PHYS 218sec. 517-520ReviewChap. 12GravitationWhat you have to know• Newton’s law of gravitation• Gravitational potential energy• Motion of satellites• Kepler’s three laws• We skip from section 12.6 through section 12.8N. Copernicus G. GalileiT. BraheJ. KeplerI. NewtonNewton’s law of gravity12211 2 2: gravitational constant6.67 10 N m / kggmmFGrGG−=≈ × ⋅Always attractive1m2mrO1rG1m2m2rG12rr−GGIn vector form,the gravitational force exerted by m2on m1is12, 2 on 1 1221212 1 2ˆ,ˆwhere is the unit vector in the direction of gmmFGrrrrr= −−GGG, 2 on 1gFGThe negative signmeans that it is an attractive force.Gravity for spherically symmetric bodies122 is still valid, where is the distance measured from the center of massgmmFG rr=For an object which has spherically symmetric mass distribution: concentrate all the mass of the object at its center.rEarthof mass mEm2EgmmFGr=ER2If the mass lies on the surface of the earth, EgEmmmFGR=Ex 12.2rAcceleration due to gravity120.01 kg, 0.5 kg, 0.05 mmmr===2m1m1aG2aG()()()1211 2 212 212221182the acceleration of due to 6.67 10 N m / kg 0.5 kg10.05 m1.33 10 m/sgmmFGm m Gmammr r−−× ⋅ ×== = ==×()()()2111 2 212 122222210 2the acceleration of due to 6.67 10 N m / kg 0.01 kg10.05 m2.66 10 m/sgmmFGm m Gmammr r−−× ⋅ ×== = ==×1221amam=Ex 12.3Superposition of gravitational forcesGravitational force is a vector.OMMmddxy30 30128.00 10 kg, 1.00 10 kg, 2.00 10 mMmd=× =×=×The gravitation force exerted on m= vector sum of two forces1F2FF21222The distance between and upper is 2 ,thus, . Also, .2mMdmM mMFG FGdd==1The position of is (0,0) and the position of upper is ( , ).11The unit vector for the direction of is , .22mMddr⎛⎞⎟⎜⎟⎜⎟⎟⎜⎝⎠111122226 2512 2 1222 26This leads to11,,,022111.81 10 N, 4.72 10 N221.87 10 N, =arctan 14.6xyxyxxx yyyyxyxFFFFFFFFFF FF FF F FFFFFFθ====⇒ =+= += × =+ = = ×⎛⎞⎟⎜⎟⇒ =+=× =°⎜⎟⎜⎟⎜⎝⎠WeightWeight of a body: the total gravitational force exerted on the body by ALL other bodies in the universeAt the surface of the Earth, we can neglect other stellar objects.2: weight of mass at the Earth's surfaceEgEGm mwF mR==Radius of the Earth2Since , we can express as: acceleration due to gravityEEwmg gGmgR==Mass of the EarthEx 12.4Gravity on Mars623Earth weight of the Mars lander: 3920 N.Mars: radius 3.40 10 m, mass 6.42 10 kgEMMwRm==× =×Use this information to know the mass of the Mars lander23920 NSince , the mass of the lander is 400 kg9.8 m/swwmg mg====At d = 6000000 m above the surface of Mars()()()()()11 2 2 2322262Weight of the lander6.67 10 N m / kg 6.42 10 kg 400 kg 194 N9.4 10 macceleration due to the gravity of Mars194 N0.48 m/s400 kgMMgMgMmm mmFG GrRdFgm−× ⋅ ×== = =+×== =Gravitational potential energyWhen the gravitational acceleration is constantUmgh=In general, the gravitational acceleration depends on r1r2rGravitational forcedisplacement221112221When the mass moves from to , the work done by thegravity isrrEEEgrav rrrmrrGm m Gm m Gm mW F dr drrrr==− = −∫∫EgravGm mUr= −Earth of mass EmReference point: at , 0gravrU= ∞ =Gravitational potential energy IIGravitational force is conservative0gFdr=∫vAt the surface of the Earth2If , (where )1111EEEEgrav EEEEEEEEEErR h R hGm m Gm mhUGmmRh R hR R RGm m GmmhRR=+⎛⎞⎟⎜⎟= − = −≈−−⎜⎟⎜⎟⎜++⎝⎠= − += constant, so can be droppedg=Therefore, gravUmgh=()()1cf. the Taylor expansion11 when 111nxnxxxx−+ ≈ ++ ≈−Ex 12.5From the earth to the moon1ErR=Earth: mass and radius EEmR22ErR=Muzzle speed needed to shoot the shellfrom REto 2RETo obtain the speed, we use energy conservation.111 2 2122112121What we need to obtain is the speed at . By energy conservation,.The minimum speed can be obtained when 0.102Now use and 2 ,EEEErKU KUvvGm m Gm mmvrrrR r Rv+= +=⎛⎞⎛⎞⎟⎟⎜⎜⎟⎟+ − =+−⎜⎜⎟⎟⎜⎜⎟⎟⎜⎜⎝⎠⎝⎠==()()21211 2 2 24162226.67 10 N m / kg 5.97 10 kg7900 m/s6.38 10 mEE EE EEEEEEGm Gm Gm Gm Gmrr RRRGmvR−⎛⎞⎛⎞⎟⎟⎜⎜⎟⎟= − = − =⎜⎜⎟⎟⎜⎜⎟⎟⎜⎜⎝⎠⎝⎠× ⋅ ×⇒ == =×Ex 12.5bFrom the earth to the infinity1ErR=Earth: mass and radius EEmR2r = ∞Muzzle speed needed to shoot the shellfrom REto infinity111 2 2122112121What we need to obtain is the speed at . By energy conservation,.The minimum speed can be obtained when 0.102Now use and ,EEErKU KUvvGm m Gm mmvrrrR rv+= +=⎛⎞⎛⎞⎟⎟⎜⎜⎟⎟+ − =+−⎜⎜⎟⎟⎜⎜⎟⎟⎜⎜⎝⎠⎝⎠==∞⇒ =()()11 2 2 244626.6710 Nm/kg 5.9710 kg21.12 10 m/s6.38 10 mEEGmR−× ⋅ ×==××This is called the escape speedIndependent of the mass of the objectMotion of satellitesIn projectile motion,if the speed of the projectile is small, it will return to the surface of the earthif the speed of the projectile is large enough but smaller than the escape speed it becomes a⇒⇒ satelliteif the speed of the projectile is larger than the escape speed, it never returns and travels farther away from the earth⇒Closed orbitsOpen orbitsSatellites: circular orbitsThe only force acting on a satellite is the gravity due to the earth. The gravity is the centripetal force The satellite is in a circular orbit⇒⇒The radius of the circular orbit of the satellite is determined by its speed.22Since the gravity is the centripetal force,Therefore, once is given, the radius of its orbit is fixed,or to have the radius , the satellite should have above.EEGm m Gmmvvrr rvrrv= ⇒ =Independent of the satellite massSatellites: circular orbits3/223Period of the circular motion222,EErrrTr TrvGmGmπππ== = ⇒∝For a given radius, satellite speed is determined, so is its energy211222EEEEGm m Gm Gm mEKU mv mrrrGm mr⎛⎞⎛⎞⎟⎟⎜⎜=+= +− = −⎟⎟⎜⎜⎟⎟⎜⎜⎝⎠⎝⎠= −Ex 12.6From the earth to the infinitySatellite of mass 1000 kg in a circular orbit 300 km above the earth's surfacem =ErR h=+Speed, period, acceleration()()611 2 2 24622the radius of the orbit: 6380 km 300 km 6.68 10 m6.67 10 kg m / kg 5.97 10 kgthe orbital speed: 7720 m/s6.68 10 m2period: 5440 sradial acceleration: 8.92 m/sEEradrR hGmvrrTvvarπ−=+= + = ×× ⋅ ×== =×====Ex 12.6Cont’dThe work needed to place this satellite in orbit2112111,where is the total energy on the launch pad and is the energy in orbitThe satellite is at rest on the
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