\r\n| 300,000 kg<\/td>\r\n | 100,000 kg<\/td>\r\n | 3 m<\/td>\r\n | 0.222 N<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n \r\n\r\nNote that when we double the mass of one object, the force doubles as well. \u00a0Tripling the mass triples the force. \u00a0This is because force varies directly with mass. \u00a0On the other hand, when the distance doubles, the force decreases by a factor of 4, and when the distance is tripled, we find the force is only one-ninth of what it was before! \u00a0For this reason, the Universal Law of Gravitation is sometimes referred to as an inverse square law. \u00a0Other natural phenomena, such as electrostatic force, also work according to an inverse square law. \u00a0(The force between two charged particles is governed by Coulomb\u2019s Law<\/strong>.)\r\n\r\n![\"Image](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/03\/17185922\/comet-602281_1920-200x300.jpg\") Let\u2019s take another look at the Universal Law of Gravitation. \u00a0Imagine a comet is zooming through space near Earth. \u00a0Our planet attracts the comet towards it, but as long as the comet is far enough away and travelling at a high enough speed, it should whiz right by without striking the Earth. \u00a0Because the masses of the Earth and the comet remain relatively constant, the only variable in Newton\u2019s Law is distance ([latex]d[\/latex]). The Earth has a mass of roughly [latex]6\\times10^{24}[\/latex] kg, while a typical comet has a mass on the order of about [latex]10^{13}[\/latex] kg. Therefore, the formula that gives the force of gravity exerted by the Earth on the comet is:\r\n[latex]F={\\large\\frac{\\left(6.67\\times10^{-11}\\right)\\left(6\\times10^{24}\\right)\\left(10^{13}\\right)}{d^2}\\approx\\frac{4\\times10^{27}}{d^2}}[\/latex]<\/p>\r\nThis is a rational function<\/em> in the variable [latex]d[\/latex], in which the numerator is a constant (degree 0) and the denominator has degree 2. \u00a0What happens to the gravitational force as the comet speeds by and gets further and further away? \u00a0In other words, what is the end behavior of [latex]F[\/latex] as [latex]d \\rightarrow \\infty[\/latex]?\r\n\r\nSince the degree of the denominator (2) is greater than the degree of the numerator (0), this function has a horizontal asymptote at [latex]y=0[\/latex]. \u00a0That is, [latex]f \\rightarrow 0[\/latex] as [latex]d \\rightarrow \\infty[\/latex]. \u00a0You can take comfort in the fact that the gravitational force pulling on the comet will become virtually zero if the comet manages to get far enough away.\r\n\r\nFrom gravity to electromagnetism, rational functions and variation seem to be at the heart of how our universe works. \u00a0If you can analyze rational functions and understand direct and inverse variation, then the sky\u2019s the limit!","rendered":"Isaac Newton (1642-1726) discovered the Universal Law of Gravitation, which explains how massive objects like planets exert attractive forces on one another. \u00a0Let [latex]m_1[\/latex] and [latex]m_2[\/latex] be the masses of two objects which are separated by a distance [latex]d[\/latex]. \u00a0Then the force [latex]F[\/latex] acting on each object varies directly with both [latex]m_1[\/latex] and [latex]m_2[\/latex]\u00a0and inversely with the square of [latex]d[\/latex]. \u00a0In other words, there is a certain constant, which we call the universal gravitational constant [latex]G[\/latex], such that:<\/p>\n [latex]F={\\Large\\frac{G\\cdot m_1\\cdot m_2}{d^2}}[\/latex]<\/p>\n The value of [latex]G[\/latex] has been determined to remarkable precision. \u00a0In standard units,<\/p>\n [latex]G\\approx6.67408\\times10^{-11}\\ \\frac{Nm^2}{kg^2}[\/latex]<\/p>\n Don\u2019t worry about the exact value of this constant or the strange combination of units following it. \u00a0What is important is that [latex]G[\/latex] is a constant. \u00a0Basically, because [latex]G[\/latex] is so incredibly tiny, the force due to gravity is negligible unless the masses of the objects are extremely large and the distances are not too great.