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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Given that the furthest the earth gets from the sun is approximately 94.56 million miles, and the eccentricity of earth's orbit is approximately 0.017, what is the closest approach of the earth to the sun.

$\begin{array}{1 1}91.4\; million\; miles \\94.56\; million\; miles \\ 92.96\; million\; miles \\ 93\; million\; miles \end{array} $

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According to Kepler, the planets follow an elliptical orbit, and the Sun is at the focus of the ellipse.
From the diagram above, the Sun is at one of the Foci, which means that the distance from the farthest point from Sun on the $x$-axis is $a + c = 94.56..... (1)$
By definition, eccentricity $e = \large\frac{c}{a} $$\rightarrow 0.017 = \large\frac{c}{a}$$ \rightarrow c = 0.017a$
Substituting in $(1)$, we get $1.017a = 94.56 \rightarrow a = 92.98$
$\Rightarrow c = 0.017a = 0.017 \times 92.98 = 1.58$.
Therefore the closest approach to the sun must be on the other side, closest to the focus, which is $92.98 - 1.58 = 91.4$ million miles.
answered Apr 8, 2014 by balaji.thirumalai
 

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