Gravitation Kepler s Laws

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1 Gravitation Kepler s Laws Lana heridan De Anza College Mar 15, 2015

2 Overview Newton s Law of Universal Gravitation Gravitational field Kepler s Laws

3 Gravitation The force that massive objects exert on one another. Newton s Law of Universal Gravitation F G = Gm 1m 2 r 2 for two objects, masses m 1 and m 2 at a distance r. G = Nm 2 kg 2.

4 r to generate a value for G., each of mass m, fixed to the er Gravitation or thin metal wire as illusf mass M, are placed near the larger spheres causes the rod brium orientation. The angle eam reflected from a mirror.1 is often referred to as an e varies as the inverse square examples of this type of force in vector form by defining a rected from particle 1 toward is (13.3) 5 k/x, where k is a constant. A direct pro- Figure 13.1 Cavendish apparatus for measuring gravitational forces. F 21 m 1 Consistent with Newton s third law, F 21 F 12. rˆ12 r F 12 m 2 Figure 13.2 The gravitational force between two particles is attractive. The unit vector r^12 is directed from particle 1 toward particle 2. r 2 ˆr 1 2 F G,1 2 = Gm 1m 2 for two objects, masses m 1 and m 2 at a distance r. G = Nm 2 kg 2.

5 Gravitational Potential Energy Remember from Chapter 7, F x = du dx ; U = F (x) dx ince W = F (r) dr, this tells us that the work done by gravity on an object is equal to minus the change in potential energy.

6 Gravitational Potential Energy Remember from Chapter 7, F x = du dx ; U = F (x) dx ince W = F (r) dr, this tells us that the work done by gravity on an object is equal to minus the change in potential energy. rf U = r i rf F(r) dr = Gm 1m 2 r i r 2 dr ( 1 = Gm 1 m 2 1 ) r f r i

7 Gravitational Potential Energy It is useful to pick a reference point to set the scale for gravitational potential energy. What would be a good point?

8 Gravitational Potential Energy It is useful to pick a reference point to set the scale for gravitational potential energy. What would be a good point? Infinite distance! For r i =, U(r i ) = 0. Then we can define: U(r) = Gm 1m 2 r This will always be a negative number.

9 Gravitational Potential Energy (13.14) center of the Earth particles inside the U is always negative th system, a similar is, the gravitational es m 1 and m 2 sepa- (13.15) for any pair of par- 1/r 2. Furthermore, and we have chosen finite. Because the do positive work to y the external agent Gravitational potential energy of the Earth particle U(r) = Gm system 1m 2 r O GM E m R E U M E Earth The potential energy goes to zero as r approaches infinity. R E Figure Graph of the grav- r

10 Acceleration due to Gravity This force in that it gives objects weight, F g. For an object of mass m near the surface of the Earth: and where F g = mg g = GM E R 2 E M E = kg is the mass of the Earth and R E = m is the radius of the Earth. The force F g acts downwards towards the center of the Earth.

11 Acceleration due to Gravity The acceleration due to gravity, g, can vary with height! F G = GM ( ) Em GME r 2 = m r 2 = mg Depends on r the distance from the center of the Earth. uppose an object is at height h above the surface of the Earth, then: g decreases as h increases. g = GM E (R E + h) 2

12 Acceleration due to Gravity The acceleration due to gravity, g, can vary with height! F G = GM ( ) Em GME r 2 = m r 2 = mg Depends on r the distance from the center of the Earth. uppose an object is at height h above the surface of the Earth, then: g decreases as h increases. g = GM E (R E + h) 2 g is the not just the acceleration due to gravity, but also the magnitude of the gravitational field.

13 Acceleration due to Gravity This force in that it gives objects weight, F g. For an object of mass m near the surface of the Earth: and where F g = mg g = GM E R 2 E M E = kg is the mass of the Earth and R E = m is the radius of the Earth. The force F g acts downwards towards the center of the Earth.

14 Acceleration due to Gravity The acceleration due to gravity, g, can vary with height! F G = GM ( ) Em GME r 2 = m r 2 = mg Depends on r the distance from the center of the Earth. uppose an object is at height h above the surface of the Earth, then: g decreases as h increases. g = GM E (R E + h) 2

15 Acceleration due to Gravity The acceleration due to gravity, g, can vary with height! F G = GM ( ) Em GME r 2 = m r 2 = mg Depends on r the distance from the center of the Earth. uppose an object is at height h above the surface of the Earth, then: g decreases as h increases. g = GM E (R E + h) 2 g is the not just the acceleration due to gravity, but also the magnitude of the gravitational field.

