General Physics Chapter 2: Planetary Motion and Gravitation PDF
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This document covers planetary motion and gravitation, exploring the behavior of objects in rotating bodies and simple kinematics relating to rotational motion, the behavior of attraction between objects with mass experiencing linear acceleration, and elliptical planetary orbits around the sun. It includes discussion of scenarios and examples.
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Chapter 2 Planetary Motion and Gravitation In the previous chapter, we discussed the different characteristics of rotating bodies, the behavior of objects upon rotating bodies, and simple kinematics regarding rotational motion. In this chap...
Chapter 2 Planetary Motion and Gravitation In the previous chapter, we discussed the different characteristics of rotating bodies, the behavior of objects upon rotating bodies, and simple kinematics regarding rotational motion. In this chapter, we will discuss the celestial applicability of rotational kinematics. Particularly, the behavior of attraction betweentwoobjectswithmassexperiencinglinearaccelerationandtheelliptical orbit of our planets around the sun. Before proceeding to this chapter, please take note of the following reminders: 1. The universal gravitational constant 𝐺 is completely different with the acceleration due to gravity𝑔 2. In the formula for universal lawofgravitation, 𝑟 isthedistancebetween themass concentrationof the objects, not their surfaces 3. An increase in𝑟 will rapidly decrease the gravitational force 4. Thegravitationalforceexperiencedbytwoobjectsareequalregardlessof mass 5. The“orbitalareas”inKepler’s2ndlawisthetheareathattheplanethas swept (the lines connected to the planet and to the sun) and not the circumference of the ellipse 6. Perihelion and aphelion are actually distances from the centers of mass between two celestial bodies Chapter 2.1. Newton’s Law of Universal Gravitation SCENARIO 1 Haveyoueverwonderedwhythemoonwon'tcrashuponusbutinstead,itkeeps an almost steady orbit around the Earth? The same can also be said upon the planets in our solar system. Figure 17. Planets in their respective orbit As discussedinChapter1.4,thereisacentripetalforceactingupontheplanets that’s why the moon keeps asteadyorbitaroundEarth.Well,analmoststeady one since Earth is not a perfect sphere. This centripetal force, according to Newton, was somewhat equivalent to the gravitational force experienced by fallingobjectsonEarth.Becauseofthis,Newtonstatedthattheplanetswerein astateoffreefallaroundthesunbut,insteadofcrashinguponthem,theyarein orbit around the sun. Newton then strengthened this finding by conducting a thought experiment. Thisexperimentinvolvedacannonballlaunchedintheairatanincrediblylarge forcethatitperfectlycounteractsgravity->theobject’sspeedisequalorgreater than the curvature speed of Earth. Since it perfectly counteracts gravity, the cannonball will orbit around the planet -> it will not hitthegroundanditwill enter a continuous state of free fall unless frictional forces (air drag) are present. Figure 18. A cannonball in orbit SCENARIO 2 Imagine an apple with a certain distance from Earth. Obviously, Earth will attract the apple but why is it that the apple won’t attract Earth? Figure 19. Gravitation attraction between an apple and the Earth Remember that the apple and Earth exerts an equal gravitational force upon each other which complies to Newton’s Third Law. Given equal exerted gravitational forces, the apple is being pulled because it experiences greater acceleration and this can be proved mathematically. Assumethattheapplehasmass 1. 