Physics 111. Tuesday, November 9, Universal Law Potential Energy Kepler s Laws. density hydrostatic equilibrium Pascal s Principle

Transcription

1 ics Tuesday, ember 9, 2004 Ch 12: Ch 15: Gravity Universal Law Potential Energy Kepler s Laws Fluids density hydrostatic equilibrium Pascal s Principle

2 Announcements Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468

3 Announcements This week s lab will be another physics workshop - on fluids this time. No quiz this week.

4 Gravity We extend our set of forces to include Newton s Universal Law of Gravity. We also look at how this new conservative force results in a new Gravitational Potential energy. Finally, we look briefly at Kepler s Laws.

5 Ch 5: Applying Newton s Laws We ve mentioned last week that any two bodies that have gravitational mass exert an attractive force on one another. So far, we ve only looked at the form this force takes near the surface of the Earth, namely Fg = W = m g

6 Ch 5: Applying Newton s Laws But this form has been derived from a more general form known as Newton s Universal Law of Gravity: F = Gm 1 m 2 g 2 1 Universal Gravitational Constant r 2 ˆr 2 1 G = Nm 2 where m 1 and m 2 are the masses of the two objects, r is the distance between them, and is a unit vector pointing from m 2 to m 1. ˆr 2 1 kg 2

7 What s this ˆr? Well, we ve seen unit vectors for the directions in a Cartesian coordinate system. Remember ˆx ŷ ẑ The unit vector in the x-direction The unit vector in the y-direction The unit vector in the z-direction

8 What s this ˆr? x z r ˆr y It s the unit vector that points radially out from the origin of the coordinate system to the point of interest in space. The quantity r (the radial distance from the origin to the point of interest) is one of three coordinates in the spherical coordinate system.

9 Ch 5: Applying Newton s Laws F g 2 1 F g1 2 m 1 m 2 Notice, these two forces are equal in magnitude and opposite in direction. In fact, they ARE a 3rd Law Pair! We also note that these forces act at a distance: that is, the two objects have no direct physical contact with one another.

10 Worksheet #1

11 Tues Worksheet #1 Two satellites A and B have the same mass and go around planet Earth in concentric orbits. The distance of satellite B from Earth s center is twice that of satellite A. What is the ratio of the gravitational force of the Earth on satellite B to the gravitational force of the Earth on satellite A? 1. 1/8 2. 1/4 3. 1/ PI, Mazur (1997)

12 Ch 5: Applying Newton s Laws Such action-at-a-distance forces are termed field forces. This means that we can define a new physical quantity known in this case as the gravitational field. G = Gm 2 2 r 2 ˆr 2 Every mass has an associated gravitational field around it.

13 Ch 5: Applying Newton s Laws G = Gm 2 2 r 2 If we put a mass m 1 in the field G 2 created by the presence of mass m 2, it feels a force given by ˆr 2 F = m G g

14 Ch 5: Applying Newton s Laws How do we reconcile these two forms? F = m G g True universally! G = Gm 2 2 r 2 ˆr 2 W = mg True only for experiments near the surface of planet Earth. Let s see what value G 2 has near the surface of Earth.

15 Ch 5: Applying Newton s Laws G = Gm 2 2 r 2 ˆr 2 Let s see what value G 2 has near the surface of Earth. m 2 will be the mass of the Earth and r will be the radius of the Earth. Since our experiments are at the surface of the Earth, the distance from our objects to the center of the Earth applies. m = kg r = m G = ( Nm )( kg) kg 2 ( m) 2 = 9.83 m s 2

16 Worksheet #2

17 Tues Worksheet #2 Two satellites A and B have the same mass and go around planet Earth in concentric orbits. The distance of satellite B from Earth s center is twice that of satellite A. What is the ratio of the tangential speed of B to that of A? 1. 1/ PI, Mazur (1997)

18 We ve seen that near the Earth s surface, the function for gravitational potential energy takes the form U m g y g = But we noted that this form is correct only for problems that take place at or near the Earth s surface. For problems in outer-space we need to use a more general form of this function...

