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# Gravitation

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### Gravitation

1. 1. FORCES AND NEWTON’SLAW OF GRAVITATIONStandard Competency:Analyze the nature phenomenon and its regularity within thescope of particle’s MechanicsBase Competency:Analyze the regularity of planetary motion within the universebase on Newton’s LawLearning Objectives:After completing this chapter, students should be able to[1] Analyze the relation between gravitational force and object’s masses with their distance[2] Calculate the gravitational forces resultant on particles within a system[3] Compare the gravitational acceleration at different positions[4] Formulate the quantitiy of potential gravity at a point due to several object’s masses[5] Analyze the planetary motion within a universe base on Keppler’s Law[6] Calculate the speed of satellite and its terminal velocityReferences:[1] John D Cutnell and Kenneth W. Johnson (2002). Physics 5th Ed with Compliments. John Wiley and Sons, Inc. pp[2] Douglas C. Giancolli (1985). Physics: Principles with Applications, 2nd Edition. Prentice Hall, Inc. pp[3] Sunardi dan Etsa Indra Irawan (2007). Fisika Bilingual SMA/MA untuk SMA/MA Kelas XI. CV Yrama Widya pp[4] http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l3d.html
2. 2. WHAT IS GRAVITY Gravity is a force that exist between the earth and any object surround it. The force of gravity tends to attract objects. It attracts the objects with same force. The name of gravity is proposed by both Galileo Galileo and Newton. Newton showed that the same force exists between all objects.GRAVITATIONAL FIELDDEFINITION: A space around an object where the gravitational force exist and has a certain value in its every point Gravitational field is a concept introduced by Newton to describe a force acting on objects throught out a distance in a space. According Newton, there seems to be outof mind if there are two object that can interact each otherwithout being physically interact.Field, a distance space, is then “to be created” to bridge howremote objects (distance separated objects) could interactingeach other.Base on Newton, FIELD is a quantification ofdistance acting force. Recall that F = m a(Newton’s II Law) is a type of contact force.Also FIELD is a vector quantity so it could bevisualized by an arrow which shows themagnitude and direction of grafivitationalfield.Representation of fields as arrangement arrows is well knownas force line
3. 3. GRAVITATION PHENOMENONBase on Newton’s Observations of how an apple could fallfrom its tree:[1] All objects which are on certain height will always fall freely toward the earth surface[2] A cannon ball will move in archery trajectory while shot in a certain angle and initial force[3] Moon is always stay in its trajectory while evolves the Earth and Earth is always stay in its trajectory while evolves the SunNewton Proposed:[1] There is a force whose kept all the remote objects downward to the earth surface[2] Such force works on two different objects which is separated in a certain distance[3] Such force has a property of being attractive[4] Such force works without any physical contact between those objects[5] Such force apply to all objects in universe, hence it is universal
4. 4. Gravitational Force FormulaBase on Newton’s Proposes, there is an attractive force whichworks on two different object separated in a certain distance.The force is propotional to two masses F ≈ m1 m2The force is propotional to inverse square of the distancebetween those masses 1 F ≈ r2 m1 m2 m1 m2 F ≈ → F =G r2 r2F = the attractive force G = a constant,m1 = mass of first particle UNIVERSAL GRAVITATIONALm2 = mass of second particle CONSTANTr = linier distance of both particles = 6.67 x 10−11 N.m2/kg2 (measured by Henry Cavendish)GRAVITATION FIELD FORCEGravitation field force is a force experience by an object dueto gravitational attraction per unit mass F m g= → g=G m r2Direction of gravitational field is goes to thecenter of object’s mass.If an object is under gravitational forceinfluenced by several objects, the gravitationalfield force on the object is the sum of eachgravitational field another objects g = (g1 )2 + (g1 )2 + 2(g1 )(g2 ) cos θ