24𝑘𝑔 andthedistancebetweentheirmasses is 200𝑘𝑚. The gravitational force between them would be approximately . Using this force, we can calculate the acceleration of both objects 12348. 3𝑁 using the universal formula for force𝐹 = 𝑚𝑎. 12348. 3𝑁 = 1. 24𝑘𝑔(𝑎) [Apple] 2 𝑎 = 9958. 31𝑚/𝑠 24 12348. 3𝑁 = 5. 972×10 𝑘𝑔(𝑎) [Earth] −21 2 𝑎 = 2. 07 × 10 𝑚/𝑠 2 TheappleacceleratesontoEarthby9958. 31𝑚 /𝑠 andtheEarthacceleratesonto −21 2 theappleby2. 07 × 10 𝑚/𝑠 whichisincrediblysmallandisoftenanegligible quantity. The gravitational force between two objects with mass 𝑚1 and 𝑚2 respectively and their masses are𝑟 meters apart can be calculated as follows: 𝑚1𝑚 𝐹𝑔 = 𝐺 2 2 [2.1] 𝑟 Figure 20. Parts of Gravitational Attraction between two objects with mass Newton’s formula however, was not complete since the value of 𝐺 was not defined. However, Henry Cavendish completed his formula by giving the first direct measurement of gravitational attraction 71 years after Newton passed away. Cavendish provided and named the constant as the universal −11 2 2 gravitational constantwhich has a value of6. 67 × 10 𝑁 · 𝑘𝑔 /𝑚 . Example 2.1. Calculate the force of attractionbetweentwofriendswithmasses 45. 92𝑘𝑔 and48. 85𝑘𝑔 if the distance of their masses is5𝑚. Solution: A. Since the problem directly states the given values, we can already substitute them into the formula. −11 2 2 (45.92𝑘 𝑔)(48.85𝑘𝑔) 𝐹 = (6. 67 × 10 𝑁 · 𝑘𝑔 /𝑚 ) × 2 (5𝑚) −9 𝐹 = 5. 98 × 10 𝑁 −19 Therefore, the best friends exert 5. 98 × 10 𝑁 of gravitational force in each other.Yes,youhearditright!Peoplealsoexertgravitationalforceoneachother but this force is negligible because of the gravitational pull we experience on Earth. Example 2.2. Two spheres are at rest on a table. One sphere has a mass of 3 1. 92𝑘𝑔 with a volumeof 338. 37𝑚 .Theotheronehasamassof 2. 22𝑘𝑔 anda 3 volume of 401. 89𝑚 . If their surfacesare 12𝑚 apartfromeachother,calculate the gravitational force exerted by the spheres to each other. Solution: A. The value for 𝑟 is not directly given in the problem since only their surfaces are 12𝑚 apart. To determine the value of 𝑟, we need to take account the radius of each sphere which canbecalculatedbyusingtheir volumes. Recall that the formula for the volume of a sphere is 4 3 𝑉𝑠𝑝ℎ𝑒𝑟𝑒 = 3 π𝑟 . 3 For the 1st sphere (𝑉 = 338. 37𝑚 ): 3 4 3 338. 37𝑚 = 3 π𝑟 3 338.37𝑚 3 4π = 𝑟 3 𝑟 = 8. 99𝑚 3 For the 2nd sphere (𝑉 = 401. 89𝑚 ): 3 4 3 401. 89𝑚 = 3 π𝑟 3 401.89𝑚 3 4π = 𝑟 3 𝑟 = 9. 8𝑚 B. Addthesevaluestothedistancebetweenthespheres’surfacesof12𝑚 and we will obtain the distance between their masses of30. 79𝑚. C. We are now ready for substitution. Use 𝑚1 = 1. 92𝑘𝑔, 𝑚2 = 2. 22𝑘𝑔, 𝑟 = 30. 79𝑚. −11 2 2 (1.92𝑘𝑔)(2.22𝑘𝑔) 𝐹 = (6. 67 × 10 𝑁 · 𝑘𝑔 /𝑚 ) × 2 (30.79𝑚) −13 𝐹 = 3 × 10 𝑁 Chapter 2.2. Kepler’s Laws of Planetary Motion The Copernican revolution, which placed the sun as the center of the solar system,wasamajoreventduringtherenaissanceperiod.Itwasoneamongthe many testaments of scientific flourishment. One important figure in such an event was Tycho Brahe, who was known for his instruments for celestial observation, which paved the way to theCopernicanmodelofthesolarsystem. Duringhistime,Copernicusthoughtthattheplanetswereinaperfectlycircular orbit around the sun until Johannes Kepler introduced the data he gathered. Kepler discovered that the planets orbit their stars in an ellipticalmanner.An ellipse is a conic section where there are two foci and these foci, when added, sums up to a constant. Using this data, Keplerwasabletocomeupwiththree laws of planetary motion which are as follows: 1. Kepler’s 1st Law states that the orbit of every planet is anellipse with the sun acting as one of its foci. The point where the planet