Answer
$(a)\space 7.56\space km/s$
$(b)\space 5807\space s $
Work Step by Step
(a) Here we use the equation $V=\sqrt {\frac{GM_{E}}{r}}$ where, V - speed of the Hubble space telescope, $M_{E}$ - the mass of the earth, G - universal gravitational constant, r - radius of the orbit.
$V=\sqrt {\frac{GM_{E}}{r}}$; Let's plug known values into this equation
$V=\sqrt {\frac{6.67\times10^{-11}\space Nm^{2}/kg^{2}\times5.97\times10^{24}\space kg}{596\times10^{3}\space m+6.37\times10^{6}\space m}}$
$V=7561\space m/s=7.561\space km/s$
(b) Here we use the equation $T=\sqrt {\frac{4\pi^{2}r^{3}}{GM_{E}}}$ to find the orbital period.
$T=\sqrt {\frac{4\pi^{2}r^{3}}{GM_{E}}}$ ; Let's plug known values into this equation.
$T= 2\pi \sqrt {\frac{r^{3}}{GM_{E}}}$
$T= 2\pi\sqrt {\frac{(596\times10^{3}\space m+6.37\times10^{6}\space m)^{3}}{6.67\times10^{-11}\space Nm^{2}/kg^{2}\times5.97\times10^{24}\space kg}}$
$T=2\pi \sqrt {\frac{3.4\times10^{20}}{39.8\times10^{13}}}s=5807\space s$
Orbital period = 5807 s
