Chapter 12 problems

1. Chapter 12 problems

2. Chapter 12 problems

3. 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!]

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

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

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

7. 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].

8. Chapter 13 problems

9. 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

10. Chapter 13 problems

12. Principle of equivalence

13. Curved Space

14. 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!!

15. 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

16. Pascal’s Vases (from UIUC) http://demo.physics.uiuc.edu/lectdemo/scripts/demo_descript.idc?DemoID=229

17. 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).

18. 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?

19. 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

20. 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.

21. 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!!

22. Pascal’s Principle (hydraulic systems) LARGE force out Small force in Small force in

23. Chapter 14 problems

24. Equation of Continuity(mass in must = mass out) A1v1=A2v2 Assuming that r is constant (i.e. an incompressible fluid)

25. E.G. with the Equation of Continuity(mass in must = mass out) What is the flow through the unmarked pipe?