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. [27 no answer; about half had a rough idea, but didn’t nail it succinctly, the other half were confused] Gravity is related to mass thoguh the equation F=Gm1m2/r^2 and mass is related to density by the equation Density=mass/volume [True, but what does knowing G really give you?] (13-11) g =GME/RE since we know g=9.80m/s2; knowing G gives us ME., and RE had been well-known since the time of the greeks.

Cavendish Experiment Artist’s conception of the original Cavendish experiment to “Weigh the Earth” http://en.wikipedia.org/wiki/Cavendish_experiment

Cavendish Experiment Size of the angle Artist’s conception of the original Cavendish experiment to “Weigh the Earth” Size of the angle Change is greatly exaggerated in this cartoon; it’s hard to measure (tiny)! http://en.wikipedia.org/wiki/Cavendish_experiment

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Chapter 13 problems

Chapter 13 problems (c) What is its potential energy at launch? (d) What is its kinetic energy at launch?

Chapter 13 problems ME (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. (19 correct; 5 made errors, some way off, some silly; 26 no answer; 2 were confused). Basically, for the most part the class got this. Schwarzchild radius=2949.62 m; v^2=(2GM)/R; v=3*10^8 m/s, G=6.67*10^-11, M=1.99*10^30 kg. [right idea, but an ESTIMATE with 6 sig figs??] Using the escape speed formula, I found the Schwarzschild radius for the Sun to be approx 2950m. My only concern with my answer is whether or not the gravitational constant in the equation should be 6.67x10^-11 or is there a separate gravitational constant specifically for the Sun. [YES, the same G works for everything: “Universal”]

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!!

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

P = Po + rgh 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

How about a non-uniform Fluid?? 28 no answer 19 same 1 decrease 6 increase What happens to the pressure at the bottom when the salad dressing separates into oil (top) and vinegar (bottom)? The pressure on the bottom of the vessel remains the same because the total weight of the fluid above it has not changed even though the lighter fluid has moved to the top. [THE TOTAL WEIGHT IS NOT THE RELEVANT issue; think of along pipe in a barrel] The pressure on the bottom of the vessel remains the same according to Pascal's Principle, which states that the pressure is transmitted uniformly to all portions of the fluid. [BUT THIS ONLY HOLDS FOR HOMOGENEOUS FLUIDS; as we shall soon see] 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?? 28 no answer 19 same 1 decrease 6 increase What happens to the pressure at the bottom when the salad dressing separates into oil (top) and vinegar (bottom)? The pressure on the bottom of the vessel remains the same because the total weight of the fluid above it has not changed even though the lighter fluid has moved to the top. [THE TOTAL WEIGHT IS NOT THE RELEVANT issue; think of along pipe in a barrel] The pressure on the bottom of the vessel remains the same according to Pascal's Principle, which states that the pressure is transmitted uniformly to all portions of the fluid. [BUT THIS ONLY HOLDS FOR HOMOGENEOUS FLUIDS; as we shall soon see] 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!!

+L The mass (and weight) of water above the bottom is much less on the left than for that of a cylinder of height H+L and constant diameter D, but the pressure at the bottom is the same for both vessels!

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?