Newton's Law of Universal Gravitation (PDF)

Newton’s Law of Universal Gravitation Weight is the force of gravity, but gravity also keeps the moon in orbit around the Earth, and the Earth in orbit around the sun. Gravitational acceleration decreases as a mass gets further from the center of the Earth (we can assume it’s 9.8 ms-2 to a h...

Newton’s Law of Universal Gravitation Weight is the force of gravity, but gravity also keeps the moon in orbit around the Earth, and the Earth in orbit around the sun. Gravitational acceleration decreases as a mass gets further from the center of the Earth (we can assume it’s 9.8 ms-2 to a height of about 20 km above Earth’s surface). Any two masses will exert a mutual force of gravity on each other (equal in magnitude and opposite in direction). G = 6.673 x 10-11 Nm2/kg2 m1 and m2 are the two masses between G m1m2 which the gravitational force is acting. Fg = r2 r is the distance between the centers of the two masses involved; it is also the radius of the orbit if there is an orbit. Properties of Gravity: The force of gravity is attractive; it pulls two mass toward each other. Each will experience the same amount of force. The force of gravity will always act along a straight line between the two masses for which it is being considered. The gravitational force on a system of multiple masses can be considered to act on the center of mass of the system. The force of gravity is directly proportional to the product of the two masses (m1 m2). The force of gravity is inversely proportional to the distance between the two masses squared (1/r2). m1 m2 G m mE r F = r2 F F The acceleration of gravity which is called the GRAVITATIONAL FIELD is the effect in a region of space due to the presence of a mass; no second mass needs to be at that point. By definition the GRAVITATIONAL FIELD is the amount of force per unit mass that is available at a particular point in space due to the presence of another mass. Where m is the mass of the object F = m ag F/m = ag and ag is the gravitational field of the Earth. 9.81 m/s2 = 9.81 N/kg G m mE G mE F = m ag = ag = r2 r2 The equation above tells us the acceleration of a mass that is caused by the mass of Earth. The gravitational acceleration is caused by the object NOT ACCELERATING. Note: The gravitational field at the surface is called surface gravity Mass comes in two varieties that are really the same: Inertial mass that resists forces [ F = m a ] Gravitational mass that creates/feels gravity [ F = G m1m2/r2 ] Since a = G m / r2, we can determine the mass of an object if we know the radius and its gravitational field strength (acceleration) at a given point (often the surface). The radius of the Earth is 6.38 x 106 m, and the acceleration of gravity near the Earth’s surface is 9.81 m/s2. Use this information to determine the mass of the Earth. ag = G m r2 ag m = r2 G ( 6.38 x 106 m )2 ( 9.81 ms-2 ) m = 5.98 x 1024 kg m = G A fully suited astronaut on Earth weighs 1400 N. The astronaut goes to the moon which has a mass of 7.35 x 1022 kg and a radius of 1740 km. How much would the fully suited astronaut weigh on the moon? m = W / g = 1400 N / 9.81 ms-2 = 142.7 kg G m 1m 2 6.67 x 10-11 Nm2/kg2 ( 142.7 kg ) ( 7.35 x 1022 kg ) Fg = = r2 ( 1 740 000 m )2 About 1/6 as much Fg = 231 N = 230 N The dwarf planet Ceres has a mass of 9.38 x 1020 kg and a radius of 4.70 x 105 m. What is the surface gravity (acceleration) on Ceres? G m 6.67 x 10-11 Nm2/kg2 ( 9.38 x 1020 kg ) ag = 2 = = 0.283 