<\/p>\n Let\u2019s explore the effect on gravitational force when mass or distance between the objects varies. \u00a0In the table below mass is measured in kilograms (kg), distance in meters (m), and force in Newtons (N).<\/p>\n \n \n\n\n| [latex]m_1[\/latex]<\/td>\n | [latex]m_2[\/latex]<\/td>\n | [latex]d[\/latex]<\/td>\n | [latex]F={\\large\\frac{\\left(6.67\\times10^{-11}\\right)\\cdot m_1\\cdot m_2}{d^2}}[\/latex]<\/td>\n<\/tr>\n | \n| 100,000 kg<\/td>\n | 100,000 kg<\/td>\n | 1 m<\/td>\n | 0.667 N<\/td>\n<\/tr>\n | \n| 200,000 kg<\/td>\n | 100,000 kg<\/td>\n | 1 m<\/td>\n | 1.334 N<\/td>\n<\/tr>\n | \n| 300,000 kg<\/td>\n | 100,000 kg<\/td>\n | 1 m<\/td>\n | 2.001 N<\/td>\n<\/tr>\n | \n| 300,000 kg<\/td>\n | 100,000 kg<\/td>\n | 2 m<\/td>\n | 0.500 N<\/td>\n<\/tr>\n | \n| 300,000 kg<\/td>\n | 100,000 kg<\/td>\n | 3 m<\/td>\n | 0.222 N<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n <\/p>\n Note that when we double the mass of one object, the force doubles as well. \u00a0Tripling the mass triples the force. \u00a0This is because force varies directly with mass. \u00a0On the other hand, when the distance doubles, the force decreases by a factor of 4, and when the distance is tripled, we find the force is only one-ninth of what it was before! \u00a0For this reason, the Universal Law of Gravitation is sometimes referred to as an inverse square law. \u00a0Other natural phenomena, such as electrostatic force, also work according to an inverse square law. \u00a0(The force between two charged particles is governed by Coulomb\u2019s Law<\/strong>.)<\/p>\nLet\u2019s take another look at the Universal Law of Gravitation. \u00a0Imagine a comet is zooming through space near Earth. \u00a0Our planet attracts the comet towards it, but as long as the comet is far enough away and travelling at a high enough speed, it should whiz right by without striking the Earth. \u00a0Because the masses of the Earth and the comet remain relatively constant, the only variable in Newton\u2019s Law is distance ([latex]d[\/latex]). The Earth has a mass of roughly [latex]6\\times10^{24}[\/latex] kg, while a typical comet has a mass on the order of about [latex]10^{13}[\/latex] kg. Therefore, the formula that gives the force of gravity exerted by the Earth on the comet is:<\/p>\n [latex]F={\\large\\frac{\\left(6.67\\times10^{-11}\\right)\\left(6\\times10^{24}\\right)\\left(10^{13}\\right)}{d^2}\\approx\\frac{4\\times10^{27}}{d^2}}[\/latex]<\/p>\n This is a rational function<\/em> in the variable [latex]d[\/latex], in which the numerator is a constant (degree 0) and the denominator has degree 2. \u00a0What happens to the gravitational force as the comet speeds by and gets further and further away? \u00a0In other words, what is the end behavior of [latex]F[\/latex] as [latex]d \\rightarrow \\infty[\/latex]?<\/p>\nSince the degree of the denominator (2) is greater than the degree of the numerator (0), this function has a horizontal asymptote at [latex]y=0[\/latex]. \u00a0That is, [latex]f \\rightarrow 0[\/latex] as [latex]d \\rightarrow \\infty[\/latex]. \u00a0You can take comfort in the fact that the gravitational force pulling on the comet will become virtually zero if the comet manages to get far enough away.<\/p>\n From gravity to electromagnetism, rational functions and variation seem to be at the heart of how our universe works. \u00a0If you can analyze rational functions and understand direct and inverse variation, then the sky\u2019s the limit!<\/p>\n\n\t\t\t | |