Calculate the time needed for a spacecraft to make a Hohmann transfer from Earth to Mars
We need to insert the appropriate constants in Equation 8.58.
$$
\begin{aligned}
\frac{m}{k} & =\frac{m}{G m M_{\text {Sun }}}=\frac{1}{G M_{\text {Sun }}} \\
& =\frac{1}{\left(6.67 \times 10^{-11} \mathrm{~m}^3 / \mathrm{s}^2 \cdot \mathrm{kg}\right)\left(1.99 \times 10^{30} \mathrm{~kg}\right)} \\
& =7.53 \times 10^{-21} \mathrm{~s}^2 / \mathrm{m}^3
\end{aligned}
$$
Because $k / m$ occurs so often in solar system calculations, we write it as well.
$$
\begin{aligned}
\frac{k}{m} & =1.33 \times 10^{20} \mathrm{~m}^3 / \mathrm{s}^2 \\
a_t & =\frac{1}{2}\left(r_{\text {Earth-Sun }}+r_{\text {Mars-Sun }}\right) \\
& =\frac{1}{2}\left(1.50 \times 10^{11} \mathrm{~m}+2.28 \times 10^{11} \mathrm{~m}\right) \\
& =1.89 \times 10^{11} \mathrm{~m} \\
T_t & =\pi\left(7.53 \times 10^{-21} \mathrm{~s}^2 / \mathrm{m}^3\right)^{1 / 2}\left(1.89 \times 10^{11} \mathrm{~m}\right)^{3 / 2} \\
& =2.24 \times 10^7 \mathrm{~s} \\
\end{aligned}
$$