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- 1. FORCES AND NEWTON’S LAW OF GRAVITATION Standard Competency: Analyze the nature phenomenon and its regularity within the scope of particle’s Mechanics Base Competency: Analyze the regularity of planetary motion within the universe base on Newton’s Law Learning 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 velocity References: [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. 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 FIELD DEFINITION: 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 out of mind if there are two object that can interact each other without being physically interact. Field, a distance space, is then “to be created” to bridge how remote objects (distance separated objects) could interacting each other. Base on Newton, FIELD is a quantification of distance 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 be visualized by an arrow which shows the magnitude and direction of grafivitational field. Representation of fields as arrangement arrows is well known as force line
- 3. GRAVITATION PHENOMENON Base on Newton’s Observations of how an apple could fall from 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 Sun Newton 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. Gravitational Force Formula Base on Newton’s Proposes, there is an attractive force which works on two different object separated in a certain distance. The force is propotional to two masses F ≈ m1 m2 The force is propotional to inverse square of the distance between those masses 1 F ≈ r2 m1 m2 m1 m2 F ≈ → F =G r2 r2 F = the attractive force G = a constant, m1 = mass of first particle UNIVERSAL GRAVITATIONAL m2 = mass of second particle CONSTANT r = linier distance of both particles = 6.67 x 10−11 N.m2/kg2 (measured by Henry Cavendish) GRAVITATION FIELD FORCE Gravitation field force is a force experience by an object due to gravitational attraction per unit mass F m g= → g=G m r2 Direction of gravitational field is goes to the center of object’s mass. If an object is under gravitational force influenced by several objects, the gravitational field force on the object is the sum of each gravitational field another objects g = (g1 )2 + (g1 )2 + 2(g1 )(g2 ) cos θ
- 5. NEWTON’S LAW OF UNIVERSAL GRAVITATION In 1666 Isaac Newton determined that the same force that kept the planets in motion must also exist between every object. 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 to calculate the magnitude of electrical force between two charged bodies. Both are inverse-square laws, in which forse is inversely proportional te the square of the distance between the bodies. Coulomb’s law has the product of two charges in place of the product of the masses, and the electrostatic constant in place of the gravitational constant. What was Newton stated is now become a Universal Law of Gravitation. It is universal because it works for all kinds, types and sizes of object. No matter how small or big the objects 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. The law of gravitation is universal and very fundamental. It can be used to understand the motions of planets and moons, determine the surface gravity of planets, and the orbital motion 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. PROBLEM SOLVED What is the magnitude of the gravitational force between the earth (m = 5.98 x 1024 kg) and a 60-kg man whose stand on 6.38 x 106 m away SOLUTION 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. PROBLEM SOLVED What is the magnitude of the gravitational force that acts on each bicycle’s tyres where m1 = 12 kg and m2 = 25 kg and 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. Cavendish and the Value of G The value of G was not experimentally determined until nearly a century later (1798) by Lord Henry Cavendish using a torsion balance. Cavendish's 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 more twist of the wire, the more the system pushes backwards to restore itself towards the original position. Cavendish had calibrated his instrument to determine the relationship between the angle of rotation and the amount of torsional force. Cavendish then brought two large lead spheres near the smaller spheres attached to the rod. Since all masses attract, the large spheres exerted a gravitational force upon the smaller spheres and twisted the rod a measurable amount. Once the torsional force balanced the gravitational force, the rod and spheres came to rest and Cavendish was able to determine the gravitational force of attraction between the masses. By measuring m1, m2, d and Fgrav, the value of G could be determined. Cavendish's measurements resulted in an experimentally 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. GRAVITATIONAL ACCELERATION Symbolized by letter g, gravitational acceleration is the magnitude of gravitational field F Mm F =mg → g= ; F =G m r2 Mm G g= r2 = G M m r2 where 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 source The value of gravitational acceleration g is depend on the location of the object. If M is mass of earth, then r is the distance of object from the center of the earth (the location of object)
- 11. WEIGHT: Acceleration Due to Gravity WEIGHT is a measurement of the force on a object caused by gravity trying to pull the object down. The weigth of an object of on the earth is the gravitational force that earth exerts 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 Force An object has weight whether or not it is resting on the earth’s surface Gravitational force is acting even when the distance r is much bigger than the radius of the earth R Mearth mobject W =G r2 Mearth W =G mobject r2 W =mg
- 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 N Potential Energy and Potential Gravities Potential Energy Gravity (Ep) is a work that needed to displace an object from one position to infinity. Mearth mobject Ep = − G r Potential Energy Gravity Potential Gravity (V) = object' s mass Mearth V = −G r
- 13. The Law of Energy Conservation on Gravity The 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 2r The 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 r2 Escape (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. 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 R3 Keppler’s law state that ratio of square any planet’s revolution period revolves the sun and triple rank of average distance between planet and sun is always constant KEPPLER’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. 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