is closest to itssuniscalledtheperihelionandthepointwheretheplanet is furthest to its sun is called the aphelion. The distancesbetweenthe aphelionandperihelionoftheplanetsareapproximatelyequalduetothe low eccentricity of their elliptical orbits. Figure 21. Perihelion and Aphelion 2. Kepler’s 2nd Law states thataplanetsweepsoutequalareasover the ellipse at equal time intervals. To state this in a clearer way, a planet’s orbital speed depends uponitsdistancefromthesun.Itisfaster when it is nearer but it is slower when it is further. Figure 22. Demonstration of Kepler’s 2nd Law on Earth’s Orbit IfwelookatFigure22,assumingthattheshadedareas(A)areequalareas,the planetwillalsotravelfrompointsCtoDandpointsNtoMatthesameamount of time because of Kepler’s 2nd law. The planet travels faster in points C to D and the planet travels slower in points M to N. 3. Kepler’s 3rd Law states that the square of any planet’s orbital period will be proportional to the cube of the semi-major axis. Figure 23. The semi-major axis of an elliptical orbit The semi-major axis 𝑎 of an ellipse is defined as its longest radius.Incelestial context,thisismeasuredinastronomicalunits.Theorbitalperiodofaplanet𝑃, on the other hand, is measured in years. These two variables are directly proportional and it can be expressed mathematically as follows: 2 3 𝑃 = 𝑎 [2.2] Since these perihelions and aphelions for most planets are extremely precise, theirorbitalperiodsandsemi-majoraxescanbemeasuredusingtheexactsame properties of other planets. This can be expressed mathematically as follows: 𝑃 2 𝑎 3 ( 𝑃1 ) = ( 𝑎1 ) [2.3] 2 2 Kepler’s laws were a breakthrough in the scientific community becauseitgave birthtocelestialmechanics,especiallytherelativedistancesbetweentheplanets fromthesunandthemathematicalformulastodescribethemovementfromthe heavens. Example2.3.Aplanettravelsatfourdistinctpointsonitsorbitconsecutivelyas shown in the diagram below. The distance between points A to B to the sun respectively are 3. 2𝐴𝑈 and 1. 8𝐴𝑈 and the distance from C to D to the sun respectively are 1. 48𝐴𝑈 and 2. 81𝐴𝑈. The velocity of the planet from point B andpointArespectivelyare66𝑘𝑚/𝑠 and92𝑘𝑚/𝑠.Findthevelocityoftheplanet at point D assuming that areas E and F are equalandtheplanetspeedsupto 109𝑘𝑚/𝑠 at point C. Figure 24. The elliptical orbit of a planet in Example 2.3 Solution: A. We will use Kepler’s 2nd law and the Law of Conservation of Angular Momentum to solve this problem. We will first start by the problem assumesequalareasbetweenEandF.Sincethisisthecase,accordingto thelawofconservationofangularmomentum,thedistancesofthepoints to the foci𝑟 is inversely proportional to the planet’svelocity𝑣. B. Area E includes points A and B.Therefore,wecanestablishanequation that relates the𝑟 and𝑣 of the planet at pointsA and B which is as follows: 𝑟𝐴𝑣 = 𝑟𝐵𝑣 𝐴 𝐵 We can also establish for points C and D: 𝑟𝐶𝑣 = 𝑟𝐷𝑣 𝐶 𝐷 C. DuetoKepler’s2ndLaw,thetimeittakesfortheplanettotravelfromA to B is equal from C to D. Therefore, we will set the ratio between the planet’s status from A to B equal to C to D. Therefore, the established formula is: 𝑟𝐴𝑣 𝑟𝐶𝑣 𝐴 𝑟𝐵𝑣 = 𝐶 𝑟𝐷𝑣 [2.4] 𝐵 𝐷 D. Substituteallgivenvaluesfromtheproblemandyoucannowsolveforthe velocity at point D (𝑣𝑑). (3.2𝐴𝑈)(66𝑘𝑚/𝑠) (1.48𝐴𝑈)(109𝑘𝑚/𝑠 ) (1.8𝐴𝑈)(92𝑘𝑚/𝑠) = (2.81𝐴 𝑈)(𝑣𝑑) (1.48𝐴 𝑈)(109𝑘𝑚/𝑠 ) (1.8𝐴𝑈)(92𝑘𝑚/𝑠 ) 𝑣𝑑 = (3.2𝐴𝑈)(66𝑘 𝑚/𝑠 ) (2.81𝐴𝑈) 𝑣𝑑 = 45. 01𝑘𝑚/𝑠 We can see that the velocity at point D decreased because the planet is going away from the sun. 1 Example 2.4. Planet A can complete 4 of its orbit around its star by6.5years 3 while Planet B can complete 8 of its orbit at the same star by 7.7 years. In planetA,thedistancesofitsaphelionandperihelionrespectivelyare1. 