I can’t give a numeric answer without the satellite’s orbital period or its orbital radius/altitude — please provide the orbital period $T$ (in minutes) or the orbital radius $r$ (in km or m).
Explanation
- Angular speed for a circular orbit (in radians per minute) is
$$\omega=\frac{2\pi}{T}$$
where $T$ is the orbital period in minutes.
- Distance traveled in time $t$ (e.g., $t=60\text{ min}$ for 1 hour) along a circular orbit of radius $r$ is
$$s=r\theta=r\omega t=r\left(\frac{2\pi}{T}\right)t.$$
Equivalently, the fraction of an orbit covered in $t$ minutes is $t/T$, so
$$s=\frac{t}{T}\cdot 2\pi r.$$
Example (typical low Earth orbit): if $T=90\text{ min}$ and altitude $\approx400\text{ km}$, then $r\approx6371+400=6771\text{ km}$.
- $\omega=\dfrac{2\pi}{90}\approx0.0698\ \text{rad/min}.$
- In one hour ($t=60\text{ min}$), fraction of orbit $=60/90=2/3$, so
$$s=\frac{60}{90}\cdot2\pi\cdot6771\text{ km}\approx2.836\times10^4\text{ km}.$$
If you give me the actual $T$ or $r$ (or altitude), I’ll compute the exact $\omega$ and the distance for 1 hour.