Explanation:
Let the length of perihelion be [tex]p[/tex]
Let the length of aphelion be [tex]a[/tex]
Let the length of semi major axis of the elliptical path be [tex]s[/tex]
As we know that semi major axis is the average of perihelion and aphelion.
So,[tex]s=\frac{p+a}{2} =\frac{1558.14}{2}=779.07[/tex][tex]millionKm[/tex]
Let the eccentricity of the elliptical path be [tex]e[/tex]
[tex]e=1-\frac{p}{s}=1-\frac{741.52}{779.07}=0.048[/tex]
