Answer
To find the period, we use Eq. $13-34$ but replace $r$ with $a .$ The result is
$$T=\sqrt{\frac{4 \pi^{2} a^{3}}{G M}}=\sqrt{\frac{4 \pi^{2}\left(6.63 \times 10^{6} \mathrm{m}\right)^{3}}{\left(6.67 \times 10^{-11} \mathrm{m}^{3} / \mathrm{s}^{2}+\mathrm{kg}\right)\left(5.98 \times 10^{24} \mathrm{kg}\right)}}$$$$=5.37 \times 10^{3} \mathrm{s} \approx 89.5 \mathrm{min} .$$
Work Step by Step
To find the period, we use Eq. $13-34$ but replace $r$ with $a .$ The result is
$$T=\sqrt{\frac{4 \pi^{2} a^{3}}{G M}}=\sqrt{\frac{4 \pi^{2}\left(6.63 \times 10^{6} \mathrm{m}\right)^{3}}{\left(6.67 \times 10^{-11} \mathrm{m}^{3} / \mathrm{s}^{2}+\mathrm{kg}\right)\left(5.98 \times 10^{24} \mathrm{kg}\right)}}$$$$=5.37 \times 10^{3} \mathrm{s} \approx 89.5 \mathrm{min} .$$
