Answer
(a) $C = 1.987 \times {10^{ - 10}}$ $day\cdot k{m^{ - 3/2}}$
(b) ${T_S} = 10744.9$ days
(c) The perihelion and the aphelion of Saturn is $1.35 \times {10^9}$ km and $1.51 \times {10^9}$ km, respectively.
Work Step by Step
(a) For the earth, we have ${T_E} = 365$ days. So,
$C = \frac{T}{{{a_E}^{3/2}}} = \frac{{365}}{{{{\left( {150 \times {{10}^6}} \right)}^{3/2}}}} = 1.987 \times {10^{ - 10}}$ $day\cdot k{m^{ - 3/2}}$
(b) For the planet Saturn, we have ${a_S} = 1.43 \times {10^9}$ km. By Kepler's Third Law, $C$ is constant. So,
$C = \frac{{{T_S}}}{{{a_S}^{3/2}}} = 1.987 \times {10^{ - 10}}$
${T_S} = \left( {1.987 \times {{10}^{ - 10}}} \right)\cdot{\left( {1.43 \times {{10}^9}} \right)^{3/2}} = 10744.9$ days
(c) 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.
By Kepler's First Law, the Sun is at one focus. As in Exercise 63, let the Sun be located at the origin, ${F_1}$.
From part (b) we get the semimajor axis of Saturn: ${a_S} = 1.43 \times {10^9}$ km. So, the distance between $A$ and $A'$ is
$\frac{{de}}{{1 + e}} + \frac{{de}}{{1 - e}} = 2 \times 1.43 \times {10^9}$
$de\left( {\frac{1}{{1 + e}} + \frac{1}{{1 - e}}} \right) = 2 \times 1.43 \times {10^9}$
Since $e=0.056$, we get
$de\left( {\frac{1}{{1 + 0.056}} + \frac{1}{{1 - 0.056}}} \right) = 2 \times 1.43 \times {10^9}$
$de \simeq 1.43 \times {10^9}$
Using the table, we get
${x_A} = \frac{{de}}{{1 + e}} = \frac{{1.43 \times {{10}^9}}}{{1 + 0.056}} \simeq 1.35 \times {10^9}$ km.
Thus, the perihelion of Saturn is $1.35 \times {10^9}$ km
${x_{A'}} = - \frac{{de}}{{1 - e}} = - \frac{{1.43 \times {{10}^9}}}{{1 - 0.056}} \simeq - 1.51 \times {10^9}$ km.
Thus, the aphelion of Saturn is $1.51 \times {10^9}$ km.
