Transcript P221_2008_week11

Chapter 12 problems
Chapter 12 problems
•
The first experimental determination of the universal Gravitational
Constant (G), which appears in Newton’s law for gravitational force was
derived (some 100 years after the fact) from experiments by Henry
Cavendish using equipment he had inherited (and modified) from Rev.
John Michell. When Cavendish published his results in the Philosophical
Transactions of the Royal Society of London, his article actually had the
title “Experiments to determine the Density of the Earth”. Explain how one
might make a connection between the determination of G and the
determination of this quantity. [19 no answer; 6 correct; 16 confused]
•
Density is mass per unit volume, and so if there is an area that is
more dense than another location, it follows that there is more
mass concentrated in one area than in the other. The gravitational
constant therefore would be larger in the denser area than the
less dense area. [Many were like this, but keep in mind “G” is a
constant (it does not depend on anything as far as we know; on
the other hand: “g” does depend on ME and RE {AND “G”})]
G depends on mass, and mass depends on density
First, Newton's law of universal gravitation would be used. If G is
determined, then the mass of Earth could also be determined
from the 9.8m/s^2 gravitational acceleration on the Earth surface,
another known mass, and the distance between Earth and the
other known mass. [ALMOST PERFECT!]
•
•
Cavendish Experiment
Artist’s conception of the original Cavendish experiment to
“Weigh the Earth”
http://en.wikipedia.org/wiki/Cavendish_experiment
Cavendish Experiment
Artist’s conception of the original Cavendish experiment to
“Weigh the Earth”
http://en.wikipedia.org/wiki/Cavendish_experiment
Size of the angle
Change is greatly
exaggerated in this
cartoon; it’s hard
to measure (tiny)!
Chapter 13 problems
(c) What is its potential energy at launch?
(d) What is its kinetic energy at launch?
•
The Schwarzschild radius of an astronomical object is approximately
equal to be that radius for which a sphere of the mass in question has
an escape velocity equal to the speed of light. Estimate the
Schwarzschild radius for our Sun. If you could compress the mass of
our Sun into a sphere of this radius, it would form a black hole.
•
(15 correct; 5 made errors, some way off, some silly;
19 no answer; 3 were confused)
•
The Schwarzchild radius is equivalent to the radius that produces an
escape velocity equal to the speed of like. Thus, 3x10^8 m/s = v = sqr
root(2GM/R). M = 1.99x10^30 and G = 6.67x10^-11 Nm^2/kg^2. So,
3x10^8 m/s = sqr root((6.67x10^-11)(1.99x10^30)/R). So, R =
6.78x10^56 m . (where did this come from?)
Rs=2GM/c^2 G=6.67x10-11 NxM^2/kg^2 c= 3.00x108 m/s M=1.989 x
10^30 kg Schwarzschild radius = 2.94814 km [right idea, but an
ESTIMATE with 6 sig figs??]
Sun (mass (kg) = 1.99x10^30, radius (m) = 6.96x10^8, Escape Speed
(km/s) = 618) Speed of light (m/s) = 2.998x10^8
2.998x10^8=((2*6.673x10^-11* 1.99x10^30)/R)^1/2 Schwarzschild
radius=2.97x10^3 m [ i.e. about 3 km].
•
•
Chapter 13
problems
Kepler’s second Law
(equal areas in equal times)
This is equivalent to
saying that the
angular momentum
of the planet must be
conserved
throughout the orbit.
l = m(r x v)
http://www.windows.ucar.edu/tour/link=/the_universe/uts/kepler2.html&edu=elem
Chapter 13 problems
Principle of equivalence
Curved Space
Hydrostatic Pressure
The magnitude of the force experienced
by such a device does not depend
on its orientation! It depends on
the depth, g, surface pressure and
area (DA). Pressure does not have
a direction associated with it
it is in all directions at once!!
Manometer as a P gauge
The height difference can be used
as a measure of the pressure difference
(assuming that the density of the
liquid is known). Hence we have
Pressures measured in “inches of Hg”
or “mm of Hg” (i.e. Torr).
P = Po + rgh
Pascal’s Vases (from UIUC)
http://demo.physics.uiuc.edu/lectdemo/scripts/demo_descript.idc?DemoID=229
Hydrostatic Pressure
If the fluid is of uniform density, then the
pressure does NOT depend on the shape
of the container (be careful for cases where
the density is not constant however!! See the
next slide).
Hydrostatic Pressure
If the fluid is of uniform density, then the
pressure does NOT depend on the shape
of the container (be careful for cases where
the density is not constant however!!).
Why is the pressure
at the bottom of
these two containers
the same?
Hydrostatic Pressure
If the fluid is of uniform density, then the
pressure does NOT depend on the shape
of the container (be careful for cases where
the density is not constant however!!).
The walls in the first
container provide the same
forces provided by the
extra fluid in the second
container
How about a non-uniform
Fluid??
17 no answer
16 same
2 decrease
7 increase
What happens to the pressure at the bottom when the salad
dressing separates into oil (top) and vinegar (bottom)?
• Pressure on the bottom of the vessel would remain the same. This is
because the total mass above the bottom and the position of the
bottom has not changed. Depth is the major factor in pressure and
this factor is not affected by a phase change above.
• I think the pressure on the bottom of the vessel would just stay the
same even after the phase separation because Pascal's principle
states that when pressure is applied to a confined liquid, this
pressure is transmitted evenly throughout the entire liquid.
• as the two seperate, the molecules begin to settle down, an exert
less pressure onto the container. The top however may increase as
the oild is pushed upward into a smaller area. [ Meaning what?]
• THIS IS NOT AN EASY QUESTION, BUT IT IS A GOOD ONE! You
have to think about WHY the pressure in the homogeneous case
does not depend on the shape of the container.
How about a non-uniform
Fluid??
17 no answer
16 same
2 decrease
7 increase
What happens to the pressure at the bottom when the salad
dressing separates into oil (top) and vinegar (bottom)?
• Pressure on the bottom of the vessel would remain the same. This is
because the total mass above the bottom and the position of the
bottom has not changed. Depth is the major factor in pressure and
this factor is not affected by a phase change above.
• I think the pressure on the bottom of the vessel would just stay the
same even after the phase separation because Pascal's principle
states that when pressure is applied to a confined liquid, this
pressure is transmitted evenly throughout the entire liquid.
• as the two seperate, the molecules begin to settle down, an exert
less pressure onto the container. The top however may increase as
the oild is pushed upward into a smaller area. [ Meaning what?]
• THIS IS NOT AN EASY QUESTION, BUT IT IS A GOOD ONE! You
have to think about WHY the pressure in the homogeneous case
does not depend on the shape of the container. The downward
force from the slanted portion of the vessel is reduced because
the density of the fluid at the top has decrease after
separation!!-> PRESSURE AT BOTTOM WILL DECREASE!!
Pascal’s Principle (hydraulic
systems)
LARGE
force out
Small force
in
Small force
in
Chapter 14
problems
Equation of Continuity
(mass in must = mass out)
A1v1=A2v2
Assuming that r is constant
(i.e. an incompressible fluid)
E.G. with the Equation of Continuity
(mass in must = mass out)
What is the flow through the unmarked pipe?