Spacecraft will be moving in 1700 m/s.
Option C
The diameter of the moon is 3500 km and the free fall acceleration at the surface is given as [tex]1.6\ \mathrm{m} / \mathrm{s}^2[/tex]
The radius will be half of the diameter of the moon that can be written as:
[tex]r_{\text {moon }}=1.75 \times 10^{6}[/tex]
By the application of the equation for orbit speed, we get
[tex]\begin{aligned}&v_{\text {orbit}}=\sqrt{r \times q}\\&v_{\text {orbit}}=\sqrt{\left(1.75 \times 10^{6}\right) \times\left(1.6\ \mathrm{m} / \mathrm{s}^{2}\right)}\\&v_{\text {orbit}}=1700\ \mathrm{m} / \mathrm{s}\end{aligned}[/tex]
