Answer:
[tex]a=5.16\times 10^{7}\ km[/tex]
Explanation:
It is given that,
The planet Mercury travels in an elliptical orbit with eccentricity 0.206, e = 0.206
The minimum distance from the sun, [tex]b=4.6\times 10^7\ km=4.6\times 10^{10}\ m[/tex]
The relation between the minimum and the maximum distance from the sun is given by :
[tex]e=\sqrt{1-\dfrac{b^2}{a^2}}[/tex]
a is the maximum distance from the sun
[tex]a^2=\dfrac{b^2}{1-e}[/tex]
[tex]a^2=\dfrac{(4.6\times 10^{10})^2}{1-0.206}[/tex]
[tex]a=5.16\times 10^{10}\ m[/tex]
or
[tex]a=5.16\times 10^{7}\ km[/tex]
So, the maximum distance from the sun is [tex]5.16\times 10^{7}\ km[/tex]. Hence, this is the required solution.
