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Q. 53

Expert-verified

Calculus

Found in: Page 772

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Measurements indicate that Earth’s orbital eccentricity is $0.0167$ and its semimajor axis is $1.00000011$ astronomical units.
(a) Write a Cartesian equation for Earth’s orbit.
(b) Give a polar coordinate equation for Earth’s orbit, assuming that the sun is the focus of the elliptical orbit.

Part (a) The cartesian equation is $\frac{x^{2}}{(1.00000011)^{2}} + \frac{y^{2}}{(0.9998606552)^{2}} = 1$ .

Part (b) The polar equation is $r = \frac{1.00000011}{1 + 0.0167 \cos θ}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part (a) Step 1. Given information.

The eccentricity is $0.0167 .$

The semi-major axis is role="math" localid="1649526544989" $1.00000011 astronomical units.$

Part (a) Step 2. Explanation.

Now,

$e = \frac{\sqrt{A^{2} - B^{2}}}{A} A = 1.00000011, e = 0.0167 0.0167 = \frac{\sqrt{(1.00000011)^{2} - B^{2}}}{1.00000011} 0.0167 (1.00000011) = \sqrt{(1.00000011)^{2} - B^{2}} 0.0167000018 = \sqrt{1.00000022 - B^{2}} (0.0167000018)^{2} = {(\sqrt{(1.00000022) - B^{2}})}^{2} 0.00027889006 = (1.00000022) - B^{2} B^{2} = 0.99972133 A^{2} = (1.00000011)^{2} = 1.00000022 T h e r e f o r e, \frac{x^{2}}{1.00000022} + \frac{y^{2}}{0.99972133} = 1 or \frac{x^{2}}{(1.00000011)^{2}} + \frac{y^{2}}{(0.9998606552)^{2}} = 1$

Part (b) Step 1. Explanation.

Now, the standard form,

$r = \frac{e u}{1 + e \cos θ} T h e r e f o r e, r = \frac{1.00000011}{1 + 0.0167 \cos θ}$

Most popular questions for Math Textbooks

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