Answer
The aphelion is $4.5$ billion miles.
Work Step by Step
Since the trajectory is an ellipse, we may use the results of Exercise 60, which corresponds to Figure 24:
$\begin{array}{*{20}{c}}
{Point}&{A'}&{{F_2}}&C&{{F_1}}&A\\
{x - coordinate}&{ - \frac{{de}}{{1 - e}}}&{ - \frac{{2d{e^2}}}{{1 - {e^2}}}}&{ - \frac{{d{e^2}}}{{1 - {e^2}}}}&0&{\frac{{de}}{{1 + e}}}
\end{array}$
where $A$ and $A'$ are the focal vertices, ${F_1}$ and ${F_2}$ are the foci, $C$ is the center of the ellipse.
According to Kepler's First Law, the Sun is at one focus. So, let the Sun be located at the origin, ${F_1}$. From Figure 24 we see that $A$ is the perihelion. Thus, ${x_A} = \frac{{de}}{{1 + e}} = 2.7$ billion miles. We have,
$de = 2.7\left( {1 + e} \right)$
From the table we see that the aphelion is at $A'$. So,
${x_{A'}} = - \frac{{de}}{{1 - e}}$
Substituting the term $d e$ in ${x_{A'}}$ gives
${x_{A'}} = - \frac{{de}}{{1 - e}} = - \frac{{2.7\left( {1 + e} \right)}}{{1 - e}}$
Since $e \simeq 0.25$, we get
${x_{A'}} = - \frac{{2.7\left( {1 + 0.25} \right)}}{{1 - 0.25}} = - 4.5$
Hence, its aphelion is $4.5$ billion miles.
