Slide 1 / 78 Slide 2 / 78 AP Physics C - Mechanics Universal Gravitation 2015-12-04 www.njctl.org Slide 3 / 78 Table of Contents Click on the topic to go to that section Newton's Law of Universal Gravitation · · Gravitational Field Gravitational Potential Energy · · Orbital Motion · Total Energy and Escape Velocity · Kepler's Laws
Slide 4 / 78 Newton's Law of Universal Gravitation Return to Table of Contents Slide 5 / 78 Newton's Law of Universal Gravitation The law of gravitation states that every mass in the universe exerts an attractive force on every other mass directly proportional to the product of the two masses and inversely proportional to the square of the distance between their centers of mass. Sir Isaac Newton published this law in 1687 in Principia . The force is along a line connecting the two masses. Slide 6 / 78 Newton's Law of Universal Gravitation Since the gravitational force is attractive, we define F 12 as the force that m 1 exerts on m 2, r is the magnitude of the distance between the two masses and r 12 is a unit vector directed from m 1 to m 2 . The negative sign shows that each mass exerts an attractive force on the other. The second equation is for the force that m 2 exerts on m 1 . The universal gravitational constant, G, was calculated in a 1798 experiment by Henry Cavendish, using a torsion balance.
Slide 7 / 78 Newton's Law of Universal Gravitation v Moon The origin of the Law of Universal Gravitation, like a m most great scientific apple advances, is properly shared a a by multiple people, and is also shrouded in the Earth confusion of history. Brahe's and Kepler's work led to it. Robert Hooke corresponded with Newton about it. Edmond Halley questioned Newton about it. Sir Isaac Newton put it in its most complete form and published it in his Principia , one of the most famous mathematics and physics books of all time. Slide 8 / 78 Newton's Law of Universal Gravitation v Newton connected the idea Moon that objects, like apples, a m fall towards the center of apple Earth, with the idea that the a a moon orbits around the Earth while also falling Earth towards the center of the Earth. The moon remains in a circular orbit because it has a velocity perpendicular to its acceleration. Slide 9 / 78 Newton's Law of Universal Gravitation v Moon The apple part of this story a m was first related to Newton's apple a a friend William Stukeley, and separately to John Conduitt, the husband of his niece - but Earth fifty years after he wrote the law! So, it's unclear what role the apple played. Probably, Brahe, Kepler, Hooke and Halley had a greater influence.
Slide 10 / 78 1 Two objects of mass m are originally separated by a distance r and the force acting on each one is F. What is the new force if the distance between them is cut in half? A F/4 B F/2 m m C F m m D 2F E 4F Slide 11 / 78 2 Two objects, one of mass m, and the other of mass 2m are separated by a distance r. Which is true about the force between the two objects? A Mass m experiences a greater force. B Mass 2m experiences a greater force. C Both experience the same attractive force. D Both experience the same repulsive force. E The force increases as they get farther apart. Slide 12 / 78 Gravitational Field Return to Table of Contents
Slide 13 / 78 Gravitational Field Here's where history and physics are out of synch. Newton created his theory of gravitation in 1687, along with the definition of the gravitational force. It clearly works, in that the motions of the celestial objects are described by it as well as earthly phenomena. But, there is a problem - it implies "action at a distance," where objects interact with each other without contact. This was hard to accept. This was solved in 1832 by Michael Faraday when he came up with the theory of the electric field. Slide 14 / 78 Gravitational Field Field theory relies on an object creating a "field" which exists everywhere in space. Other objects interact with the field and not the object that generated it. Faraday was working with electricity and magnetism, and then field theory was applied to Gravity - most notably by Albert Einstein in General Relativity. Start with the gravitational force equation, and let M be the object that is generating the gravitational field, and m is a small "test particle." Divide the force by this test particle, and the result is the gravitational field, g. Slide 15 / 78 Gravitational Field By now, you should be familiar with g; you've been taught it is the gravitational acceleration due to a mass. And when that mass is the planet Earth, g = 9.8 m/s 2 . g is now being defined more generally as the gravitational field - it is a vector field and is illustrated below. A similar diagram will be seen again in the electricity and magnetism chapters of this course. The lines are a mathematical abstraction - the arrows point in the direction of the field (towards the mass M generating it), and the closer the lines are to each other imply a stronger field.