19 Gm m U = 1 2 g r This form is derived using calculus and the relationship between force and potential energy (also from calculus) with the assumption that U = 0 when r = Formula for gravitational pot l

20 If we have a system that involves several masses, we can compute the total potential energy of the system as the sum of the potential energies between each pair of masses in the system. So, for a 3-mass system... m 1 m r 3 12 r 23 U tot = U 12 + U 13 + U 23 = G m m m m m m 2 3 r 12 r 13 m 2 r 13 r 23

21 U tot = U 12 + U 13 + U 23 = G m m m m m m 2 3 r 12 r 13 r 23 Notice that this result is simply the sum of the energy changes that result when each mass is brought from infinity to its final location. Mass 1 is free. m 1 m r 3 12 r 23 m 2 r 13 Bringing up mass 2 in the presence of mass 1 results in the 1st term.

22 U tot = U 12 + U 13 + U 23 = G m m m m m m 2 3 r 12 r 13 r 23 The last two terms result from bringing up mass 3 in the presence of masses 2 and 1. Mass 1 is free. m 1 m r 3 12 r 23 m 2 r 13 Bringing up mass 2 in the presence of mass 1 results in the 1st term.

23 How much energy is required to move a 1000-kg mass from the Earth s surface to a distance that is twice the Earth s radius away from the center of the Earth? Use for the Earth a mass of 6 X kg and a radius of 6400 km. Worksheet #3 ΔU g = U f U i = Gm E m sat r sat Gm E m sat r E ΔU = J Enough energy to power a 100-W light bulb for 10 years!

24 The planets in our solar system move around the Sun in roughly circular orbits. What force is responsible? This means that some force must be acting on the planets causing a centripetal acceleration.

25 Given that the Earth (m = 6 X kg) orbits the Sun (m = 2 X kg) in roughly a circular orbit (r = 1.5 X m) once per year, calculated the mean orbital speed of the Earth. Worksheet #4 F r = F G m E v 2 r ES = Gm E m S r 2 ES v 2 = Gm S r ES v = Gm r ES S

26 Given that the Earth (m = 6 X kg) orbits the Sun (m = 2 X kg) in roughly a circular orbit (r = 1.5 X m) once per year, calculated the mean orbital speed of the Earth. v = Gm S r ES = ( Nm 2 kg 2 )( kg) m v = 29.8 km/s Spaceship Earth travels through the Cosmos at a surprisingly high speed!

27 The planets do not move around the Sun in perfectly circular orbits. The first person to figure out the correct shape of the orbits was Johannes Kepler. Kepler s First Law says that the planets move around the Sun in elliptical orbits, with the Sun at one focus.

28 foci The sum of the distance from any point on the ellipse to each of the two foci is constant Major axis center

29 Kepler s Second Law says a line drawn connecting the Sun to a planet will sweep out an equal area in the ellipse in equal time intervals. The law tells us how fast a planet moves at various points in its orbit: close to the Sun the planet will have a greater speed than far from the Sun. Makes sense since gravity goes as 1/distance 2

30 A B D C If it takes one month for the planet to go from A to B... It will also take one month to go from C to D, if the areas of the blue & red triangles are the same.

31 Skip derivation Kepler s Third Law says the square of the period of the orbit of a planet is proportional to the cube of the length of the semi-major axis of the orbit. For a circular orbit, the semi-major axis is simply the radius of the circle (the diameter being the major axis).

32 Let s demonstrate this law for the case of a circular orbit. Simply use Newton s 2nd Law where the Radial Force is the Gravitational Force. Gm Sun m pl. 2 F G = F r 2 = m pl v pl r Sun pl r Sun pl

33 Gm m m Sun pl. 2 = plv pl r Sun pl The planet will complete one orbit in one period. The circumference of the orbit is the distance the planet will travel in one period. So... 2 r Sun pl v pl = C T = 2πr Sun pl T Now plug this value of velocity into the above equation...

34 Gm m m Sun pl. 2 = plv pl r Sun pl 2 r Sun pl Gm m m (2πr / T )2 Sun pl. = pl r 2 r Sun pl Sun pl Gm Sun = 4π 2 r 2 T 2 = r T 2 Sun pl 4π 2 Gm Sun r 3

35 We can simplify this expression by calculating the constants... T 2 = K r 3 Sun where... K Sun = 4π 2 Gm Sun = s 2 m 3

36 T 2 = K a 3 a = semi-major axis Kepler s Laws apply to the orbits of planets about the Sun moons about a planet satellites about a planet comets about a star.you name it! If it s in orbit, consult Kepler! Naturally, the constant K depends upon the body being orbited!

