**Topic**

The first artificial satellite was launched from Earth by the Soviet Union (former) in 1957. The satellite’s orbit is an ellipse with the center of the Earth as a focal point. The satellite is measured at 583 miles from the Earth’s surface at its nearest and 1,342 miles at its farthest (1 mile is approximately 1,609 km). Find the eccentricity of that orbit, knowing the radius of the Earth is approximately 4,000 miles

**Detailed explanation**

Choose the coordinate system so that the center of the Earth coincides with the focal point \({F_1}\) of the ellipse

Then the ellipse has the equation \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{b^2}}} = 1\) \((0 < b < a)\)

According to the problem, we have: the nearest satellite is 583 miles from the Earth’s surface and the farthest is 1342 miles, and the Earth’s radius is approximately 4000 miles, so the nearest satellite is 583 + 4000 = 4583 miles and the furthest is 1342 + 4000 = 5342 miles.

Assume the satellite has coordinates \(M\left( {x;y} \right)\).

Then the distance from the satellite to the center of the Earth is: \(M{F_1} = a + \frac{c}{a}x\)

Since \( – a \le x \le a\) \(a – c \le M{F_1} \le a + c\)

So the minimum and maximum distances from the satellite to the center of the Earth are a – c and a + c, respectively.

\( \Rightarrow \left\{ \begin{array}{l}a – c = 4,583\\a + c = 5,342\end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = 4,962.5\\c = 379.5\end{array} \right \Rightarrow e = \frac{c}{a} = \frac{{379.5}}{{4,962,5}} \ approx 0.076\)

So the eccentricity of this orbit is approximately 0.076.