Answer
It takes Mercury to travel $5.74$ days longer from $A'$ to $B'$ than from $B'$ to $A$.
Work Step by Step
From Exercise 41, we obtain the area swept out by the planet from $A'$ to $B'$ is
${A_{A'B'}} = ab\left( {\frac{1}{4}\pi + \frac{1}{2}e} \right)$
Therefore, the area swept out by the planet from $B'$ to $A$ is
${A_{B'A}} = \frac{1}{2}\pi ab - ab\left( {\frac{1}{4}\pi + \frac{1}{2}e} \right) = \frac{1}{4}\pi ab - \frac{1}{2}abe$
${A_{B'A}} = ab\left( {\frac{1}{4}\pi - \frac{1}{2}e} \right)$
Using Kepler's Second Law we find the time required to travel from $B'$ to $A$:
$\frac{{{A_{B'A}}}}{A} = \frac{{{t_{B'A}}}}{T}$
$\frac{{ab\left( {\frac{1}{4}\pi - \frac{1}{2}e} \right)}}{{\pi ab}} = \frac{{{t_{B'A}}}}{T}$
${t_{B'A}} = \left( {\frac{1}{4} - \frac{e}{{2\pi }}} \right)T$
From Exercise 41, we obtain the time required to travel from $A'$ to $B'$ is
${t_{A'B'}} = \left( {\frac{1}{4} + \frac{e}{{2\pi }}} \right)T$
Thus,
${t_{A'B'}} - {t_{B'A}} = \left( {\frac{1}{4} + \frac{e}{{2\pi }}} \right)T - \left( {\frac{1}{4} - \frac{e}{{2\pi }}} \right)T$
${t_{A'B'}} - {t_{B'A}} = \frac{e}{\pi }T$
We have for Mercury $T=88$ days and $e=0.205$. So,
${t_{A'B'}} - {t_{B'A}} = \frac{{0.205}}{\pi }88 \simeq 5.74$
Hence, it takes Mercury to travel $5.74$ days longer from $A'$ to $B'$ than from $B'$ to $A$.
