Topic
The Earth moves around the Sun in an elliptical orbit of which the Sun is the focal point. Given that this ellipse has a semi-major axis \(a \approx 149.598.261\) km and an eccentricity \(e \approx 0.017\). Find the minimum and maximum distances between the Earth and the Sun (results rounded to the nearest unit)
Solution method – See details
For ellipse(E): \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) \((0 < b < a)\)
+ The length of the radius through the focal point of the point \(M(x,y)\) on (E) is: \(M{F_1} = a + \frac{c}{a}x;M{F_2} = a – \frac{c}{a}x.\)
Detailed explanation
Choose the coordinate system so that the Sun coincides with the focal point \({F_1}\) of the ellipse. Then, applying the focal radius formula, we have the distance between the Earth and the Sun as:
\(M{F_1} = a + ex\) where \(x\) is the coordinate representing the Earth as \( – a \le x \le a\)
Since \( – a \le x \le a\) \(a + e\left( { – a} \right) \le a + \frac{c}{a}x \le a + e\left( a \right) \Leftrightarrow a – ae \le a + \frac{c}{a}x \le a + ae\)
With the semi-major axis \(a \approx 149\;598\;261\) km and the eccentricity \(e \approx 0.017\), we have:
\(147\;055\;090 \le M{F_1} \le 152\;141\;431\)
So the minimum and maximum distances between the Earth and the Sun are 147 055 090 km and 152 141 431 km, respectively.