Spring 2010 - HW11_Solution

# Spring 2010- - banuelos(ojb93 – Homework 11 – florin –(58140 1 This print-out should have 23 questions Multiple-choice questions may continue

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: banuelos (ojb93) – Homework 11 – florin – (58140) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A satellite circles planet Roton every 6 h in an orbit having a radius of 7 . 3 × 10 6 m. If the radius of Roton is 2 . 847 × 10 6 m, what is the magnitude of the free-fall acceleration on the surface of Roton? Correct answer: 4 . 06112 m / s 2 . Explanation: Basic Concepts: Newton’s law of gravi- tation F g = G m 1 m 2 r 2 . Kepler’s third law T 2 = parenleftbigg 4 π 2 GM parenrightbigg r 3 . The free-fall acceleration a on the surface of the planet is the acceleration which a body in free fall will feel due to gravity F g = G M m R 2 = ma, where M is the mass of planet Roton. This acceleration a is a = G M R 2 , (1) the number which is g on Earth. Here, how- ever, the mass M is unknown, so we try to find this from the information given about the satellite. Use Kepler’s third law for the period of the orbit T 2 = parenleftbigg 4 π 2 GM parenrightbigg r 3 . (2) By multiplying both sides with R 2 and com- paring to equation (1), we can identify our a in the right hand side T 2 R 2 = parenleftbigg 4 π 2 a parenrightbigg r 3 . If we solve for a , we obtain a = parenleftbigg 4 π 2 T 2 R 2 parenrightbigg r 3 = 4 . 06112 m / s 2 which is our answer. Although identifying a in this way is a “quick” way of solving the problem, we could just as well have calculated the planet mass M explicitly from equation (2) and inserted into equation (1). 002 10.0 points Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical orbit of planet B is two times larger than the semi- major axis of the elliptical orbit of planet A. What is the ratio of the orbital period of planet B to that of planet A? 1. T B T A = 1 8 2. T B T A = radicalbigg 1 8 3. T B T A = √ 8 correct 4. T B T A = 2 5. T B T A = √ 2 6. T B T A = 1 2 7. T B T A = 1 4 8. T B T A = 1 9. T B T A = 8 10. T B T A = radicalbigg 1 2 Explanation: Basic Concept: Kepler’s Third Law is T 2 = parenleftbigg 4 π 2 GM S parenrightbigg a 3 = K S a 3 , banuelos (ojb93) – Homework 11 – florin – (58140) 2 where K S = 4 π 2 GM S = 2 . 97 × 10 − 19 s 2 / m 2 , and where a is the semi-major axis of the elliptical orbit of the planet ( a = r the radius of a planet in a circular orbit). Solution: According to Kepler’s third law, the square of the orbital period is proportional to the cube of the semi-major axis “ a ” of the elliptical orbit. Therefore, T 2 A a 3 A = T 2 B a 3 B . Therefore, T B T A = parenleftbigg a B a A parenrightbigg 3 2 = 2 3 / 2 = √ 8 ....
View Full Document

## This note was uploaded on 04/11/2010 for the course PHYSICS 58100 taught by Professor Florin during the Spring '10 term at University of Texas at Austin.

### Page1 / 14

Spring 2010- - banuelos(ojb93 – Homework 11 – florin –(58140 1 This print-out should have 23 questions Multiple-choice questions may continue

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

View Full Document
Ask a homework question - tutors are online