Answer
The eccentricity of the orbit: $e \simeq 0.0488$
Work Step by Step
Let the sun be located at the origin. By Kepler's First Law, we see that the perihelion and the aphelion are the vertices of the ellipse. Thus,
$2a = 740 \times {10^6} + 816 \times {10^6} = 1556 \times {10^6}$
$a = 778 \times {10^6}$,
where $a$ is the semimajor axis.
The distance between the two foci is $2c$. Since the distance between each focus and its closest vertex is the same, therefore
$2c = 2a - 2\cdot740 \times {10^6}$
$2c = 1556 \times {10^6} - 2\cdot740 \times {10^6} = 76 \times {10^6}$
$c = 38 \times {10^6}$
By Theorem 4, we have
$e = \frac{c}{a} = \frac{{38 \times {{10}^6}}}{{778 \times {{10}^6}}} \simeq 0.0488$.