5. 5. NEWTON’S LAW OF UNIVERSAL GRAVITATIONIn 1666 Isaac Newton determined that the same force thatkept the planets in motion must also exist between everyobject. Newton’s law of gravitation is an empirical physical law which is describing the gravitional attraction between bodies with their masses. He stated that every object in the universe attracts every other object in the universe. Newton’s law gravitation resembles Coulomb’s S Isaac N ton ir ew law of electrical forces, which is used tocalculate the magnitude of electrical force between twocharged bodies. Both are inverse-square laws, in which forseis inversely proportional te the square of the distance betweenthe bodies. Coulomb’s law has the product of two charges inplace of the product of the masses, and the electrostaticconstant in place of the gravitational constant.What was Newton stated is now become a Universal Law ofGravitation. It is universal because it works for all kinds,types and sizes of object. No matter how small or big theobjects are, attraction between those objects is apply.Newton’s Law of Universal Gravitation For two particles, which have masses m1 and m2 and are separated by a distance r, the force that exerts on the other is directed along the line joining the particles
6. 6. The law of gravitation is universal and very fundamental. Itcan be used to understand the motions of planets and moons,determine the surface gravity of planets, and the orbitalmotion of artificial satellites around the Earth Some consequences on Gravitation Formula: masses distance Force (F) (m1 and m2) (r) ) bigger constant stronger Condition smaller constant weaker constant smaller stronger constant bigger weaker The very small value of G is affectly effective for massive mass
7. 7. PROBLEM SOLVEDWhat is the magnitude of the gravitational force betweenthe earth (m = 5.98 x 1024 kg) and a 60-kg man whosestand on 6.38 x 106 m awaySOLUTION m1 m2 F =G r2 (5.98 x 1024 kg)(60 kg) = (6.67 x 10 −11 N.m2 /kg2 ) (6.38 x 106 m)2 = 587.9 N
8. 8. PROBLEM SOLVEDWhat is the magnitude of the gravitational force that actson each bicycle’s tyres where m1 = 12 kg and m2 = 25 kgand r = 1.2 m?(approximately the mass of a bicycle)SOLUTION m1 m2 F =G r2 (12 kg)(25 kg) = (6.67 x 10 −11 N.m2 /kg2 ) = 1.4 x 10 − 8 N (1.2 m)2 (this force is extremely small and could be neglected and this is why your bicycle will not bended by itself) PROBLEM SOLVED What is the magnitude of the gravitational force due to moon (m = 7.35 x 1022 kg) on a 50-kg man where moon- earth is 3.84 x 108 m away SOLUTION m1 m2 F =G r2 −11 2 2 (7.35 x 1022 kg)(50 kg) = (6.67 x 10 N.m /kg ) (3.84 x 108 m)2 = 1.67 x 10 −3 N = 0.00167 N Gravitational force due to moon on earth could be ignored
9. 9. Cavendish and the Value of GThe value of G was not experimentally determined until nearlya century later (1798) by Lord Henry Cavendish using atorsion balance. Cavendishs apparatus involved a light, rigid rod about 2-feet long. Two small lead spheres were attached to the ends of the rod and the rod was suspended by a thin wire. When the rod becomes twisted, the torsion of the wire begins to exert a torsional force which is proportional to the angle of rotation of the rod. The moretwist of the wire, the more the system pushes backwards torestore itself towards the original position. Cavendish hadcalibrated his instrument to determine the relationshipbetween the angle of rotation and the amount of torsionalforce.Cavendish then brought two large lead spheres near thesmaller spheres attached to the rod. Since all masses attract,the large spheres exerted a gravitational force upon thesmaller spheres and twisted the rod a measurable amount.Once the torsional force balanced the gravitational force, therod and spheres came to rest and Cavendish was able todetermine the gravitational force of attraction between themasses.By measuring m1, m2, d and Fgrav, the value of G could bedetermined. Cavendishs measurements resulted in anexperimentally determined value of 6.75 x 10-11 N m2/kg2.Today, the currently accepted value is 6.67259 x 10-11 N m2/kg2.
