Transcript Dimensional Analysis

CAPSTONE Lecture 3
Gravity, Energy and their uses
07.06.2010
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Gravitational Energy,
The Cavendish Experiment
The Virial Theorem
Computation
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Big G and little g
Newton’s force law, F=-GmM/r2, reduce to
F=mg, where g=GM/r2, M is the mass of
the Earth and r is the radius of the Earth
(onto which a body is falling).
Galileo (~1580 – 1642) derived the last
equation for bodies falling on the Earth,
and found g=980 cm/sec2.
Newton’s law is a generalization of
Galileo’s result, for bodies of any M and r.
G and M always appear together. To
measure M, one must know G.
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Newton had to show, mathematically, that for perfect,
uniform sphere, one can treat the entire mass of the
Earth as if it is a point source at the center of the Earth.
(For objects like the Earth, this rule is a very good
approximation, even though the Earth is not perfectly
homogeneous.)
He had to assume that the nature of the material of
Earth did not affect the value of G or the (1/r2) form of
the force law. (No one has ever been able to show this
assumption to be wrong.
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THE CAVENDISH EXPERIMENT
The modern version of the Cavendish experiment is
described here and we will do it in class. You will
do the experiment yourself later in the term.
a) Suspend, with a torsion ribbon, a dumbbell with
two small weights of known mass, separated by a
few centimeters in a plastic case (to isolate the
balls from air currents.
b) On a movable arm, outside the case, place to large
balls, at the same separation as the small balls
from each other. Place the large balls as close as
possible to the small ones and let the system
equilibrate (it takes all night for it to reach
equilibrium.)
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c) Rotate the movable arm so the big balls are
as close as possible to the opposite small balls.
d) A force has been removed from one side of
each ball and added to the other side of each
small ball. The net force change is 2GmM/r2,
where r is the separation of the large and small
balls. The small balls will move under this force.
A mirror attached to the torsion string reflects a
laser beam onto the wall. The acceleration of the
laser spot can be measured and the force thus
determined. Since m, M and r are known, G can
be detected. G=6.67 x 10-8 cm3/g-sec2.
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Since the balls move on a plane parallel to the
surface of the Earth, the effect of the Earth’s
mass is the same on both balls and
perpendicular to the force of the large mass
balls. The Earth has been “removed” from the
experiment.
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Kinetic and Potential Energy
K=(1/2)mv2, defined as kinetic energy of motion
U=-GMm/r, the potential energy of a system of two
interacting gravitational forces.
Consider two bodies in an otherwise empty Universe,
very far apart. Then, K=0 (nothing is making them move
fast) and U=0 (r is very large).
Energy is conserved, so
Total energy at the start of the fall of the two bodies to
each other is K+U=0, as for the point when they are
closest together.
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Free fall velocity and escape velocity
(1/2)m(vend)2-GmM/rend = 0.
(1/2)m(vend)2 = GmM/rend
Solve for v at the end of the fall,
v=(2GM/r (closest approach))1/2
If two bodies are close to each other to start, and
we want to remove them to a large distance
(infinity) from each other, we must make them
both move just as fast as they were moving when
they fell from infinity.
Thus, v(free-fall)=- v(escape)
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Virial theorem
F= - GmM/r2= - mv2/r
(1/2)(-GmM/r2)=(1/2) (-mv2/r)
(1/r)(1/2)(GmM/r)=(1/r)(1/2)mv2
-(1/2)(-GmM/r)=(1/2)mv2
-U/2=K
K=-U/2
This theorem applies for objects in stable,
bound orbits.
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Units
Here are the relations of the
above quantities in terms of
units. Use [x] to mean “the units
of x.”
[m] = grams, or g
[x] = cm (centimeters)
[v] = cm/sec
[a] = cm/sec2
[F] = [ma] = (g-cm)/sec2
[K] = [mv2] =( g-cm2)/sec2 = ergs
[U] = [GmM/r] = [G] g2/cm; since
U is energy, it has the same units
as K. Then,
[G] =cm3/(g-sec2)
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Dimensional Analysis
Two sample problems, to demonstrate the use
of dimensional analysis:
1. A ball accelerates at 980 cm/sec2 and has
mass 100 grams. What force is acting on the
ball?
F=ma=100 g x 980 cm/sec2= 9.8 x 104 gcm/sec2.
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2. A planet orbits at 1 AU from the Sun.
How fast is it moving? M is the mass of
the Sun and r is the separation of the
planet and the Sun.
v = (GM/r)1/2 = [{6.67 x 10-8 cm3/(g-sec2)}
x {2 x 1033 g}/(1.5 x 1013 cm)] 1/2
v=[ {(6.67 x 2)/1.5} x {(10-8 x 1033)/1013} x
{gxcm3/(g-sec2)}/ cm}] 1/2
v = [9 x 1012 cm2/sec2] 1/2
v = 3 x 106 cm/sec = 30 km/sec (1
km/sec = 105 cm)
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