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Q. 54

Expert-verified
Found in: Page 772

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Measurements indicate that the orbital eccentricity of Mars is$0.0935$ and its semimajor axis is $1.517323951$ astronomical units. (a) Write a Cartesian equation for the orbit of Mars. (b) Do $x$ and $y$ have the same meaning as in Exercise 53? (c) Give a polar coordinate equation for the orbit of Mars, assuming that the sun is the focus of the elliptical orbit.

Part (a) The cartesian equation is $\frac{{x}^{2}}{2.230227197}+\frac{{y}^{2}}{2.28214493}=1\text{.}$

Part (b) The coordinates $x,y$have the same meaning as given for the earth's orbit.

Part (c) The polar equation is $r=\frac{1.517323951}{1+0.0935\mathrm{cos}\theta }$.

See the step by step solution

## Part(a) Step 1. Given information.

The eccentricity is $0.0935$

Semi-major axis is $\text{1.517323951}$

## Part (a)  Step 2. Explanation.

Now,

$e=\frac{\sqrt{{A}^{2}-{B}^{2}}}{A}\phantom{\rule{0ex}{0ex}}A=1.517323951,e=0.0935\phantom{\rule{0ex}{0ex}}0.0935=\frac{\sqrt{\left(1.517323951{\right)}^{2}-{B}^{2}}}{1.517323951}\phantom{\rule{0ex}{0ex}}0.141869789=\sqrt{\left(1.517323951{\right)}^{2}-{B}^{2}}\phantom{\rule{0ex}{0ex}}\left(0.141869789{\right)}^{2}={\left(\sqrt{\left(1.517323951{\right)}^{2}-{B}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}0.020127037-2.30227197=2.30227197-{B}^{2}-2.30227197\phantom{\rule{0ex}{0ex}}{B}^{2}=2.28214493\phantom{\rule{0ex}{0ex}}{A}^{2}=\left(1.517323951{\right)}^{2}=2.230227197$

Therefore, equation becomes,

$\frac{{x}^{2}}{2.230227197}+\frac{{y}^{2}}{2.28214493}=1.$

## Part (b) Step 1. Explanation.

The coordinates $x,y$have the same meaning as given for the earth's orbit.

## Part (c) Step 1. Explanation.

The polar equation will be,

$r=\frac{eu}{1+e\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}r=\frac{1.517323951}{1+0.0935\mathrm{cos}\theta }$