Chapter 5 Centripetal Force and Gravity
v Centripetal Acceleration v Velocity is a Vector v It has Magnitude and Direction v If either changes, the velocity vector changes. Tumble Buggy Demo
v Centripetal Acceleration (a c ) v Always Directed Toward the Center of Rotation a c = v t r 2 r v t v t
Derivation Distance traveled in one revolution = 2πr Time elapsed in one revolution = T v t = 2πr T ω = 2π T T = 2πr v t r v t a c = 2πv t T T = 2πv t a c 2πr v t = 2πv t a c v t a c = v 2 t r
Circular Motion An object moving in a circle must have a force acting on it; otherwise it would move in a straight line. The direction of the force is towards the center of the circle.
Circular Motion This force may be provided by the tension in a string, the normal force, or friction, among others.
v Centripetal Force F c = ma c = mv t r 2
Problem A fast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 75.0 mph. The 0.190 kg ball is 52.0 cm from the pivot point at her shoulder. Just before the ball leaves her hand, what is its centripetal force? 75.0 mph = 33.52 m/s a c = v t ²/r = 33.52²/0.52 = 2.16 X 10 3 m/s² F c = ma c = (0.190)(2160.75) = 411 N
Newton s Law of Universal Gravitation Newton s insight: The force accelerating an apple downward is the same force that keeps the Moon in its orbit.
Newton s Law of Universal Gravitation The gravitational force is always attractive, and points along the line connecting the two masses: The two forces shown are an action-reaction pair.
Newton s Law of Universal Gravitation G is a very small number; this means that the force of gravity is negligible unless there is a very large mass involved (such as the Earth).
Gravitational Attraction of Spherical Bodies Gravitational force between a point mass and a sphere: the force is the same as if all the mass of the sphere were concentrated at its center.
Gravitational Attraction of Spherical Bodies The acceleration of gravity decreases slowly with altitude:
Gravitational Attraction of Spherical Bodies Once the altitude becomes comparable to the radius of the Earth, the decrease in the acceleration of gravity is much larger:
Gravitational Attraction of Spherical Bodies The Cavendish experiment (1798) allows us to measure the universal gravitation constant:
Tides Usually we can treat planets, moons, and stars as though they were point objects, but in fact they are not. When two large objects exert gravitational forces on each other, the force on the near side is larger than the force on the far side, because the near side is closer to the other object. This difference in gravitational force across an object due to its size is called a tidal force.
Tides This figure illustrates a general tidal force on the left, and the result of lunar tidal forces on the Earth on the right.
Kepler s Laws of Orbital Motion Johannes Kepler made detailed studies of the apparent motions of the planets over many years, and observed that planets follow elliptical orbits, with the Sun at one focus of the ellipse.
Ø Problem Gravitation Assume that you have a mass of 50.0 kg. Earth has a mass of 5.97x10 24 kg and a radius of 6.38x10 6 m. What is the force of gravitational attraction between you and Earth?
Gravitation Ø Solution Ø m 1 = 5.97x10 24 kg Ø m 2 = 50.0kg Ø r = 6.38x10 6 m F = G m m 1 r 2 2 24 11 (5.97x10 kg)(50.0kg) F = ( 6.67x10 ) = 489N 6 2 (6.38x10 m)
Ø Problem Gravitation A satellite orbits the earth in a circular motion. The radius of the earth is 6.38x10 6 m and the satellite s altitude is 35,786 km. If the mass of the earth is 5.97x10 24 kg and the mass of the satellite is 500 kg, what is the satellite s orbital velocity?
Ø Solution Ø m 1 = 5.97x10 24 kg Ø m 2 = 500kg Gravitation m 2 v t 2 Ø r = 35,786,000m + 6.38x10 6 m r = G m 1m 2 r 2 2 (500kg)v t 42,166, 000m = (6.67x10 11 ) (5.97x1024 kg)(500kg) (42,166, 000m) 2 v = 3.07 km/s
Homework p. 165 Multiple Choice 5-17 (odd) pp. 167-169 7, 19, 25, 31, 37, 45, 63 (for 63, Lunar Orbiter s mass is 1870 kg, Moon s mass is 7.35 X 10 22 kg)