Wednesday, Nov. 27, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #21Wednesday, Nov. 27, 2002Dr. JaehoonYu1. Gravitational Field2. Energy in Planetary and Satellite Motions3. Escape Speed4. Fluid and Pressure5. Variation of Pressure and Depth6. Absolute and Relative PressureToday’s homework is homework #21 due 6:00pm, Friday, Dec. 6!!Wednesday, Nov. 27, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Announcements• Remember the Term Exam on Monday, Dec. 9 in the class– Covers all material in chapters 11 – 15– Review on Wednesday, Dec. 4• Happy Thanksgiving!!!Wednesday, Nov. 27, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3The Gravitational FieldThe force exists every point in the space.The gravitational force is a field force.If one were to place a test object of mass m at any point in thespace in the existence of another object of mass M, the test object will fill the gravitational force, , exerted byM.gmFg=In other words, the gravitational field at a point in space is the gravitational force experienced by a test particle placed at the point divided by the mass of the test particle.Therefore the gravitational field g is defined as mFgg≡So how does the Earth’s gravitational field look like?gWhere is the unit vector pointing outward from the center of the EarthrˆEFar away from the Earth’s surfaceClose to the Earth’s surfacemFg=rRGMEEˆ2−=Wednesday, Nov. 27, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4The Gravitational Potential EnergyWhat is the potential energy of an object at the height y from the surface of the Earth?No, it would not. Because gravitational force is a central force, and a central force is a conservative force, the work done by the gravitational force is independent of the path.The path can be looked at as consisting of many tangential and radial motions. Tangential motions do not contribute to work!!!UDo you think this would work in general cases?Why not?Because this formula is only valid for the case where the gravitational force is constant, near the surface of the Earth and the generalized gravitational force is inversely proportional to the square of the distance.OK. Then how would we generalize the potential energy in the gravitational field?REmmriFgrfFgmgy=Wednesday, Nov. 27, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5More on The Gravitational Potential EnergySince the gravitational force is a radial force, it only performed work while the path was radial direction only. Therefore, the work performed bythe gravitational force that depends on the position becomesPotential energy is the negative change of work in the pathdWU∆Since the Earth’s gravitational force is ( )2rmGMrFE−=So the potential energy function becomesifUU−Since potential energy only matters for differences, by taking the infinite distance as the initial point of the potential energy, we getrmGMUE−=For any two particles?rmGmU21−=The energy needed to take the particles infinitely apart.For many particles?∑=jijiUU,,rdF ⋅=()drrF= →path wholeFor theW( )∫=firrdrrFifUU −=( )∫−=firrdrrF∫=firrEdrrmGM2−−=ifErrmGM11Wednesday, Nov. 27, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Example 14.6A particle of mass m is displaced through a small vertical distance ∆ynear the Earth’s surface. Show that in this situation the general expression forthe change in gravitational potential energy is reduced to the ∆U=mg∆y.Taking the general expression of gravitational potential energyThe above formula becomesU∆U∆Since the situation is close to the surface of the EarthEfEiRrRr ≈≈ and Therefore, ∆U becomes2EERymGMU∆−=∆Since on the surface of the Earth the gravitational field is 2EERGMg =The potential energy becomes ymgU∆−=∆−−=ifErrmGM11()ififErrrrmGM−−=ifErrymGM∆−=Wednesday, Nov. 27, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7Energy in Planetary and Satellite MotionsConsider an object of mass m moving at a speed vnear a massive object of mass M (M>>m).ESystems like the Sun and the Earth or the Earth and the Moon whose motions are contained within a closed orbit is called Bound Systems.2rmGMEFor a system to be bound, the total energy must be negative.Assuming a circular orbit, in order for the object to be kept inthe orbit the gravitational force must provide the radial acceleration. Therefore from Newton’s second law of motionMvrWhat’s the total energy?The kinetic energy for this system is221mvTherefore the total mechanical energy of the system isESince the gravitational force is conservative, the total mechanical energy of the system is conserved.UK+=rGMmmv −=221ma=rvm2=rmGME2=UK+=rGMm2−=Wednesday, Nov. 27, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Example 14.7The space shuttle releases a 470kg communication satellite whilein an orbit that is 280km above the surface of the Earth. A rocket engine on the satellite boosts it into a geosynchronous orbit, which is an orbit in which the satellite stays directly over a single location on the Earth, How much energy did the engine have to provide?What is the radius of the geosynchronous orbit?From Kepler’s 3rdlawT2TTherefore the geosynchronous radius isBecause the initial position before the boost is 280kmirThe total energy needed to boost the satellite at the geosynchronous radius is the difference of the total energy before and after the boost Where KEisEKGSrE∆32EKT=()()m731424314241023.41089.91064.81089.91064.8×=××=××=−−mRE51080.2 ×+=m61065.6 ×=−−=iGSsErrmGM 112J106724111019.11065.611023.4124701098.51067.6×=×−×××××−=−EGM2=π43214/1089.9 ms−×=3GSErK=sday41064.81×==Wednesday, Nov. 27, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9Escape SpeedConsider an object of mass m is projected vertically from the surface of the Earth with an initial speed vi and eventually comes to stopvf=0 at the distance rmax.ESolving the above equation for vi, one obtainsTherefore if the initial speed viis known, one can use this formula to compute the final height hof the object.Because the total energy is conservedIn order for the object to escape Earth’s gravitational field completely, the initial speed needs to beREmhMEvivf=0 at h=rmaxhivescvThis is called the escape speed. This formula is valid for any planet or large mass objects. How does this depend on the mass of the escaping
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