Answer:
Approximately [tex]6.59\times 10^{3}\; \rm m\cdot s^{-1}[/tex].
Explanation:
Look up the radius of the earth: approximately [tex]6.371 \times 10^{3}\; \rm km[/tex].
The radius of the orbit of this satellite is the of the radius of the earth (at ground level) plus the height of the satellite relative to ground level:
[tex]\begin{aligned}r &\approx 6.371 \times 10^{3}\; {\rm km} + 550\; {\rm km} \\ &= 6.921 \times 10^{3}\; \rm km\end{aligned}[/tex].
Convert the units of both distance and time to standard units:
Orbital radius:
[tex]\begin{aligned}r &\approx 6.921 \times 10^{3}\; {\rm km} \\ &= 6.921 \times 10^{3}\; {\rm km} \times \frac{10^{3}\; \rm m}{1\; \rm km} \\ &= 6.921 \times 10^{6}\; \rm m\end{aligned}[/tex].
Orbital period:
[tex]\begin{aligned}t &= 110\; \text{minute} \\ &= 110\; \text{minute} \times \frac{60\; \text{second}}{1\; \text{minute}} \\ &= 6.6 \times 10^{3}\; \text{second}\end{aligned}[/tex].
Calculate the circumference of this orbit:
[tex]\begin{aligned}& 2\, \pi\, r \\ \approx\; & 2 \, \pi \times 6.921 \times 10^{6}\; {\rm m} \\ \approx\; & 4.35 \times 10^{7}\; \rm m\end{aligned}[/tex].
Calculate the orbital speed (tangential) of this satellite:
[tex]\begin{aligned}v &= \frac{2\, \pi\, r}{t} \\ &\approx \frac{4.35 \times 10^{7}\; \rm m}{6.6 \times 10^{3}\; \rm s} \\ &\approx 6.6 \times 10^{3}\; \rm m \cdot s^{-1}\end{aligned}[/tex].
