This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Overton, Mays Homework 29 Due: Apr 15 2005, 4:00 am Inst: Turner 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The radius of the satellite around the center of the Earth is r = 6 R . Hint: You may find it useful to take into account that the gravitational force is a con- servative force. Hint: The universal gravitational force law is ~ F = G M m r 2 r . Earth Satellite R 6 R Caution: Neglect the rotational kinetic en- ergy due to the Earths rotation. Find the energy required to launch a satel- lite from Earth into the circular orbit at the specified radius r = 6 R . 1. E = 13 12 G M m R 2. E = 10 9 G M m R 3. E = 5 6 G M m R 4. E = 12 11 G M m R 5. E = 7 6 G M m R 6. E = 11 12 G M m R correct 7. None of these 8. E = 3 4 G M m R 9. E = 10 11 G M m R 10. E = 4 3 G M m R Explanation: Basic Concepts: Force of gravity between two masses m 1 and m 2 at a distance r F g = G m 1 m 2 r 2 and U g = G m 1 m 2 r , where G is the gravitational constant. Centripetal acceleration for circular motion is a r = v 2 r , for the radial acceleration, where v is the tangential speed. Solution: Since the gravitational force sup- plies the centripetal force, m a r = m v 2 r = G m M r 2 so that m v 2 = G m M r = G m M 6 R . Then the kinetic energy of the satellite in orbit is found to be K r = 1 2 m v 2 = G m M 2 (6 R ) . Conservation of energy gives us U R + E = U r + K r E = U r- U R + K r =- G M m 6 R + G M m R + G M m 12 R = G M m R - 2 12 + 12 12 + 1 12 = 11 12 G M m R . 002 (part 1 of 2) 10 points A satellite with mass m is orbiting around Overton, Mays Homework 29 Due: Apr 15 2005, 4:00 am Inst: Turner 2 the Earth on a circular path with a radius r . Denote the mass and the radius of the Earth by M and R , respectively, and the gravitational acceleration at the surface of the Earth by g . The magnitude of the centripetal accelera- tion of the satellite is given by 1. a c = 1 2 g r R 2 2. None of these 3. a c = g 2 4. a c = g R r 2 correct 5. a c = 1 2 g R r 6. a c = g R r 7. a c = g r R 2 8. a c = g 9. a c = 1 2 g R r 2 Explanation: Basic Concepts: The Universal Law of Gravity is F = G m 1 m 2 r 2 . Gravitational potential energy is U =- G m 1 m 2 r and if we set U = 0 at r = . Solution: The force of gravity acts as cen- tripetal force, F c = F G , or m a c = G M m r 2 so a c = G M r 2 . (1) One might already recall the expression for g ; however, if we dont, for an object of mass m on the surface of the Earth m g = G M m R 2 so that g = G M R 2 . (2) (The actual value, 9.8 m/s 2 , differs somewhat from this value due to other effects, such as the rotation of the Earth)....
View Full Document