Answer: 5,640 s (94 minutes)
Explanation:
the tangential speed of the HST is given by
[tex]v=\frac{2\pi r}{T}[/tex] (1)
where
[tex]2\pi r[/tex] is the length of the orbit
r is the radius of the orbit
T is the orbital period
In our problem, we know the tangential speed: [tex]v=7,750 m/s[/tex]. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:
[tex]r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m[/tex]
So, we can re-arrange equation (1) to find the orbital period:
[tex]T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s[/tex]
Dividing by 60, we get that this time corresponds to 94 minutes.
