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Chapter 12 apple In chapter 2 we saw that close to the surface of the earth the gravitational force Fg is: a. Constant in magnitude Fg = mg and b. Is directed towards the center of the earth mg C . earth m1 m2 F12 F21 r Gravitation In this chapter we will give the general form of the gravitational force between two masses m1 and m2 . This force explains in detail the motion of the planets around the sun and of all celestial bodies. This is a law truly at a cosmic level (12-1) Final Exam: Friday 5/2/03 8:00-11:00 am Rm. 215 NSC 3 long problems from chapters 9,10,11,12, and 13 6 mini problems from chapters 1-13 Bring with you: 2 pens 1 page with equations (both sides) Your ID Celestial objects are divided into two categories: 1. Stars They have fixed positions with respect to each other That is the reason why we can group hem into constellations that maintain the same shape 2. Planets They follow complicated paths among the stars An example is given in the picture. Note: All stars rotate around the star polaris every 24 hours (12-2) Rotation Axis of the Celestial sphere Polaris Star N earth S Celestial sphere Geocentric System 1. The earth is at the center of the universe 2. The stars are fixed on a sphere (known as the “celestial sphere”) which rotates about its axis every 24 hours. The axis connects the center of the earth with the star polaris. All stars move on circular paths on the celestial sphere and complete a rotation every 24 hours (12-3) Ptolemaic System Ptolemy in the 2nd century AD used the following geocentric model to describe the complicated motions of the planets. The planets and the sun move on small circular paths called the epicycles. The centers of the epicycles move around the earth on larger circles called deferents Ptolemy’s system gives a reasonable description of the motion of the planets and it was accepted for 1400 years (12-4) The Heliocentric System In 1543 Copernicus introduced the heliocentric system. According to this scheme the sun is at the center of the solar system. The planets and the earth rotate about the sun on circular orbits. The immobility of the stars was ascribed to their great distance. The heliocentric system was not accepted for almost a century (12-5) Nicolaus Copernicus (1473-1543) He discussed his ideas in the book: De Revolutionibus Orbitum Coelestium published after his death Thycho Brache in Denmark constructed sophisticated astronomical instruments and studied in detail the motion of the planets with a accuracy of ½ minute (1 minute = 1/60 degree) Brache died in 1601 before he had a chance to analyze his data. This task was carried out by his assistant Johannes Kepler for the next 20 years. His conclusions are summarized in the form of three laws that bear his name (Kepler’s laws) (12-6) t A A t Kepler’s first law Planets move on elliptical paths (orbits) with the sun at one focus (12-7) Kepler’s second law During equal time intervals the vector r that points from the sun to a planet sweeps equal areas A constant t Kepler’s third law (12-8) If T is the time that it takes for a planet to complete one revolution around the Sun and R is half the major axis of the ellipse then: P1 R1 Sun T2 C 3 R R2 C is constant for all planets of the solar system If the planet moves on a circular path then R is simply the orbit radius. For the two planets in the figure Kepler’s third law can be written as: 2 2 P2 T1 T2 3 3 R1 R2 Isaac Newton: Newton had formulated his three laws of mechanics. It was natural to check and see if they apply beyond the earth and be able to explain the motion of the planets around the sun. Newton after studying Kepler’s laws came to the following conclusions: 1. Kepler’s second law implies that the attractive force F exerted by the sun on the planets is a central force 2. If he assumed that the magnitude of the attractive force F has the form: k F 2 r (k is a constant) and applied Newton’s second law and calculus (which he had also discovered) he got orbits that are conic sections k T2 C 3. A force F 2 gave Kepler’s third law: 3 r R (12-9) Conic Section is one of the four possible curves (circle, ellipse, parabola, hyperbola) we get when we cut the surface of a cone with a plane, as shown in the figure below (12-10) v Kepler’s third law for circular orbits P m F M Sun R k mv 2 F 2 (eqs.1) , F (eqs.2) R R Eliminate F from eqs.1 and eqs.2 2 2 k 2 R 4 R v2 (eqs.3) Period T T2 mR v v2 2 3 4 Rm 2 Substitute v from eqs.3 into the last equation: T k T 2 4 2 m constant C Note: We know that C 3 k R does not depend on m. The answer lies in the constant k mv 2 k 2 R R (12-11) Newton’s Law of Gravitational Attraction m1 m2 F12 F21 Gm1m2 F12 F21 r2 r Two masses m1 and m 2 placed at a distance r exert an attractive force