kepler

# Kepler - Math 32A Lecture 3(Rogawski Notes on Kepler's Laws These notes contain some challenging mathematics Read them carefully and try to put all

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

Math 32A, Lecture 3 (Rogawski): Notes on Kepler’s Laws These notes contain some challenging mathematics. Read them care- fully and try to put all the pieces together. Here are the headings: 1. Newton’s Second Law 2. A Simple but Important Result 3. Angular Momentum 4. Area Swept Out by the Radial Vector 5. The Planetary DiFerential Equation 6. Law of Equal Areas in Equal Times 7. The Velocity Circle 8. Grand ±inale: The Law of Ellipses 9. Summary 1. Newton’s Second Law Let r ( t ) denote the path of a particle of mass m in three-dimensional space. The velocity and acceleration vectors are v ( t ) = r ( t ) , a ( t ) = r ′′ ( t ) When convenient, we drop the reference to t and write r , v , a , but it is understood these are vector-valued functions of time. Newton’s Second Law of Motion is often stated in scalar form: F = ma (force equals mass times acceleration). However, both force and acceleration are vectors, and the Law of Motion is true as a vector equation: F ( r ) = m a When we write the law in this form, we are implicitly assuming that the force depends only the particle’s position r and not on time. ±or example, gravitational force depends on position but not time. If the force did depend on time (for example, the force on an electron due to a varying electric ²eld), it would be necessary to write F ( t, r ) in place of F ( r ). 2. A simple but important result Theorem 1. For any vector-valued function h ( t ) , d dt h ( t ) × h ( t ) = h ( t ) × h ′′ ( t )

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

View Full Document
2 Proof. By the Product Rule for cross products, d dt h ( t ) × h ( t ) = h ( t ) × h ( t ) b ² this is zero + h ( t ) × h ′′ ( t ) = h ( t ) × h ′′ ( t ) Note that h ( t ) × h ( t ) = 0 because v × v = 0 for any vector v . s Actually, this result is not quite as simple as it seems. Suppose that h ( t ) = a x ( t ) , y ( t ) , 0 A . Then h ( t ) × h ( t ) = v v v v v v i j k x y 0 x y 0 v v v v v v = ( xy x y ) k and the derivative simpliFes due to cancellation: d dt h ( t ) × h ( t ) = d dt ( xy x y ) k = ( xy ′′ + x y x y xy ′′ ) k = ( xy ′′ xy ′′ ) k This is equal to h ( t ) × h ′′ ( t ) as claimed since h ( t ) × h ′′ ( t ) = v v v v v v i j k x y 0 x ′′ y ′′ 0 v v v v v v = ( xy ′′ x ′′ y ) k 3. Angular Momentum The angular momentum of a moving particle is the vector m J where
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 08/19/2009 for the course MATH 32A taught by Professor Gangliu during the Spring '08 term at UCLA.

### Page1 / 6

Kepler - Math 32A Lecture 3(Rogawski Notes on Kepler's Laws These notes contain some challenging mathematics Read them carefully and try to put all

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

View Full Document
Ask a homework question - tutors are online