ms-2 r ( 470 000 m )2 If an astronaut on Ceres dropped her hammer, how long would it take to fall 1.20 m? x = ½ a t2 1.20 m = ½ (0.283 ms-2) t2 t = 2.91 s When Newton developed his equation for gravity, he was relating the motion of an apple falling to Earth to the motion of the moon. As the apple falls, it speeds up. As the moon orbits, its speed stays essentially the same (it does speed up and slow down some since its orbit is not a perfect circle). How the force of gravity account for the near constant speed of the moon? The force of gravity is almost perpendicular to the velocity of the moon, so it mainly changes direction! If the force is not perfectly perpendicular, it might change the speed and direction. Since the moon’s orbit is almost a perfect circle, the change in speed is minimal. 970 m/s – 1082 m/s ≈ 10% variation Johannes Kepler discovered the geometry of this concept (1609) before Newton was even born (1642) !  KEPLER’S 3 LAWS of PLANETARY MOTION Kepler Brahe The First Law: The planets move around the sun in elliptical orbits with the sun at one focus. Though they are elliptical, most of the orbits are nearly circular. The Second Law: A line from a planet to the sun will sweep out equal areas in equal amounts of time. The planet must move faster when it is closer to the sun and slower when it is further away. The shaded areas are related to the distance the planet moved during a specific amount of time. When the planet is closer to the sun, it must travel a greater distance in the given time to insure that the area is the same. A greater distance in the same time means the planet is moving faster. The Third Law: The ratio of a planet’s average distance cubed to the square of its period is the same for all of the planets. R3 / T2 = constant or R13 / T12 = R23 / T22 The Earth moves around the sun in 1.00 year at a radius of 1.00 astronomical units (AU). Pluto orbits at a radius of 39.48 AU. Determine the period of Pluto’s orbit. RE3 / TE2 = RP3 / TP2 TE2 / RE3 = TP2 / RP3 (1.00 yr)2 / (1.00 AU)3 = (TP)2 / (39.48 AU)3 (39.48 AU)3 (1.00 yr)2 / (1.00 AU)3 = (TP)2 TP = 248 yrs If we started looking at Pluto in 1776, it would complete one orbit by 2024. It was discovered in 1930, so it won’t complete one orbit since its discovery until 2178. Earlier we used the two equations for acceleration and derived an equation for mass: Gm v2 = r2 r 4 p2 r3 m = G t2 If we rearrange this equation, it can be written as: Gm r3 = 4 p2 t2 This is Kepler’s 3rd Law! R3 / T2 = constant m was the mass of the Sun which was constant for all the planets. The moon orbits the Earth once every 27.3 days. The radius of the orbit is 384,000 km. A geostationary satellite orbits the Earth once every day. What is the radius of the orbit of a geostationary satellite? RM3 / TM2 = RS3 / TS2 (384 000 km)3 / (27.3 days)2 = (RS)3 / (1.00 day)2 (5.6623 x 1016 km3) / (745.29 days2) = (RS)3 / (1.00 day2) 7.59746 x 1013 km3 / days2 = (RS)3 / (1.00 day)2 RS = (7.59746 x 1013 km3)1/3 RS = 42 353 km = 42 400 km Things orbiting Earth Combining equations for gravity and circular motion: ag = ac v = d / t = 2pr / t Gm v2 = 4 p2 r2 r2 r v2 = t2 Gm 4 p2 r2 = r2 r t2 r2 Gm 4 p2 r r23 = G r2 t2 G 4 p2 r3 m = G t2 The Dawn spacecraft (ion propelled!) was launched on September 27, 2007 to visit the two asteroids, Vesta & Ceres ( in 2011 & 2015 ). These are the two largest objects in the asteroid belt (Ceres is a dwarf planet). The 1210 kg spacecraft will orbit the 2.7 x 1020 kg