Slide 16 / 78 Slide 17 / 78 3 An object of mass 10.0 kg is sitting on the surface of the earth. What is its weight? A 10 N B 49 N C 98 N D 116 N E 980 N Slide 18 / 78 4 An object of mass 10 kg is sitting on the surface of Mars. What is the value of g for Mars? What is the weight of the object? M Mars = 6.4x10 23 kg R Mars = 3.4x10 6 m G = 6.67x10 -11 Nm 2 /kg 2 A g = 3.4 m/s W = 34 N B g = 3.4 m/s W = 340 N C g = 6.8 m/s W = 68 N D g = 6.8 m/s W = 680 N E g = 9.8 m/s W = 98 N
Slide 19 / 78 Gravitational Field of a Sphere The gravitational field outside a solid sphere is measured from the center of mass of the object - it is the same as if all of the object's mass was concentrated at its center (assuming a uniform mass density within the sphere, or if it changes only with the distance from the center). Newton wanted to prove why this was true - and this was one of the drivers for his invention of calculus. First, lets start with the gravitational field due a uniform thin spherical shell (it's hollow). Slide 20 / 78 Gravitational Field of a Shell Using integral calculus, the gravitational field due to a uniform thin spherical shell is found to be (the derivation can be found in more advanced college physics texts): for r > R Let's now show a geometric for r < R explanation of why g = 0 within the shell. R Slide 21 / 78 Gravitational Field of a Shell Consider a point mass, m, somewhere A b within a uniform thin spherical shell. m b r b Extend two cones, of lengths r a and r b , from the point mass that end up on the surface of the shell, subtending r a m surface areas of A a and A b . A a m a The masses of the surface segments are proportional to their surface areas, which are proportional to the squares of the radii:
Slide 22 / 78 Gravitational Field of a Shell Rearranging the equation: A b m b r b The gravitational forces between m and the two shell segments decrease r a m as the square of the radius vectors. A a m a Therefore, the force from segment A a on the point mass m is equal to the force on the point mass from A b , resulting in zero net force on mass m. Since there is no force on the point mass, there is no gravitational field within the shell. Slide 23 / 78 Gauss's Law for Gravity A b m b for r > R r b for r < R r a m A a m a This result - gained both by calculus and illustrated by geometry is identical to a result obtained by Gauss using vector calculus. Gauss's Law for Gravity makes the same statement that Newton's Universal Law of Gravitation does! And as we'll see later, this law is mathematically similar to describing the behavior of electric and magnetic fields . Slide 24 / 78 Gravitational Field of a Solid Sphere Time to calculate the gravitational field of a uniform solid sphere. This is more appropriate for planetary and cosmological applications - planets and stars for the most part are not hollow. By building up an infinite number of infinitely thin concentric uniform spherical shells, each of which's gravitational field is the same as if its mass was concentrated at its center, we arrive at a solid sphere where: for r > R R But what about the gravitational field within the sphere? It is NOT equal to zero as in the case of the uniform thin spherical shell.
Slide 25 / 78 Gravitational Field Inside a Solid Sphere The gravitational field within a uniform thin spherical shell is zero. It feels no force from any mass outside the shell. If a sphere with radius r is inscribed within the sphere of radius R, only the mass within the smaller sphere contributes to its gravitational field. R r Slide 26 / 78 Gravitational Field Inside a Solid Sphere The mass within the sphere of radius r is defined as m. The mass of the sphere of radius R is defined as M. The gravitational field at r is due solely to the mass within the smaller sphere. R r The gravitational field for r ≤ R is then: Slide 27 / 78 Gravitational Field of a Solid Sphere We can now show the gravitational fields inside and outside a solid sphere. The field increases linearally as the radius increases, and the field outside decreases as the square of the distance from the center of mass. R for r ≤ R for r > R r
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