Thursday, June 8, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #7Thursday, June 8, 2006Dr. Jaehoon Yu• Kepler’s Laws• Motion in Accelerated Frames• Work done by a constant force• Scalar Product of Vectors• Work done by a varying force• Work and Kinetic Energy Theorem• Potential EnergyToday’s homework is HW #4, due 7pm, Monday, June 12!!Thursday, June 8, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu2Announcements• Mid-term exam– 8:00 – 10am, Thursday, June 15, in class– CH 1 – 8 or 9?Thursday, June 8, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu3Kepler’s Laws & EllipseKepler lived in Germany and discovered the law’s governing planets’movement some 70 years before Newton, by analyzing data.Newton’s laws explain the cause of the above laws. Kepler’s third law is a direct consequence of law of gravitation being inverse square law.F1F2bcaEllipses have two different axis, major (long) and minor (short) axis, and two focal points, F1& F2a is the length of a semi-major axisb is the length of a semi-minor axis1. All planets move in elliptical orbits with the Sun at one focal point.2. The radius vector drawn from the Sun to a planet sweeps out equal area in equal time intervals. (Angular momentum conservation)3. The square of the orbital period of any planet is proportional to the cube of the semi-major axis of the elliptical orbit.Thursday, June 8, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu4The Law of Gravity and Motions of Planets•Newton assumed that the law of gravitation applies the same whether it is the apple on the surface of the Moon or of the Earth.•The interacting bodies are assumed to be point like particles.Therefore the centripetal acceleration of the Moon, aM,isNewton predicted that the ratio of the Moon’s acceleration aMto the apple’s acceleration g would be gaMREMoonApplegaMv234/1070.280.91075.2 smaM−−×=××=Newton also calculated the Moon’s orbital acceleration aMfrom the knowledge of its distance from the Earth and its orbital period, T=27.32 days=2.36x106sMaThis means that the distance to the Moon is about 60 times that of the Earth’s radius, and its acceleration is reduced by the square of the ratio. This proves that the inverse square law is valid. ()()22/1/1EMRr=2⎟⎟⎠⎞⎜⎜⎝⎛=MErR42861075.21084.31037.6−×=⎟⎟⎠⎞⎜⎜⎝⎛××=Mrv2=()MMrTr2/2π=24TrM2=π()23268/1072.21036.21084.34sm−2×=×××=π()26080.9≈Thursday, June 8, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu5Kepler’s Third LawIt is crucial to show that Keper’s third law can be predicted from the inverse square law for circular orbits.Since the orbital speed, v, of the planet with period T isSince the gravitational force exerted by the Sun is radiallydirected toward the Sun to keep the planet on a near circular path, we can apply Newton’s second law2rMGMPsTrvπ2=The above can be writtenThis is Kepler’s third law. It’s also valid for the ellipse for r being the length of the semi-major axis. The constant Ksis independent of mass of the planet. Msssvr2rMGMPsSolving for T one can obtain 2TsKandrvMp2=()rTrMp2/2π=324rGMs⎟⎟⎠⎞⎜⎜⎝⎛=π3rKs=⎟⎟⎠⎞⎜⎜⎝⎛=sGM24π3219/1097.2 ms−×=Thursday, June 8, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu6Example of Kepler’s Third LawCalculate the mass of the Sun using the fact that the period of the Earth’s orbit around the Sun is 3.16x107s, and its distance from the Sun is 1.496x1011m.Using Kepler’s third law.The mass of the Sun, Ms, is2TsM324rGMs⎟⎟⎠⎞⎜⎜⎝⎛=π3rKs=2324rGTπ⎛⎞=⎜⎟⎝⎠()()2311211 741.496 106.67 10 3.16 10π−⎛⎞⎜⎟=××⎜⎟×× ×⎝⎠kg301099.1 ×=Thursday, June 8, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu7Kepler’s Second Law and Angular Momentum ConservationSince the gravitational force acting on the planet is always toward radial direction, it is a central forceConsider a planet of mass Mpmoving around the Sun in an elliptical orbit.τBecause the gravitational force exerted on a planet by the Sun results in no torque, the angular momentum L of the planet is constant. This is Keper’s second law which states that the radius vector from the Sun to a planet sweeps out equal areas in equal time intervals. dATherefore the torque acting on the planet by this force is always 0.Since torque is the time rate change of angular momentum L, the angular momentum is constant.τLSBADCrdrSince the area swept by the motion of the planet is dtdArF=׈rFr=×0=dLdt=0=Lconst=rp=×prMv=×pMrv=×const=12rdr=×12rvdt=×dtMLp2=pML2=const=Thursday, June 8, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu8Motion in Accelerated FramesNewton’s laws are valid only when observations are made in an inertial frame of reference. What happens in a non-inertial frame?Fictitious forces are needed to apply Newton’s second law in an accelerated frame.This force does not exist when the observations are made in an inertial reference frame.What does this mean and why is this true?Let’s consider a free ball inside a box under uniform circular motion.We see that the box has a radial force exerted on it but none on the ball directlyHow does this motion look like in an inertial frame (or frame outside a box)?rFrHow does this motion look like in the box?The ball is tumbled over to the wall of the box and feels that it is getting force that pushes it toward the wall.Why?According to Newton’s first law, the ball wants to continue on its original movement but since the box is turning, the ball feels like it is being pushed toward the wall relative to everything else in the box.vThursday, June 8, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu9Non-Inertial FrameExample of Motion in Accelerated FramesA ball of mass m is hung by a cord to the ceiling of a boxcar that is moving with an acceleration a. What do the inertial observer at rest and the non-inertial observer traveling inside the car conclude? How do they differ? mThis is how the ball looks like no matter which frame you are in.F=∑Inertial FrameθHow do the free-body diagrams look for two frames?Fg=mgmθTFg=mgmθTFficacHow do the motions interpreted in these two frames? Any differences?For an inertial frame observer, the forces being exerted on the ball are only T and Fg. The acceleration of the ball is the same as that of the box car and is provided by the x component of the
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