16 Fields field A field is any kind of physical quantity that has values specified at every point in space and time.

17 Fields field A field is any kind of physical quantity that has values specified at every point in space and time. Fields were first introduced as a calculation tool. A force-field can be used to identify the force a particular particle will feel at a certain point in space and time based on the other objects in its environment that it will interact with. We do not need a description of the sources of the field to describe what their effect is on our particle. Gravitational force: Electrostatic force: F G = m( GMˆr r 2 ) = mg F E = qe

18 Fields Gravitational force: Electrostatic force: F G = m( GMˆr r 2 ) = mg F E = qe Gravitational field: g = F G m Electric field: E = F E q The field tells us what force a test particle of mass m (in the gravitational case) or charge q (in the electrostatic case) would feel at that point in space and time.

19 e perpendicular to the electric field lines passing through them. t the end of ection 25.2, the equipotential surfaces associated ectric Examples field consist of of Fields a family of planes perpendicular to the 25.11a shows some representative equipotential surfaces for this ld produced f charge Fields are drawn with lines showing the direction of force that a test particle will feel at that point. The density of the lines at that point in the diagram indicates the approximate magnitude of the A spherically symmetric electric field produced by a point charge force at that point. An electric field produced by an electric dipole tential surfaces (the dashed blue lines are intersections of these surfaces with the page) and elecq E b c

20 Examples of Fields The gravitational field caused by the un-earth system looks something like: 1 Figure from

21 nce of the field and measure oting Gravitational the force exerted Field of on the it. Earth ject (in this case, the Earth) rce that would be present if a a b Near the surface of the Earth: le in a field analysis model. in an area of space in which d a property of the particle, l version of the particle in a ational, and the property of mass m. The mathematical icle in a field model is Equa- (5.5) e particle in a field model. In lts in a force is electric charge: ure 13.4 (a) The gravitational were placed in the field. The magnitude of the field vector at any location is the magnitude of the free-fall acceleration at that location. Farther out from the Earth: a

22 Gravitational Field of the Earth a Uniform g: e particle establishes a gravitad by measuring the force on a e a particle of mass m is placed at it experiences a gravitational b (5.5) gure 13.4 (a) The gravitational s ld theory vectors of gravitation in the in vicinity of a g m F g mg continued A test mass m experiences a force F g = mg, where g is the field vector.

23 Gravitational Potential (Not gravitational potential energy!) We can define a new quantity gravitational potential. Usually written Φ or V. The change in gravitational potential is equal to the integral of the field along a path (with a minus sign). Φ = g ds (1) Notice: this is very similar to what we had for the relation between force and potential energy: U = F ds (2) In fact, eq (2) = m eq (1)

24 Gravitational Potential Φ = g ds and so, the radial component of g can be found by: g r = dφ dr For the gravitational field around a point-like mass M, g = GM r 2 ˆr, Φ = GM r F = GMm r 2 ˆr, U = GMm r

25 Gravitational Potential 1 Figure from

26 Gravitational Potential A uniform field, as near the surface of the Earth. (a) OTENTIAL + (b) Equipotential surface Field line The blue lines represent the gravitational field. The orange dashed lines are surfaces of equal gravitational potential.

27 Motion of the Planets The planets in our solar system orbit the un. (As planets in other systems orbit their stars.) This is called a heliocentric model.

28 Motion of the Planets The planets in our solar system orbit the un. (As planets in other systems orbit their stars.) This is called a heliocentric model. Nicolaus Copernicus (early 1500s A.D.) is credited with the paradigm since he developed a mathematical model and took seriously the idea that the implication was that the Earth moved around the un, but others had similar thoughts: Aristarchus of amos (c. 270 BCE) Martianus Capella (400s A.D.) Aryabhata (500s A.D.), Nilakantha omayaji (1500s A.D.) Najm al-dīn al-qazwīnī al-kātibī (1200s A.D.)

29 Motion of the Planets After Copernicus s proposal, Tycho Brahe gathered a lot of data about the positions of stars and planets. Johannes Kepler inherited Brahe s data and did the calculations to deduce a complete model. Galileo gathered additional data that supported the heliocentric model and popularized it.

30 Kepler s Laws Kepler s Three laws give simple rules for predicting stable planetary orbits. 1. All planets move in elliptical orbits with the un at one focus. 2. The radius vector drawn from the un to a planet sweeps out equal areas in equal time intervals. 3. The square of the orbital period of any planet is proportional to the cube of the semimajor axis of the elliptical orbit.

