Answer
The Hubble Space Telescope's orbital period is $97.0~minutes$
Work Step by Step
When a satellite orbits a planet, the gravitational force provides the centripetal force to keep the satellite moving in a circle. Let $M_p$ be the mass of the planet and let $M_s$ be the mass of the satellite. We can find an expression for the angular speed of a satellite:
$\frac{G~M_p~M_s}{R^2} = M_s~\omega^2~R$
$\omega = \sqrt{\frac{G~M_p}{R^3}}$
We can find an expression for a satellite's orbital period $P$:
$P = \frac{2\pi}{\omega} = 2\pi~\sqrt{\frac{R^3}{G~M_p}}$
We can find the Hubble telescope's orbital period $P$:
$P = 2\pi~\sqrt{\frac{R^3}{G~M_p}}$
$P = 2\pi~\sqrt{\frac{(6.38\times 10^6~m+6.13\times 10^5~m)^3}{(6.67\times 10^{-11}~m^3/kg~s^2)~(5.97\times 10^{24}~kg)}}$
$P = 5822.7~s = 97.0~minutes$
The Hubble Space Telescope's orbital period is $97.0~minutes$