asteroid, Vesta, at a radius of 1.7 x 105 m. What will be the force of gravity between Dawn and Vesta when orbit is attained? F g = G m 1m 2 r2 Fg = ( 6.673 x 10-11 Nm2/kg2 ) ( 1210 kg ) ( 2.7 x 1020 kg ) (1.7 x 105 m)2 Fg = 754.35 N = 750 N What is the gravitational field strength of Vesta at the orbital radius? g = F / m = 754.35 N / 1210 kg = 0.6234 N/kg g g = 0.62 N/kg What is the centripetal acceleration of Dawn’s orbit? g = ag = ac = 0.62 N/kg = 0.62 m/s2  What is the average speed of Dawn’s orbit (a = 0.62 m/s2, r = 1.7 x 105 m)? ac = v 2 / r v = r ac v = 1.7 x 105 m ( 0.62 m/s2 ) = 324.65 m/s = 320 m/s What is the period (time for one orbit) of Dawn’s orbit? v=d/t d = 2pr t=2pr/v t = 2 p ( 1.7 x 105 m ) / ( 324.65 m/s ) = 3290 s t = 3300 s = 55 minutes Alexis has landed on an asteroid with radius of 987 m. She drops her hammer and finds that it takes 8.2 s to fall 1.3 m. What is the acceleration of gravity at the surface of the asteroid? (Hint: the hammer moves in a straight line). y = vot + ½ a t 2 ( vo = 0 ) 1.3 m = ½ a ( 8.2 s )2 a = 2 ( 1.3 m ) / ( 8.2 s )2 = 0.038667 m/s2 = 0.039 m/s2  KEPLER’S 3 LAWS of PLANETARY MOTION Kepler Brahe The First Law: The planets move around the sun in elliptical orbits with the sun at one focus. Though they are elliptical, most of the orbits are nearly circular. The Second Law: A line from a planet to the sun will sweep out equal areas in equal amounts of time. The planet must move faster when it is closer to the sun and slower when it is further away. The shaded areas are related to the distance the planet moved during a specific amount of time. When the planet is closer to the sun, it must travel a greater distance in the given time to insure that the area is the same. A greater distance in the same time means the planet is moving faster. The Third Law: The ratio of a planet’s average distance cubed to the square of its period is the same for all of the planets. R3 / T2 = constant or R13 / T12 = R23 / T22 The Earth moves around the sun in 1.00 year at a radius of 1.00 astronomical units (AU). Pluto orbits at a radius of 39.48 AU. Determine the period of Pluto’s orbit. RE3 / TE2 = RP3 / TP2 TE2 / RE3 = TP2 / RP3 (1.00 yr)2 / (1.00 AU)3 = (TP)2 / (39.48 AU)3 (39.48 AU)3 (1.00 yr)2 / (1.00 AU)3 = (TP)2 TP = 248 yrs If we started looking at Pluto in 1776, it would complete one orbit by 2024. It was discovered in 1930, so it won’t complete one orbit since its discovery until 2178. Earlier we used the two equations for acceleration and derived an equation for mass: Gm v2 = r2 r 4 p2 r3 m = G t2 If we rearrange this equation, it can be written as: Gm r3 = 4 p2 t2 This is Kepler’s 3rd Law! R3 / T2 = constant m was the mass of the Sun which was constant for all the planets. The moon orbits the Earth once every 27.3 days. The radius of the orbit is 384,000 km. A geostationary satellite orbits the Earth once every day. What is the radius of the orbit of a geostationary satellite? RM3 / TM2 = RS3 / TS2 (384 000 km)3 / (27.3 days)2 = (RS)3 / (1.00 day)2 (5.6623 x 1016 km3) / (745.29 days2) = (RS)3 / (1.00 day2) 7.59746 x 1013 km3 / days2 = (RS)3 / (1.00 day)2 RS = (7.59746 x 1013 km3)1/3 RS = 42 353 km = 42 400 km Things orbiting Earth Some general circular motion & gravity concepts: Circular motion at a constant speed has an acceleration that changes the DIRECTION of the mass. a = v2 /r Centripetal acceleration = toward the center (a ┴ v) Circular motion at a constant speed has a speed v = d / t. Since the distance around a circle is 2 p r: v = 2pr / t For multiple orbits: v = N 2 p r / t (N = # orbits in time, t) For orbits, gravity is the centripetal acceleration, and gravity gets weaker as distance increases [ a = G m / r2 ]. That means orbits are slower (smaller v) when they are farther from the object being orbited. In the equation, a = G m / r2, the m is the mass that “causes” the gravity; the mass being accelerated is not in the equation. 