Explanation Given: The circular orbit r ( t ) = 〈 R cos ( ω t ) , R sin ( ω t ) 〉 . The relationship, ω 2 = k R − 3 Formula used: The differential equation referred in the question. r ' ' ( t ) = − k r ( t ) ‖ r ( t ) ‖ 3 Calculation: We have been given the equation for circular orbit as: r ( t ) = 〈 R cos ( ω t ) , R sin ( ω t ) 〉 Let us figure out its first and second derivatives. ⇒ r ' ( t ) = 〈 − R ω sin ( ω t ) , R ω cos ( ω t ) 〉 ⇒ r ' ' ( t ) = 〈 − R ω 2 cos ( ω t ) , − R ω 2 sin ( ω t ) 〉 Let us now plug all the expressions on both the sides of the equation below: ⇒ 〈 − R ω 2 cos ( ω t ) , − R ω 2 sin ( ω t ) 〉 = − k 〈 R cos ( ω ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 14.6, Problem 8E

To determine

To verify:

That the given circular orbit satisfies the differential equation $r'' (t) = - \frac{k}{{‖ r (t) ‖}^{2}} e_{r} = - \frac{k r (t)}{{‖ r (t) ‖}^{3}}$ . Then use it to deduce Kepler’s Second Law.

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Chapter 14.6, Problem 7E

Chapter 14.6, Problem 9E

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Chapter 14 Solutions

Calculus - Standalone book

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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To verify: That the given circular orbit satisfies the differential equation r ' ' ( t ) = − k ‖ r ( t ) ‖ 2 e r = − k r ( t ) ‖ r ( t ) ‖ 3 . Then use it to deduce Kepler’s Second Law.

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To verify: That the given circular orbit satisfies the differential equation r''(t)=−k‖r(t)‖2er=−kr(t)‖r(t)‖3. Then use it to deduce Kepler’s Second Law.

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Chapter 14 Solutions

To verify:

That the given circular orbit satisfies the differential equation $r'' (t) = - \frac{k}{{‖ r (t) ‖}^{2}} e_{r} = - \frac{k r (t)}{{‖ r (t) ‖}^{3}}$ . Then use it to deduce Kepler’s Second Law.