
by point masses.
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Summary of Chapter 12
•
Acceleration of gravity:
•
Mass of the Earth:
•
Kepler’s laws:
1. Planetary orbits are ellipses, Sun at one
focus
2. Planets sweep out equal area in equal time
3. Square of orbital period is proportional to
cube of distance from Sun
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Summary of Chapter 12
• Orbital period:
• Gravitational potential energy:
• U is a scalar, and goes to zero as the
masses become infinitely far apart
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Summary of Chapter 12
• Total mechanical energy:
• Escape speed:
• Tidal forces are due to the variations in
gravitational force across an extended body
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