Answer:
7.55 km/s
Explanation:
The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:
[tex]G\frac{mM}{R^2}=m\frac{v^2}{R}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}[/tex] is the gravitational constant
[tex]m=11,100 kg[/tex] is the mass of the telescope
[tex]M=5.97\cdot 10^{24} kg[/tex] is the mass of the Earth
[tex]R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m[/tex] is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)
v = ? is the orbital velocity of the Hubble telescope
Re-arranging the equation and substituting numbers, we find the orbital velocity:
[tex]v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s[/tex]
