Answer:
Approximately [tex]12.9\; \rm km \cdot h^{-1}[/tex], assuming that this orbit is circular.
Step-by-step explanation:
The question is asking for a tangential velocity with the unit [tex]\rm km \cdot h^{-1}[/tex]. The unit of the given distance is already in [tex]\rm km[/tex] as required. Convert the unit of the orbital period to hours:
[tex]\begin{aligned}T &= 1230\; \text{day} \times \frac{24\; \rm h}{1\; \text{day}}. = 29520\; \rm h\end{aligned}[/tex].
Calculate the angular velocity [tex]\omega[/tex] of this planet from its orbital period:
[tex]\begin{aligned}\omega &= \frac{2\, \pi}{T} \\ &= \frac{2\, \pi}{29520\; \rm h} \approx 2.1285 \times 10^{-4}\; \rm h^{-1}\end{aligned}[/tex].
Given the radius [tex]r[/tex] of the orbit of this planet, the tangential velocity [tex]v_{\perp}[/tex] of this planet would be:
[tex]\begin{aligned}v_{\perp} &= \omega\, r \\ &\approx 2.1285 \times 10^{-4}\; \rm h^{-1} \times 2.22\times 10^{7}\; \rm km \\ &\approx 4.73 \times 10^{3}\; \rm km \cdot h^{-1}\end{aligned}[/tex].
If the orbit of this planet is circular, the velocity of the planet would be equal to its tangential velocity: [tex]4.73\times 10^{3}\; \rm km \cdot h^{-1}[/tex].
