Homework 33

# Homework 33 - homework 33 BAUTISTA, ALDO Due: Apr 21 2006,...

This preview shows pages 1–3. Sign up to view the full content.

homework 33 – BAUTISTA, ALDO – Due: Apr 21 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 2. 10 points. Two satellites A and B orbit the Earth in the same plane. Their masses and radii have the relationships m B = 6 m A and r B = 3 r A . r B A A B What is the ratio of the orbital speeds v B v A ? 1. v B v A = 1 3 correct 2. v B v A = 3 3. v B v A = 1 3 4. v B v A = 1 2 5. v B v A = 9 6. v B v A = 1 2 7. v B v A = 2 8. v B v A = 3 9. v B v A = 2 10. v B v A = 1 9 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 , where G is the gravitational constant. Circular motion a r = v 2 r , for the radial (centripetal) acceleration. Note: v is the tangential speed. Solution: Since the force of gravity is re- sponsible for holding a satellite in its orbit, the orbital centripetal force is equal to the force of gravity F r = m v 2 r = G M m r 2 , where again M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit (from the Earth’s center). Thus the tangential speed v of an orbit at distance r is, v = r G M r . Note: The speed is independent of m . Thus the ratio v B v A = r G M r B r G M r A = r r A r B . And since r B = 3 r A , we have v B v A = r r A 3 r A = 1 3 . Question 2 Part 2 of 2. 10 points. Let the distance of the satellite A from the center of the Earth be r A = 10 R , where R is the radius of the Earth. Denote the gravitational acceleration at the surface of the Earth by g . The gravitational acceleration due to the Earth at satellite A is given by 1. g A = g 121

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
homework 33 – BAUTISTA, ALDO – Due: Apr 21 2006, 4:00 am 2 2. g A = g 81 3. g A = g 4. g A = g 11 5. g A = g 10 6. g A = g 9 7. g A = g 100 correct 8. g A = g 3 9. g A = g 10 10. g A = g 11 Explanation: Since g = G M R 2 , then at r A , g A = G M r 2 A = G M (10 R ) 2 = g 100 . Question 3 Part 1 of 1. 10 points. Given: k = 4 π 2 G M s , where M s is the mass of the Sun. Suppose that the gravitational force law between two massive objects is F g = G m 1 m 2 r 2+ ± , where ± is a small number. Which of the following would be the rela- tionship between the period T and radius r of a planet in circular orbit? 1. T 2 = k r 3 2. T 2 = k r 3 3. T 2 = k r 3+ ± correct 4. T 2 = k r 2 - 3 ± 5. T 2 = k r 2+3 ± 6. T 2 = k r 3 - ± 7. T 2 = k r 3 ± 8. T 2 = k r 3 - 2 ± 9. T 2 = k r 3+2 ± 10. T 2 = k r 3+2 Explanation: Kepler’s third law changes from its normal form if gravity is not quite an inverse square law. Let
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

### Page1 / 8

Homework 33 - homework 33 BAUTISTA, ALDO Due: Apr 21 2006,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online