31 is Kepler s the semicenter of the First Law: axis has Elliptical length b. Orbits Defining an ellipse: un, the un y the general re zero. The ection com-, c increases correspond or an ellipse length a, and the semiminor r 1 r 2 F 1 c a F 2 b x eccentricity er hand, the Figure 13.6a cury s orbit. rom a circle, a is the semimajor Each axis focus is located a 2 at = a b 2 + c 2 b is the semiminordistance axis c from the center. c is the distance from the center of the ellipse to the focus e is the eccentricity of the ellipse e = c/a Figure 13.5 Plot of an ellipse.

32 Kepler s First Law: Elliptical Orbits All planets move in elliptical orbits with the un at one focus. er 13 Universal Gravitation pe of h has ) the e of the shape comet ger The un is located at a focus of the ellipse. There is nothing physical located at the center (the black dot) or the other focus (the blue dot). un Orbit of Comet Halley un Center Orbit of Mercury Comet Halley Center a b The planets orbits are close to circular. (Mercury s is the between the planet and the un is a 1 c. At this point, called the aphelion, the planet is at its maximum distance from the un. (For an object in orbit around the least circular.) Earth, this point is called the apogee.) Conversely, when the planet is at the right end of the Halley s ellipse, the Comet distance hasbetween an orbit the planet with and a high the un eccentricity. is a 2 c. At this point, called the perihelion (for an Earth orbit, the perigee), the planet is at its minimum distance from the un. Kepler s first law is a direct result of the inverse-square nature of the gravita-

33 tional force. Circular M p and elliptical the gravitational force center. Thes K that move repeatedly around the un F are also unbound objects, g v a such as el The radius vector drawn from the the unmun to a once planetand sweeps then out never return to equal areas in equal time intervals. these objects also varies as the inve ta allowed paths for these objects inclu What does it mean? ra a th Kepler s econd Law: Equal Areas in Equal Time M p Kepler s econd Law Kepler s second law can d r be v dt shown a r un F g v angular un momentum. Consider a p elliptical orbit (Fig. 13.7a). Let s con M to be so much more da massive than th tational force exerted by the un Eo radius vector, directed toward the a this central force about an axis thro The area swept out by r in When the planet is closer to the un, it Therefore, must be moving because faster. the external a time interval dt is half the d r v dt an isolated area of system the parallelogram. for angular mom va r un planet is a constant of the motion: th K

34 un angular momentum. Consider a plan elliptical orbit (Fig. 13.7a). Let s consi to be so much more massive than the tational beige area, force da? exerted by the un on radius vector, directed toward the un this central da force = 1 about r dr an axis throu 2 Therefore, because the external t an isolated system = 1 r for vangular dt 2 mome planet is a constant of the motion: Kepler s econd Law: Equal Areas in Equal Time M a un b un M a r F g da d r dt Figure 13.7 (a) The gravitational force acting on a planet The area swept out by r in a time interval dt is half the area of the parallelogram. un F g M p da v v d r dt r v v are also unbound objects, such as a met the un once and then never return. T these objects also varies as the inverse allowed paths for these objects include Kepler s econd Law Kepler s second law can be shown to angular momentum. Consider a plan elliptical orbit (Fig. 13.7a). Let s consid to be so much more massive than the p tational force exerted by the un on t radius vector, directed toward the un this central force about an axis throug Therefore, because the external to an isolated system for angular mome planet is a constant of the motion: Evaluating L for the planet, D L 5 0 L 5 r 3 p 5 M p r We can relate this result to the follo val dt, the radius vector r in Figure 13 the area 0 r 3 d r 0 of the parallelogr the displacement of the planet in the Evaluating L for the planet, D L 5 0 da r 3 d r r 3 p 5 M p r L

