Transcript Concept Development 12-1 - Heck's Physics

Table of Contents
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13
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Concept Development 12-1
Concept Development 13-1
Concept Development 13-2
Concept Development 13-3
Concept Development 14-1
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Concept Development 12-1
Created for CVCA Physics
By
Dick Heckathorn
17 February 2K + 5
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1. Inverse Square Law
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1/4
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1/9
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1/16
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CD 12-1 p49
Inverse Square Law
At 4d
3d
2d
5d apple weighs
1/9
1/25
1/4
1/16NN
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Demo Distorted Room
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Distorted Room or . . .
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2. Inverse Square Law
25 m2
5 m from the source?
10 m from the source? 100 m2
CD 12-1 p49
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3. Inverse Square Law
Hold one hand twice as far away
as the other hand. You see…..
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CD 12-1 p49
Demo Sun-Earth-Moon
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Sun Earth Moon Allignment
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Concept Development 13-1
1.An apple that has a mass of 0.1 kg
has the same mass wherever it is.
The amount of matter that makes
up the apple does not depend
depends on
upon the location of the apple. It
has the same resistance to
acceleration wherever it is—its
inertia everywhere is the same.
different.
CD 13-1 p51
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The weight of the apple is a different
story. It may weight exactly 1 N in San
Francisco and slightly less in milehigh Denver. On the surface of the
moon the apple would weigh 1/6 N,
and far out in space it may have
almost no weight at all. The quantity
that doesn’t change with location is
weight
the quantity that may change
mass and
is its mass weight.
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CD 13-1 p51
the force due
That’s because weight, is
mass
to gravity on a body, and this force
varies with distance. So weight is the
force of gravity between two bodies,
usually some small object in contact
with the earth. When we refer to the
weight, of
mass
an object we are usually
speaking of the gravitational force that
attracts it to the earth.
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CD 13-1 p51
2. Force and Weight
The answers are:
…weigh 500 N.
…weigh 500 N
…force of 500 N,
…force of 500 N,
…weights 500 N!
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CD 13-1 p51
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CD 13-1 p52
4.a Fill in the gravitational force on
the object at R/2 and R = 0.
1/2
0
The motion would be: to and fro
(in simple harmonic motion).
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CD 13-1 p52
5. a Fill in the weights of the object
on the planet’s shrinking surface.
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1
100
b. Object is still collapsed and a
ladder it erected ‘R’ distance long
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CD 13-1 p51
Concept Development 13-2
1. Fnew = 4 Fold
2. Fnew = 4 Fold
CD 13-1 p53
3. Fnew = 16 Fold
4. Fnew = Fold no change
5. Fnew = 2/9 Fold
6. Fnew = Fold no change
CD 13-1 p54
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Concept Development 13-3
a.
force
vector
on
B.
d.
b.
c.
Why?
Is thisthe
Which
force
has
the
more,
greater
less,
acceleration?
the
sameofas
1. Draw
Consider
two
equal-mass
blobs
A?
water, A and B, initially at rest in the
moon’s gravitational field. The
vector shows the gravitational force
of the moon on A.
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Concept Development 13-3
e. Because of different accelerations,
with time A gets further ahead of B
and the distance between A and B
increases.
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Concept Development 13-3
f. If A and B were connected by a
rubber band, with time the the
rubber band would stretch.
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Concept Development 13-3
g. This stretching is due to the difference
in the moon’s gravitational pulls.
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Concept Development 13-3
their
accelerations
will consist
h. Then
The two
blobs
will eventually
crash
of changes
in direction
into
the moon.
To orbit around
only. the
moon instead of crashing into it, the
blobs should move tangentially.
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CD 13-1 p55
2. Now consider the same two
blobs located on opposite sides
of the earth.
a. Because of differences in the
moon’s pull on the blobs, they
tend to spread away from each other.
approach each other
b. This spreading produces
ocean tides.
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CD 13-1 p55
c. If earth and moon would be
closer, gravitational forces
between them would be
more, the same, less
and the difference in
gravitational forces on the near
and far parts of the ocean
would be
more, the same, less
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CD 13-1 p55
d. Because the moon’s orbit is
slightly elliptical, earth and
moon are closer in winter
months than in summer. On a
world average, then, ocean
tides are greater in
December, June no difference
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d-R
d
d+R
a. Which of the two forces:
moon on left mass (m) or
moon on right mass (m)
is stronger and why? Fd-R
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CD 13-1 p56
d-R
d
d+R
d = 3.8 x 108 m
R = 6.38 x 106 m
mM
F G
2
R
Mm = 7.34 x 1022 kg
mon w = 1 x 106 kg
Fd+R = 3.28 x 101 = 32.8 N
Fd-R = 3.51 x 101 = 35.1 N
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CD 13-1 p56
What would happen to the orbit of
the moon if the earth suddenly
disappeared?
Find force of sun on moon
4.366 x 1020 N
Find force of earth on moon
2.027 x 1020 N
CD 13-1 p56
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Satellite Motion 14-1
km/sec
<
= 82km/sec
VVV
= <<
0.05
km/sec
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2. Satellite Motion
v
v
F
F
F
F
v
CD 13-1 p57
v
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Consider the various positions of the
satellite as it orbits the planet as shown.
With respect to the planet, in which position
does the satellite have the maximum:
a. speed? A
b. velocity? A
c. momentum? A
d. kinetic energy? A
e. grav pot energy? C
f. total energy?
same-same
CD 13-1 p58
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That’s all folks!
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