25 A moon of mass 1×1020kg is in a circular orbit around a planet. The planet exerts a gravitational force of 2×1021N on the moon. The centripetal acceleration of the moon is most nearly Fg = m a g ag = F/m ag = ( 2 x 1021 N ) / ( 1 x 1020 N ) ag = 2 x 101 ms-2 = 20 ms-2 A satellite of mass m orbits a moon of mass M in uniform circular motion with a constant tangential speed of v. The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit that is 2R from the center of the moon. What is the gravitational field strength of the moon at this new distance? ac = a g gR = G m / R2 g2 = G m / (2R)2 g2 = G m / 4 R 2 = ¼ G m / R 2 g2 = ¼ gR A satellite of mass 1000 kg is in a circular orbit around a planet. The centripetal acceleration of the satellite in its orbit is 5 ms-2. What is the gravitational force exerted on the satellite by the planet? Fg = m ag = 1000 kg ( 5 ms-2 ) = 5000 N Extra Problems The International Space Station orbits the Earth at an average radius of 6.79 x 106 m. The mass of the Earth is 5.98 x 1024 kg. Determine the gravitational acceleration of the ISS. ag = G m r2 a = (6.673 x 10-11 Nm2/kg2) (5.98 x 1024 kg) / (6.79 x 106 m)2 a = 8.6553 m/s2 = 8.66 m/s2 What is the speed (v) of the ISS in this orbit? ac = v 2 / r v = ac r v = 8.6553 m/s2 ( 6.79 x 106 m ) = 7666.1 m/s = 7670 m/s What is the period of the orbit? v=d/t d = 2pr v = d/t = 2pr/t t=2pr/v t = 2 p ( 6.79 x 106 m ) / 7666.1 m/s = 5565.1 s t = 5570 s 92.8 minutes! What is the force of gravity between a 2250 kg spacecraft and an asteroid with a mass of 4.25 x 1016 kg if they are 6370 m apart? Fg = G m1m2 r2 Fg = (6.673 x 10-11 Nm2/kg2) (2250 kg) (4.25 x 1016 kg) (6370 m)2 Fg = 157.258 N = 157 N A satellite orbits the Earth at a radius of 1.45 x 107 m. The mass of the Earth is 5.98 x 1024 kg. Determine the gravitational acceleration of the satellite. a = G m g r2 a = (6.673 x 10-11 Nm2/kg2) (5.98 x 1024 kg) / (1.45 x 107 m)2 a = 1.89796 m/s2 = 1.90 m/s2 What is the centripetal acceleration of the satellite? ac = ag = 1.90 m/s2 What is the speed (v) of the satellite in this orbit (think centripetal)? ac = v2 / r 1.89796 m/s2 = v2 / (1.45 x 107 m) v2 = 2.752 x 107 m2/s2 v = 5246 m/s = 5250 m/s How much time will one orbit take (v = d / t) ? (this is called the period of the orbit) v=d/t d = 2pr = 2p (1.45 x 107 m) = 9.1106 x 107 m 5246 m/s = d / t = 9.1106 x 107 m / t t = 9.1106 x 107 m / 5246 m/s = 1.73668 x 104 s t = 17 400 s This is just under five hours. The moon orbits the Earth once every 27.3 days. The radius of the orbit is 384,000 km. An artificial satellite orbits the Earth twice per day (every 12.0 hours). What is the radius of the orbit of the artificial satellite? RM3 / TM2 = RS3 / TS2 (384 000 km)3 / (27.3 days)2 = (RS)3 / (0.500 day)2 (5.6623 x 1016 km3) / (745.29 days2) = (RS)3 / (0.250 day2) 7.59746 x 1013 km3 / days2 = (RS)3 / (0.250 day2) RS = (1.8994 x 1013 km3)1/3 RS = 26 681 km = 26 700 km  Gm v2 ag = = r2 r Gm = v2 v=d/t =2pr/t r Gm (2 p r )2 = r t2 Gm 4 p2 r 2 = r t2 4 p2 r 3 t2 = Gm 4 p2 r 3 t= Gm T is proportional to r3/2