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Gravitation November 26, 2011 Universal Gravitation In 1665 Newton at age 23 proposed universal gravitation: that all mass exerts an attractive force on all other mass. This is the rst time the concept of gravity was extended beyond the earth to include planets, moons, and stars. Through careful measurements, he was able to show that the force causing the curved path of the moon was the same as the force causing objects to fall to the earth. Newton stated that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between the centers of mass of the two objects. Although Newton was not able to measure it, he realized a proportionality constant was necessary to write an equation for the Law of Universal Gravitation: m1 m2 2 Fg = G 2 G = 6.67 × 10−11 Nm /kg2 r The Universal Gravitation Constant (G ) was not measured until 1798 when Cavendish performed his famous torsion balance / optical lever experiment. This allowed astronomers to calculate the mass of the earth, the sun, and other astronomical bodies. Principle of Superposition The gravitational force is the result of an interaction between two bodies with an equal and opposite force on each. if more the two objects are present, the net gravitational force is the vector sum of each − → − → → − → − − → individual eect due to each object present. F 1 = F 12 + F 13 + F 14 + . . . + F 1n for n particles present. F1 = n ∑ F1i i=2 If an object is a sphere or spherical shell, one can assume all its mass is concentrated at its center. For an extended object, ´ the gravitational force is the sum of the forces from each dierential bit of mass in the object. F = dF with the integral taken over the entire object. Gravity and the Earth The gravitational eld is dened the the region around a mass where other masses experience a measurable gravitational force. The magnitude of the eld is dened as: Fg − → g = m The direction of the eld is the same direction as the force and the units are newtons/kilogram. For an object near the earth's surface, Fg = G Mre2m where r is the distance from the center of the earth. The actual acceleration due to gravity may not be equal the the strength of the gravitation eld because: 1 1. The earth is not uniform density 2. The earth is not a perfect sphereit is attened at the poles and bulges at the equator 3. The earth is rotating so objects near the equator have a centripetal acceleration as well as a gravitational acceleration. The dierence between them: ag − g = ω 2 r = 0.034m/s2 Inside the earth at a distance r < Re , the gravitational force and eld depends only on the mass that is contained in a sphere of radius r. The force due to mass outside r cancels out. Gravitational Potential Energy For objects near the earth's surface, gravitational potential energy is usually calculated with the familiar equation Ug = mg∆y , with a zero point at some convenient location the ground, the oor, the table top, etc. For objects well above the earth's surface, in orbit or other celestial objects, this zero point cannot be used. The position where the potential energy is zero is set where the force is zero, at an innite distance from the earth or other celestial body. This means the gravitational potential energy is always negative. U (r) = − GM m r The gravitational potential energy belongs to both particles but if M ≫ mwe can speak in terms of only the smaller particle's potential energy, since it is the only particle which will gain kinetic energy. For systems of more than two particles, calculate U for each pair, neglecting others, and add the results. For example, for three particles: ( ) Gm1 m2 Gm1 m3 Gm2 m3 Ug = − + + r12 r13 r23 The potential energy of a particle moving from a point far away from earth to a point P at a distance r from the center of the earth is the negative of the work done by gravity to move it there. ˆr U = −W = − F (x) dx ∞ The force vector points toward earth's center and the position vector dx points outward making a 180 degree angle between them. Substituting the force equation we get: ) [ ]r ˆ r( GM m GM m GM m U =− dx = − =− 2 x x r ∞ ∞ Work done is independent of the path between the two points, depending only on the initial and nal positions, since gravity is a conservative force. Since the potential energy can be found from the force, the reverse is also true: ( ) dUg d GM m Gmm Fg = − =− − =− 2 dr dr r r Escape Speed In order to escape the gravity of the earth or any other body, the kinetic energy must be great enough to oset the negative potential energy. From conservation of energy: ( ) GM m 1 2 K + U = 0 = mve + − 2 r2 2 Solving for the escape speed: √ 2GM R is the radius of the earth or other celestial body. ve = where R Kepler's Laws Johannes Kepler was a mathematician hired by Tycho Brahe to analyze his data on the positions of the planets in an attempt to develop a new model of the solar system. After Brahe died, Kepler had access to all the data and developed his three laws of planetary motion. 1. Law of Orbits All orbits are elliptical with the sun at one focus. Elliptical orbits are described by the semi-major axis and the eccentricity. If the eccentricity is zero, the orbit is circular. The perihelion is the point nearest the sun and the aphelion is the point farthest from the sun. The semi-major axis is one-half the distance from the sun to the aphelion. 2. Law of Areas A line connecting the planet to the sun sweeps out equal areas in equal time intervals in any part of the orbit. This means the planet moves faster when it is closer to the sun than when it is farther away. This is simply a consequence of conservation of angular momentum. With no external torques, the angular momentum (mvr ) of the planet must remain constant. 3. Law of Periods The square of the orbital period of any planet is proportional to the cube of its semi-major axis (or radius for a circular orbit). Kepler used the empirical data to formulate this law, but Newton later proved it mathematically using the 2nd law of motion and the law of universal gravitation. Assuming a circular orbit: GMm F = ma so = mω 2 r r2 But also, ω = 2π T so: ( 2 T = 4π 2 GM ) r3 Where M is the mass of the sun or whatever body is being orbited. The term in the parentheses is approximately constant for all the planets. Although Kepler's laws were developed using the planets and their orbits around the sun, but they work for all satellites in circular or elliptical orbits around any planet or star. Energy and Satellites The mechanical energy of an orbiting satellite in a circular orbit is the sum of the gravitational potential energy and the kinetic energy. We have already shown that Ug = − Gmm From the 2nd law: r v2 GM m = m r2 r 3 Rearranging: GM m = mv 2 r So the kinetic energy: K= mv 2 GM m U = =− 2 2r 2 The total energy is then: Gmm GM m GM m − =− 2r r 2r Thus the total energy is the negative of the kinetic energy. For an elliptical orbit, replace semi-major axis. K +U = 4 r with a, the