10. 10. GRAVITATIONAL ACCELERATIONSymbolized by letter g, gravitational acceleration is themagnitude of gravitational field F Mm F =mg → g= ; F =G m r2 Mm G g= r2 = G M m r2where F = gravitational force m = mass of test object M = mass of source object g = gravitational field strength (gravitational acceleration) r = distance of a point to the object sourceThe value of gravitationalacceleration g is dependon the location of theobject.If M is mass of earth,then r is the distance ofobject from the center ofthe earth (the location ofobject)
11. 11. WEIGHT: Acceleration Due to GravityWEIGHT is a measurement of the force on a object caused bygravity trying to pull the object down. The weigth of anobject of on the earth is the gravitational force that earthexerts on the object.- The weight always acts downward, toward the center of the earth- On another astronomical body, the weight is the gravitational force exerted on the object by that body (appear as the consequence of Newton’s III law)- Weight is opposite to Gravitational ForceAn object has weight whether or not it is resting on theearth’s surfaceGravitational force is acting even when the distance r is muchbigger than the radius of the earth R Mearth mobject W =G r2 Mearth W =G mobject r2 W =mg
12. 12. PROBLEM SOLVED The mass of Hubble Space Telescope is 11600 kg. Determine the weight of the telescope (a) when it was resting on the earth and (b) as it is in its orbit 598 km above the earth (R-earth = 6380 km) SOLUTION ME m(a) W =G r2 (5.98 x 1024 kg)(11600 kg) = (6.67 x 10 −11 N.m2 /kg2 ) 6 2 = 1.14 x 105 N (6.38 x 10 m)(b) as in (a) but with r = 6.38 x 106 m + 598 x 103 m = 6.98 x 106 m W = 0.950 x 105 NPotential Energy and Potential GravitiesPotential Energy Gravity (Ep) is a work that needed todisplace an object from one position to infinity. Mearth mobject Ep = − G r Potential Energy GravityPotential Gravity (V) = object s mass Mearth V = −G r
13. 13. The Law of Energy Conservation on GravityThe total energy is E = E p + Ek F = ma Mm v2 M Mm G =m → v2 = G = −G + mv 2 2 1 2 r r r r Mm 1 Mm E = −G 2 + G Mm r 2 r = −G Mm 2r = −G 2rThe conservation energy is E p (1) + E k (1) = E p (2) + E k (2) Mm 1 2 Mm 1 2 −G + 2 mv1 = − G + 2 mv2 r1 r2Escape (Drift) velocity (vesc) E = E p + Ek Mm 1 2 0 = −G + 2 mv esc M R v esc = 2 G → v esc = 2 gR Mm R 1 mv 2 2 esc = G R 2 M v esc = 2 G R M v esc = 2 G R
14. 14. KEPPLER’S LAW AND NEWTON’S SYNTHESIS F = m as Mm G = m as → v = ωR R2 M v2 2π 4π 2 G 2 = = R → v 2 = 2 R2 R R T T M M 4π 2 G = v2 → G = 2 R2 R R T M 1 G = 2 R3 4π 2 T T2 4π 2 T2 = → =k R3 GM R3Keppler’s law state that ratio of square any planet’s revolutionperiod revolves the sun and triple rank of average distancebetween planet and sun is always constantKEPPLER’S LAWS OF PLANETARY MOTION First Law: The path of the planets are ellips with the center of the sun at one focus (The Law of Ellips)Second Law: An imaginary line from the sun sweeps out equal areas in equal time intervals. Thus, planets move fastest when closest to the sun, slowest when farthest away(The Law of Equals Areas)Third Law: The ratio of the squares of the periods of any two planets revolving about the sun is equal to the ratio of the cubes of their average distance from the sun(The Law of Harmonies)
15. 15. Exercises[1] Determines the gravitational force that works on a satellite (mass ms) while is orbiting the earth (mass me) in the position of height earth radius (re) from the surface[2] Three masses of objects (each has mass of m) are on the corner of a triangle which side s. Determines the gravitational force on each mass.[3] Calculate the gravitational field force on earth’s surface[4] On height h from earth’s surface the gravitational field force is known to be equal to half the gravitational field force of earth’s surface. Define the value of h in the term of earth radius (re)[5] On a point beyond earth’s surface it is known that the potential gravity is −5.12 x 107 J/kg and gravitational earth acceleration is 6.4 m/s2. If the earth radius is 6400 km, calculate the height of such point from earth surface.[6] By what minimal velocity required is in order a bullet (mass m) which fired from earth surface could reach a height of R? (R = earth radius)[7] A satellite is orbiting earth in a circle orbital. Determine the satellite periode when is (a) orbiting exactly on earth’s surface. (b) orbiting on the height of h above the earth surface