Answer
(a) 248 y (b) $d_{min} = 4.44 \times 10^{12} m, d_{max} = 7.38 \times 10^{12} m$
Work Step by Step
(a) Using Kepler's third law,
$ T = \frac{2 \pi a^{3/2}}{\sqrt{Gm}} = \frac{2 \pi (5.91 \times 10^{12})^{3/2}}{\sqrt{(6.67 \times 10^{-11})(1.99 \times 10^{30})}} = 7.84 \times 10^9 s \approx 248 y$
(b) $d_{min} = a(1-e) = (5.91 \times 10^{12})(1-0.249) = 4.44 \times 10^{12} m$
$d_{max} = a(1+e) = (5.91 \times 10^{12})(1+0.249) = 7.38 \times 10^{12} m$
