Answer
${r_{per}} \simeq 4.38\times{10^{10}}$ m
${r_{ap}} \simeq 7.2\times{10^{10}}$ m
Work Step by Step
We have the eccentricity $e=0.244$ for the orbit of Mercury.
From Exercise 13, we obtain ${r_{per}} = a\left( {1 - e} \right)$ and ${r_{ap}} = a\left( {1 + e} \right)$.
From the table in Exercise 1, the semimajor axis of Mercury is $a = 5.79 \times {10^{10}}$ m.
So,
${r_{per}} = a\left( {1 - e} \right) = 5.79 \times {10^{10}}\left( {1 - 0.244} \right)$
${r_{per}} \simeq 4.38\times{10^{10}}$ m
${r_{ap}} = a\left( {1 + e} \right) = 5.79 \times {10^{10}}\left( {1 + 0.244} \right)$
${r_{ap}} \simeq 7.2\times{10^{10}}$ m