9𝐴𝑈 and 3. 9𝐴𝑈. Calculate the major axis of Planet B. Solution: A. ThisproblemcanbesolvedusingKepler’s3rdLaw.Whentwoplanetsare involved and rotating at the same star, the ratio between their orbital periods and their semi-major axes can be described the the equation below: 𝑃 2 𝑎 3 ( 𝑃1 ) = ( 𝑎1 ) [2.5] 2 2 B. Sincetheprobleminvolvespartialorbits,wewouldwantfirsttosolvefor their full orbital periods which can be solved as follows: 6. 5 × 4 = 26𝑦𝑒𝑎𝑟𝑠 8 7. 7 × 3 = 20. 53𝑦𝑒𝑎𝑟𝑠 C. In planet A, the distances of its aphelion and perihelion respectivelyare 1. 9𝐴𝑈 and 3. 9𝐴𝑈.AphelionandPerihelionaddedmakeupthemajoraxis ofitsellipticalorbit.Tosolveforthesemi-majoraxis,wewilladdaphelion and perihelion and divide the result by 2. 1.9𝐴𝑈+3.9𝐴𝑈 2 = 2. 9𝐴𝑈 D. We can now substitute these values into ourformulainordertofindthe semi-major axis of Planet B first. 26 2 .9 3 2 ( 20.53 ) = ( 𝑎2 ) .9 2 26 2 𝑎2 = ( 20.53 ) 2.9 𝑎2 = 2 26 ( 20.53 ) 𝑎2 = 2. 48𝐴𝑈 Note that this value is just the semi-major axis of Planet B’s orbit. To calculate its major axis, we will multiply its semi-major axis by 2 which results in4. 96𝐴𝑈. Example2.5.PlanetCcancompleteoneorbitin2. 5 yearsarounditsstarwitha semi-majoraxisof 1. 7𝐴𝑈.IfPlanetD’smajoraxisis41%longerthanPlanetC’s major axis, calculate the orbital period of Planet D. Solution: A. ThisproblemcanbesolvedusingKepler’sThirdLawusingtheratioofthe orbital periods and the semi-major axes of the planets as discussed in formula2.5.Butfirst,wewillsolveforthelengthofthesemi-majoraxisof Planet D. B. The semi-major axis of planet D can be calculated by increasing the semi-major axis of planet C by 41%. 1. 7𝐴𝑈 × (1 + 0. 41) = 2. 397𝐴𝑈 C. Substitute the given values into formula2.5. .5 2 2 1.7𝐴 𝑈 3 ( 𝑃𝐷 ) = ( 2.397𝐴𝑈 ) .5 2 1.7𝐴𝑈 3 𝑃𝐷 = ( 2.397𝐴 𝑈 ) 2.5 𝑃𝐷 = .7𝐴𝑈 3 1 ( 2.397𝐴 𝑈 ) 𝑃𝐷 = 4. 19𝑦𝑒𝑎𝑟𝑠 This implies that Planet D completes one orbit in 4.19 years. Chapter 2.3. Satellites Have you ever wondered how “satellite” was defined during the junior years? Maybe some of us immediately associatethewordsatelliteforthoseman-made infrastructures wandering around space while observing and capturing the essenceofourplanet.Thisiswhatthisparticularsubchapterwilldiscuss->the definition of the word “satellite” and its various types. A satellite is a small object that revolves around a larger object in space. Yes, anythingsmallthatrevolvesaroundabigobjectisalreadyconsideredasatellite which is not limited upon man-made infrastructure.Anobjectrequirestobein orbit around another object to be a satellite. An orbit is the curved path of an object around another object. An example of satellites include our moon and the moons of the other planets, and the planets themselves since they are revolving around the sun. The two types of satellites arenaturalandartificialsatellites. 1. Natural satellites involve a smaller natural object revolving around a biggernaturalobject.Examplesofnaturalsatellitesincludethemoon(due to its orbit around Earth) and the planets (d ue to their orbit around the sun). ➔ Only Earth (1), Mars (2), Jupiter (67), Saturn (62), Uranus (27), and Neptune (14) have moons 2. Artificial satellites involve man-made objects that are designed to orbit around the Earth or other natural objects in space. These satellites are used to transmit, receive signals, and capture the essence of our planet and other planets. ➔ They are released into space through rockets or space shuttles. Rockets need enough fuel to generate enough thrust and speed to atleast 17,800 mph to stay in orbit around Earth. ➔ The principles of momentum and gravity can help a satellite stay into orbit ➔ The first artificial satellite is Sputnik-1 (Russia)
