DOC PREVIEW
UT Arlington PHYS 1443 - Lecture Notes

This preview shows page 1-2-3-4 out of 12 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 12 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Monday, 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. JaehoonYuI will be out of town this Wednesday, Oct. 8 è 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 theobject•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?gF2EERmMG=g2EERMG='mg=2rmMGE=( )2hRmMGEE+='g( )2hRMGEE+=Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu3Example for GravitationUsing the fact that g=9.80m/s2at the Earth’s surface, find the average density of the Earth.gSince the gravitational acceleration is So the mass of the Earth is GgRMEE2=Therefore the density of the Earth is ρ2EERMG=2111067.6EERM−×=EEVM=324EERGgR3=πEGRgπ43=33611/1050.51037.61067.6480.93mkg×=×××××=−πMonday, 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& F2a 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 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 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. ()( )22/1/1EMRr=2=MErR42861075.21084.31037.6−×=××=Mrv2=()MMrTr2/2π=24TrM2=π( )23268/1072.21036.21084.34sm−2×=×××=π( )26080.9≈Monday, 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 radiallydirected toward the Sun to keep the planet circle, we can apply Newton’s second law2rMGMPsTrvπ2=The above can be writtenThis is Kepler’s third law. It’s also valid for 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−×=Monday, 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, is2TsM324rGMs=π3rKs=324rGT=π( )311711210496.11016.31067.64×××××=−πkg301099.1×=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 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 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.τLSBADCrdrSince the area swept by the motion of the planet is dtdAFr×=rFrˆ×=0=dtLd=0=Lconst=pr ×=vMrp×= vrMp×=const=rdr ×=21dtvr ×=21dtMLp2=pML2=const=Monday, 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.MFθFree Body DiagramMdgMFG=NFθFWhich force did the work?=WForce FHow much work did it do?What does this mean?Physical work is done only by the component of of the force along the movement of the object.Unit?Joule)(for JmN=•Work is energy transfer!!()=⋅∑dFθcosFdMonday, Oct. 6, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu10Example of Work w/ Constant ForceA man cleaning a floor pulls a vacuum cleaner with a force of magnitude F=50.0N at an angle of 30.0owith East. Calculate the work done by the force on the vacuum cleaner as the vacuum cleaner is displaced by 3.00m to East.Does work depend on mass of the object being worked on?MF30οMd=WJW 13030cos00.30.50 =××=oYesWhy don’t I see the mass term in the work at all then?It


View Full Document

UT Arlington PHYS 1443 - Lecture Notes

Documents in this Course
Physics

Physics

30 pages

Physics

Physics

25 pages

Physics

Physics

25 pages

Waves

Waves

17 pages

Physics

Physics

16 pages

Friction

Friction

15 pages

Load more
Download Lecture Notes
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Lecture Notes and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Lecture Notes 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?