Hello, Timcago!
Kepler's First Law of Planetary Motion states that the orbit of a planet about a star is an ellipse with the star at one focus.
he point in that orbit at which the planet is closest to the star is called perihelion
and the point at which the planet is farthest from the star is called aphelion.
The eccentricity of the planet Mercury is apporximately .206.
The length of the minor axis is approximately 113,500,000 km.
Find the distance between Mercury and the Sun at perihelion and at aphelion.
You're expected to know some basics about an ellipse.
The equation is of the form: \(\displaystyle \:[1]\L\;\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\;=\;1\)
where (for this problem) \(\displaystyle a\) is the length of the semimajor axis,
\(\displaystyle \;\;\)\(\displaystyle b\) is the length of the semiminor axis.
\(\displaystyle c\) is the distance from the center to a focus . . . and: \(\displaystyle \,[2]\L\;a^2\:=\:b^2\,+\,c^2\)
The eccentricity is given by: \(\displaystyle \,[3]\L\;e\:=\:\frac{c}{a}\)
Now we're ready to bust this problem . . .
From [3], we have: \(\displaystyle \L\,c\,=\,ae\)
Substitute into [2]: \(\displaystyle \L\:a^2\;=\;b^2\,+\,(ae)^2\,=\,b^2\,+\,a^2e^2\)
We have: \(\displaystyle \L\:a^2\,-\,a^2e^2\:=\:b^2\;\;\Rightarrow\;\;a^2(1 - e^2)\:=\:b^2\;\;\Rightarrow\;\;a^2 \:= \:\frac{b^2}{1 - e^2}\)
Hence: \(\displaystyle \L\:a\;=\;\frac{b}{\sqrt{1\,-\,e^2}}\)
We are told that: \(\displaystyle \,e\:=\:0.206\,\) and \(\displaystyle \,b\:=\:\frac{113,500,000}{2}\:=\:56,750,000\)
So we have: \(\displaystyle \L\:a\;=\;\frac{56,750,000}{\sqrt{1 - 0.206^2}} \;=\;57,993,852.65\;\approx\;58,000,000\)
And: \(\displaystyle \L\:c\;=\;ae\;=\;11,946733.65\;\approx\;12,000,000\)
Code:
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* * *
* | *
* | *
* | *
| Sun
* - - - - - + - - o - - * - -
-58m | 12m 58m
* | *
* | *
* | *
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Therefore, the distance at perihelion is 46 million km,
. . . . . . . . .the distance at aphelion is 70 million km.