35 un angular momentum. Consider a plan elliptical orbit (Fig. 13.7a). Let s consi to be so much more massive than the tational beige area, force da? exerted by the un on radius vector, directed toward the un this central da force = 1 about r dr an axis throu 2 Therefore, because the external t an isolated system = 1 r for vangular dt 2 mome planet is a constant of the motion: Kepler s econd Law: Equal Areas in Equal Time M a un b un M a r F g da d r dt Figure 13.7 (a) The gravitational force acting on a planet The area swept out by r in a time interval dt is half the area of the parallelogram. un F g M p da v v d r dt r v v are also unbound objects, such as a met the un once and then never return. T these objects also varies as the inverse allowed paths for these objects include Kepler s econd Law Kepler s second law can be shown to angular momentum. Consider a plan elliptical orbit (Fig. 13.7a). Let s consid to be so much more massive than the p tational force exerted by the un on t radius vector, directed toward the un this central force about an axis throug Therefore, because the external to an isolated system for angular mome planet is a constant of the motion: D L 5 0 r F g τ Evaluating ext = 0 L = const. L for the planet, L = r p L 5 r 3 p 5 M p r L = M p r v We can relate this result to the follo val dt, the radius vector r in Figure 13 the area 0 r 3 d r 0 of the parallelogr the displacement of the planet in the Evaluating L for the planet, D L 5 0 da r 3 d r r 3 p 5 M p r L

36 un angular momentum. Consider a plan elliptical orbit (Fig. 13.7a). Let s consi to be so much more massive than the tational beige area, force da? exerted by the un on radius vector, directed toward the un this central da force = 1 about r dr an axis throu 2 Therefore, because the external t an isolated system = 1 r for vangular dt 2 mome planet is a constant of the motion: Kepler s econd Law: Equal Areas in Equal Time M a un b un M a r F g da d r dt Figure 13.7 (a) The gravitational force acting on a planet The area swept out by r in a time interval dt is half the area of the parallelogram. un F g M p da v v d r dt r v v are also unbound objects, such as a met the un once and then never return. T these objects also varies as the inverse allowed paths for these objects include Kepler s econd Law Kepler s second law can be shown to angular momentum. Consider a plan elliptical orbit (Fig. 13.7a). Let s consid to be so much more massive than the p tational force exerted by the un on t radius vector, directed toward the un this central force about an axis throug Therefore, because the external to an isolated system for angular mome planet is a constant of the motion: D L 5 0 r F g τ Evaluating ext = 0 L = const. L for the planet, L = r p L 5 r 3 p 5 M p r L = M p r v We can relate this result to the follo val dt, the radius vector r in Figure 13 the area 0 da r 3 d r 0 of the parallelogr the displacement dt = L of 2M the p planet in the Evaluating L for the planet, D L 5 0 da r 3 d r r 3 p 5 M p r L

37 un angular momentum. Consider a plan elliptical orbit (Fig. 13.7a). Let s consi to be so much more massive than the tational beige area, force da? exerted by the un on radius vector, directed toward the un this central da force = 1 about r dr an axis throu 2 Therefore, because the external t an isolated system = 1 r for vangular dt 2 mome planet is a constant of the motion: Kepler s econd Law: Equal Areas in Equal Time M a un b un M a r F g da d r dt Figure 13.7 (a) The gravitational force acting on a planet The area swept out by r in a time interval dt is half the area of the parallelogram. un F g M p da v v d r dt r v v are also unbound objects, such as a met the un once and then never return. T these objects also varies as the inverse allowed paths for these objects include Kepler s econd Law Kepler s second law can be shown to angular momentum. Consider a plan elliptical orbit (Fig. 13.7a). Let s consid to be so much more massive than the p tational force exerted by the un on t radius vector, directed toward the un this central force about an axis throug Therefore, because the external to an isolated system for angular mome planet is a constant of the motion: D L 5 0 r F g τ Evaluating ext = 0 L = const. L for the planet, L = r p L 5 r 3 p 5 M p r L = M p r v We can relate this result to the follo val dt, the radius vector r in Figure 13 the area 0 da r 3 d r 0 of the parallelogr the displacement dt = L of 2M the p planet in the da dt = constant! da r 3 d r 0 5 Evaluating L for the planet, L D L r 3 p 5 M p r

38 Kepler s Third Law: T 2 a 3 The square of the orbital period of any planet is proportional to the cube of the semimajor axis of the elliptical orbit. We will only prove this for circular orbits. For circular motion: v = 2πr T. F G = F C GM s M p r 2 = M pv 2 r GM s M p r 2 = M p r ( 4π T 2 2 = ( T 2πr GM s ) 2 ) r 3

39 Kepler s Third Law: T 2 a 3 A full derivation for the elliptical orbit case gives the same expression, but with r replaced with a: ( 4π T 2 2 = GM s ) a 3 ometimes the constant is given a name: K s = 4π2 GM s = s 2 m 3

40 ummary Gravitational force Gravitational field Kepler s Laws 4th Collected Homework! due Friday. (Uncollected) Homework erway & Jewett, Ch 13, onward from page 410. Questions: ection Qs 3, 9, 11, 15, 19

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