PHYS 1443 – Section 003 Lecture #11Free Fall Acceleration & Gravitational ForceExample for GravitationKepler’s Laws & EllipseThe Law of Gravity and Motions of PlanetsKepler’s Third LawExample of Kepler’s Third LawKepler’s Second Law and Angular Momentum ConservationWork Done by a Constant ForceExample of Work w/ Constant ForceScalar Product of Two VectorsExample of Work by Scalar ProductMonday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #11•Newton’s Law of Gravitation•Kepler’s Laws•Work Done by Constant Force•Work Done by Varying ForceMonday,Oct. 6, 2003Dr. Jaehoon YuI will be out of town this Wednesday, Oct. 8 Dr. Sosebee will give the lecture.Deadline for Homework #5 is noon, Wednesday, Oct. 8!!Deadline for Homework #6 is noon, Wednesday, Oct. 15!!Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu2Free Fall Acceleration & Gravitational ForceWeight of an object with mass m is mg. Using the force exerting on a particle of mass m on the surface of the Earth, one can get•The gravitational acceleration is independent of the mass of the object•The gravitational acceleration decreases as the altitude increases•If the distance from the surface of the Earth gets infinitely large, the weight of the object approaches 0.What would the gravitational acceleration be if the object is at an altitude h above the surface of the Earth?mgWhat do these tell us about the gravitational acceleration?gF2EERmMGg2EERMG'mg2rmMGE 2hRmMGEE'g 2hRMGEEMonday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu3Example for GravitationUsing the fact that g=9.80m/s2 at the Earth’s surface, find the average density of the Earth.gSince the gravitational acceleration is So the mass of the Earth is GgRMEE2Therefore the density of the Earth is 2EERMG2111067.6EERMEEVM324EERGgR3EGRg4333611/1050.51037.61067.6480.93mkgMonday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu4Kepler’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 the direct consequence of law of gravitation being inverse square law.1. 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.F1F2bcaEllipses have two different axis, major (long) and minor (short) axis, and two focal points, F1 & F2 a is the length of a semi-major axisb is the length of a semi-minor axisMonday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu5The Law of Gravity and Motions of Planets•Newton assumed that the law of gravitation applies the same whether it is on the Moon or the apple on the surface 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 aM to the apple’s acceleration g would be gaMREMoonApplegaMv234/1070.280.91075.2 smaMNewton also calculated the Moon’s orbital acceleration aM from the knowledge of its distance from the Earth and its orbital period, T=27.32 days=2.36x106sMaThis means that the Moon’s distance is about 60 times that of the Earth’s radius, its acceleration is reduced by the square of the ratio. This proves that the inverse square law is valid. 22/1/1EMRr2MErR42861075.21084.31037.6Mrv2 MMrTr2/224TrM2 23268/1072.21036.21084.34sm 26080.9Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu6Kepler’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 radially directed toward the Sun to keep the planet circle, we can apply Newton’s second law2rMGMPsTrv2The above can be written This is Kepler’s third law. It’s also valid for ellipse for r being the length of the semi-major axis. The constant Ks is independent of mass of the planet. Mssvr2rMGMPsSolving for T one can obtain 2TsKandrvMp2 rTrMp2/2324rGMs3rKssGM243219/1097.2 msMonday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu7Example 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, is2TsM324rGMs3rKs324rGT 311711210496.11016.31067.64kg301099.1 Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu8Kepler’s Second Law and Angular Momentum ConservationSince the gravitational force acting on the planet is always toward radial direction, it is a central force Consider a planet of mass Mp moving around the Sun in an elliptical orbit.Because 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 our 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.LSBADCrdrSince the area swept by the motion of the planet is dtdAFr rFrˆ0dtLd0Lconstpr vMrpvrMpconstrdr 21dtvr 21dtMLp2pML2constMonday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu9xyWork Done by a Constant ForceWork in physics is done only when a sum of forces exerted on an object made a motion to the object.MFFree Body DiagramMdgMFGNFFWhich force did the work?WForce FHow much work did it do?What does this
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