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Q. 53

Expert-verifiedFound in: Page 772

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

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 $r=\frac{1.00000011}{1+0.0167\mathrm{cos}\theta}$.

The eccentricity is $0.0167.$

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

Now,

$e=\frac{\sqrt{{A}^{2}-{B}^{2}}}{A}\phantom{\rule{0ex}{0ex}}A=1.00000011,e=0.0167\phantom{\rule{0ex}{0ex}}0.0167=\frac{\sqrt{(1.00000011{)}^{2}-{B}^{2}}}{1.00000011}\phantom{\rule{0ex}{0ex}}0.0167(1.00000011)=\sqrt{(1.00000011{)}^{2}-{B}^{2}}\phantom{\rule{0ex}{0ex}}0.0167000018=\sqrt{1.00000022-{B}^{2}}\phantom{\rule{0ex}{0ex}}(0.0167000018{)}^{2}={\left(\sqrt{(1.00000022)-{B}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}0.00027889006=(1.00000022)-{B}^{2}\phantom{\rule{0ex}{0ex}}{B}^{2}=0.99972133\phantom{\rule{0ex}{0ex}}{A}^{2}=(1.00000011{)}^{2}=1.00000022\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}\frac{{x}^{2}}{1.00000022}+\frac{{y}^{2}}{0.99972133}=1\text{or}\frac{{x}^{2}}{(1.00000011{)}^{2}}+\frac{{y}^{2}}{(0.9998606552{)}^{2}}=1$

Now, the standard form,

$r=\frac{eu}{1+e\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}r=\frac{1.00000011}{1+0.0167\mathrm{cos}\theta}$

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