37 The semi-major axis of the orbit of Pluto is about 4 times as great as that of the orbit of Saturn. If Saturn orbits the Sun in about 30 years, how long does it take Pluto to orbit the Sun once? 1) 30 years 2) 120 years 3) 165 years 4) 225 years 5) 240 years Worksheet #5

38 Ch 15: Fluid Mechanics Fluid Mechanics We ve spent a lot of time looking at systems of solid objects. But matter also comes in liquid and gaseous states. We can describe the motions of such substances using the extension of Newtonian mechanics known as fluid dynamics.

39 Ch 15: Fluid Mechanics Let s start by characterizing a solid mass of uniform composition. What quantities can we directly measure? If our block was made of copper, and we doubled its size, what would happen to its mass?

40 Ch 15: Fluid Mechanics 8M M V = (2L)3 V 0 = L 3 V = 8L 3 = 8V 0 L 2L Remember our scale arguments (way back in the first week of the course)!!!

41 Ch 15: Fluid Mechanics In fact, if we made a plot of the mass of our copper block versus its volume, we d find m slope = ρ V This line has a slope that characterizes the type of material from which the block is made. We define the slope of this line as the density (ρ) of the material. How would the slope of the line for a block of styrofoam compare to that for a block of lead?

42 ρ M V Tues Ch 15: Fluid Mechanics The unit of mass, the gram, was chosen to be the mass of 1 cm 3 of liquid water. So water has a density of 1 g/cm 3 or 1 kg/l or 1000 kg/m 3. The term specific gravity refers to the ratio of the density of a given substance to that of water. Objects with a specific gravity less than 1 will float in water; those greater than 1 will sink.

43 Ch 15: Fluid Mechanics ρ M V [ρ] = [ M ] [V ] [ρ] = kg m 3 Note: density is often written in grams per cubic centimeter or g/cc. There is a factor of 1000 difference between the two sets of units.

44 Ch 15: Fluid Mechanics When a force is applied over an area, we say that the object feels pressure. P F A [P] = [F] [ A] = N m 2 = Pa pascal Usually, we talk about pressure of a fluid or a gas (like the atmosphere).

45 Ch 15: Fluid Mechanics It s all around you! Our green fish is completely submerged. The water exerts a pressure on the fish. From which direction does the fish feel the pressure? In fact, the pressure is exerted in the direction normal to the body of the fish all over the fish.

46 Ch 15: Fluid Mechanics It s all around you! Think about your own experience walking around outside (when there s NOT a hard wind blowing). You don t notice any difference in pressure over the surface of your body. What would happen if there was a pressure difference across your hand? P 1 P > P 1 2 P 2

47 Ch 15: Fluid Mechanics It s all around you! Your hand would feel a net force acting to the RIGHT in this case. Therefore, your hand would start to accelerate to the right, or you would have to exert a force through your arm to counteract this force known as a pressure gradient force. F net P 2 12PP> P 1

48 Ch 15: Fluid Mechanics So, now let s look at how pressure changes with altitude. We know it s a lot harder to breathe at the top of a mountain than at sea level--there are fewer oxygen molecules and fewer molecules in general up there.

49 Ch 15: Fluid Mechanics At the Earth s surface, we sit at the bottom of an entire column of molecules in the atmosphere. These molecules exert a pressure on us at the surface of about kpa. We define this pressure to be 1 atmosphere (atm). As we go up through the atmosphere, what happens to the pressure we feel? WHY?

50 Ch 15: Fluid Mechanics Let s look at the force balance for a little layer of atmosphere as we head up the mountain. The pressure of all the molecules above our layer. P 1 A = surface area W P 2 The pressure from all the molecules below our layer. The gravitational force acting on the layer itself. P 1 < P 2

51 Ch 15: Fluid Mechanics The pressure P 1 on the top surface of area A results in a force downward of F 1 = P 1 A. P 1 A = surface area W P 2 The pressure P 2 on the bottom surface of area A results in a force upward of F 2 = P 2 A.

52 Ch 15: Fluid Mechanics If our system is in equilibrium, the net force must be 0. So... F 2 = F 1 + W P 2 A = P 1 A + mg P 1 A = surface area Δh W P 2 But what is the mass of our little layer of atmosphere? m = ρv = ρ( AΔh) ρ is the density of air.

53 Ch 15: Fluid Mechanics If our system is in equilibrium, the net force must be 0. So... P 2 A = P 1 A + mg P 2 A = P 1 A + ρ( AΔh)g P = P + ρg(δh) 2 1 ΔP = ρ g( Δh)