on each other. This force is known as the "gravitational force" and has the following characteristics: 1. F12 and F21 act along the line that connects m1 and m 2 Gm1 m2 2. F12 F21 r2 k Previously we wrote F 2 r G is the gravitational constant Thus k Gm1m2 (12-12) v Revisit Kepler’s third law P m F M Sun R k F 2 (eqs.1) R k GmM (eqs.3) We substitute k from eqs.3 into eqs.4 GmM F (eqs.2) 2 R T 2 4 2 m C (eqs.4) 3 k R T 2 4 2 m 4 2 C 3 GmM GM R T2 Thus C does not contain m. The ratio 3 does not depend R on the planet mass m. Instead it depends on the sun mass M and is the same for all the planets of the solar system. (12-13) m1 m2 F12 F21 Gm1m2 F12 F21 r2 r The gravitational constant G was measured in 1798 by Henry Cavendish. He used a balance with a quartz fiber. In order to twist a quartz fiber by an angle one has to exert a torque = c (this is very similar to the spring force F = kx). The constant c can be determined easily (12-14) Cavendish experiment: Two small masses m are placed on either arm of a quartz fiber balance. Two larger masses M are placed at a distance r from the smaller masses. The gravitational force F GmM between M and m is: F r2 The gravitational force on the two smaller masses create a torque GmM F . This torque 2 r twists the fiber by an angle given by: = c . (12-15) r F /2 /2 Fr c r 2 G mM apple M C m In his experiment Cavendish measured the mass mg of the earth! Gravitational force on apple = mg . earth R mMG Gravitational force on apple = 2 R mMG mg Solve for M 2 R gR 2 (12-16) M = 5.96 1024 kg G How does one measure the radius of the earth? This was already done by Cavendish's time by as librarian in Alexandria called Eratosthenes (around 200 BC). Eratosthenes knew that ot a particular day every year sunlight reached the bottom of a very deep well in Syene (modern Aswan). He also knew the distance between Alexandria and Syene. From this information he was able to determine R Sun’s rays Sun’s rays s (12-17) Alexandria Syene h Erathosthene’s stick well R R Ground in Alexandria C The distance s between Alexandria and Syene s R . Here is the angle between the sun's rays and the vertical in Alexandria at the time when the sun's rays in Syene reach the bottom of the well. The angle was determined using Eratosthene's walking stick and the length of the shadow it cast. tan h . He got 7 A m M m R C B Variation of g with height h Gravitational force at point A: mMG mg (h) (eqs.1) 2 ( R h) Gravitational force at point B: mMG mg (0) (eqs.2) 2 R Divide eqs.1 by eqs.2 mMG mg (h) ( R h) 2 R2 mMG mg (0) ( R h) 2 R2 g ( h) h 1 g (0) R 2 (12-18) m A 2 C B R m g ( h) h h 1 1 g (0) R R (1 x)2 1 2 x for x 1 2h g (h) g (0) 1 R Note: As h increases, g(h) decreases (12-19) m Gravitational Potential close to the surface of the earth mg h U=0 floor In chapter 7 we saw that the potential U of the gravitational force close to the surface of the earth is: U = mgh Note 1: Close to the surface of the earth the gravitational force is constant and equal to mg Note 2: The point at which U = 0 can be chosen arbitrarily we will now remove the restriction that m is close to the surface of the earth and determine U (12-20) M M m r F . . O Gravitational Potential U (12-21) m x x-axis path dx r mMG F ( x) 2 x U (r ) U () F ( x)dx r r mMGdx dx U (r ) mMG 2 2 x x r mMG 1 U (r ) mMG r x dx 1 x2 x Take U () 0 mMG U (r ) r Escape velocity is the minimum speed with which we must launch an object from the surface of the earth so that it leaves the earth for ever After Before m m r= M R M ve v=0 mve2 mMG mMG Ei , Ef 0 , Ei E f 2 R r mve2 mMG = ve 2 MGR = 11 km/s The escape velocity 2 R does not depend on m! All objects large or small must be launched at the same speed (11 km/s) to escape from the gravitation of the earth (12-22) Example (12-2) page 327 An object of mass m moves on a circular orbit of radius r around a planet of mass M. Calculate the energy E = K + U mMG F (eqs.1) 2 r mv 2 F r (eqs.2) Eliminate F mv 2 mMG between eqs.1 and eqs.2. r r2 MG v r 2 mMG mv 2 m MG U , K r 2 2 r mMG mMG E U K r 2r mMG E 2r (12-23) Gravitational force between extended spherical objects M r . C R m As long as the small sphere is outside the larger sphere the force between them is: GmM F r2 If the small sphere is inside the larger sphere the force M' . C (12-24) r m between them is given by: GmM F r2 M is the mass inside the dotted line. The surrounding shell sxerts zero force! M m r GmM F r2 M r m m M F 0 F 0 If m is outside the shell the gravitational force F is as if all the mass M of the shell is concentrated at its center and all the mass m of the sphere is concentrated at it center If m is anywhere inside the shell then the gravitational force between the shell and the sphere is zero ! Special case: The sphere and the shell are co-centric (12-25)