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Newton's Law of Gravitation

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Newton's Law of Gravitation Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation Essential idea: The Newtonian idea of gravitational force acting between two spherical bodies and the laws of mechanics create a model that can be used to calculate the motion of planets. Nature of science: Laws: Newton’s law of gravitation and the laws of mechanics are the foundation for deterministic classical physics. These can be used to make predictions but do not explain why the observed phenomena exist. Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation Understandings: • Newton’s law of gravitation • Gravitational field strength Applications and skills: • Describing the relationship between gravitational force and centripetal force • Applying Newton’s law of gravitation to the motion of an object in circular orbit around a point mass • Solving problems involving gravitational force, gravitational field strength, orbital speed and orbital period • Determining the resultant gravitational field strength due to two bodies Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation Guidance: • Newton’s law of gravitation should be extended to spherical masses of uniform density by assuming that their mass is concentrated at their centre • Gravitational field strength at a point is the force per unit mass experienced by a small point mass at that point • Calculations of the resultant gravitational field strength due to two bodies will be restricted to points along the straight line joining the bodies Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation Data booklet reference: 푀푚 퐹 = 퐺 푟2 퐹 = 푚 푀 퐹 = 퐺 푟2 Theory of knowledge: • The laws of mechanics along with the law of gravitation create the deterministic nature of classical physics. Are classical physics and modern physics compatible? Do other areas of knowledge also have a similar division between classical and modern in their historical development? Topic 6: Circular motion and gravitation 6.2 – Newton’s law of gravitation Utilization: • The law of gravitation is essential in describing the motion of satellites, planets, moons and entire galaxies • Comparison to Coulomb’s law (see Physics sub-topic 5.1) Aims: • Aim 4: the theory of gravitation when combined and synthesized with the rest of the laws of mechanics allows detailed predictions about the future position and motion of planets Field (Fundamental) Forces Gravitational Force Gravitational force is cumulative and extended to infinity. The gravitational force between two object is due to the cumulative effect of billions of billions of the atoms made up both bodies. This means that the larger the body (contain more matter), the stronger the force. But on the scale of individual particles, the force is extremely small, only in the order of 10-38 times that of the strong force. Electromagnetic Force The force is long range, in principle extending over infinite distance. However, the strength can quickly diminishes due to shielding effect. Many everyday experiences such as friction and air resistance are due to this force. This is also the resistant force that we feel, for example, when pressing our palm against a wall. This is originated from the fact that no two atoms can occupy the same space. However, its strength is about 100 times weaker within the range of 1 fm, where the strong force dominates. But because there is no shielding within the nucleus, the force can be cumulative and can compete with the strong force. This competition determines the stability structure of nuclei. Weak Nuclear Force Responsible for nuclear beta decay and other similar decay processes involving fundamental particles. The range of this force is smaller than 1 fm and is 10-7 weaker than the strong force Strong Nuclear Force This force is responsible for binding of nuclei. It is the dominant one in reactions and decays of most of the fundamental particles. This force is so strong that it binds and stabilize the protons of similar charges within a nucleus. However, it is very short range. No such force will be felt beyond the order of 1 fm (femtometer or 10-15 m). One of the most significant intellectual achievements in the history of thought. It is universal – it applies to all objects regardless of their location anywhere in the Universe. Every object in the universe attracts every other object. Gravitational force between two point masses m1 and m2 is proportional to their product, and inversely proportional to the square of their separation r. mm r F = G 12 r2 G = 6.67x10-11 Nm2/ kg2 – “Universal gravitational constant” the same value anywhere in the universe - very small value – no significant forces of attraction between ordinary sized objects. The actual value of G, the universal gravitational constant, was not known until Henry Cavendish conducted a tricky experiment in 1798 to find it. Newton’s law of gravitation The earth, planets, moons, and even the sun, have many layers – kind of like an onion: In other words, NONE of the celestial bodies we observe are point masses. Given that the law is called the universal law of gravitation, how do we use it for planets and such? Newton’s law of gravitation Newton spent much time developing integral calculus to prove that “A spherically symmetric shell of mass M acts as if all of its mass is located at its center.” M -Newton’s shell theorem. m r 2 Thus F = Gm1m2 / r works not only for point masses, which have no radii, but for any spherically symmetric distribution of mass at any radius like planets and stars. r is the distance between the centers of the masses. FYI The radius of each m1 m2 F F mass is immaterial. 12 r 21 EXAMPLE: The earth has a mass of M = 5.981024 kg and the moon has a mass of m = 7.361022 kg. The mean distance between the earth and the moon is 3.82108 m. What is the gravitational force between them? SOLUTION: F = GMm / r 2. F = (6.67×10−11)(5.981024 )(7.361022 ) / (3.82108)2 F = 2.011020 N. What is the speed of the moon in its orbit about earth? 2 SOLUTION: FC = FG = mv / r . 2.011020 = ( 7.361022 ) v 2 / 3.82108 v = 1.02103 ms-1. What is the period of the moon (in days) in its orbit about earth? SOLUTION: v = 2r / T. T = 2r / v = 2( 3.82108 ) / 1.02103 = (2.35 106 s)(1 h / 3600 s)(1 d / 24 h) = 27.2 d. Gravitational field strength Suppose a mass m is located a distance r from a another mass M. The gravitational field strength g is the force per unit mass acting on m due to the presence of M. 퐹 = 푚 The units are newtons per kilogram (N kg -1 = m s -2). Suppose a mass m is located on the surface of a planet of radius R. We know that it’s weight is F = mg. But from the law of universal gravitation, the weight of m is equal to its attraction to the planet’s mass M and equals F = GMm / R 2. mg = GMm / R 2. 퐺푀 gravitational field strength at surface of = 푅2 a planet of mass M and radius R Double the distance from the centre, g is 4 times less, and so is weight PRACTICE: A 525-kg satellite is launched from the earth’s surface to a height of one earth radius above the surface. What is its weight (a) at the surface, and (b) at altitude? SOLUTION: -2 (a) AT SURFACE: gsurface = 9.8 m s . F = mg = (525)(9.8) = 5145 N. W=5100 N -2 (b) AT ALTITUDE: gsurface+R = 2.45 m s . F = mg = (525)(2.46) = 1286.25 N W=1300N Gravitational field strength Compare the gravitational force formula F = GMm / r 2 (Force) with the gravitational field formula g = GM / r 2 (Field) Note that the force formula has two masses, and the force is the result of their interaction at a distance r. Note that the field formula has just one mass – namely the mass that “sets up” the local field in the space surrounding it. It “curves” it. The field view of the universe (spatial disruption by a single mass) is currently preferred over the force view (action at a distance) as the next slides will show. Consider the force view. In the force view, the masses know the locations of each other at all times, and the force is instantaneously felt by both masses at all times. This requires the “force signal” to be transferred between the masses instantaneously. Einstein’s special theory of relativity states unequivocally that SUN the fastest any signal can travel is at the (finite) speed of light c. Thus the action at a distance “force signal” will be slightly mdelayed in telling the orbital mass when to turn. The end result would have to be an expanding spiral mmotion, as illustrated in the following animation: We do not observe planets leaving mtheir orbits as they travel around the msun. Thus force as action at a distance doesn’t work if we are to believe special relativity. SUN And all current evidence points to mthe correctness of special relativity. So how does the field view take care of this “signal lag” problem? Simply put – the gravitational field distorts the space around the mass that is causing it so that any other mass placed at any position in the field will “know” how to respond immediately. Think of space as a stretched rubber sheet – like a drum head. Bigger masses “curve” the rubber sheet more than smaller masses.
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