Keppler's law

Use Kepler’s laws to determine the period of the Hubble Space Telescope which orbits the earth at an altitude of 610 km.

I'm using the eqation i know to find the period of an object revolving around another object

T^2=Kr^3
T^2=(4pi^2/GM)*r^3
G=6.67 x 10^-11
M=5.98× 10^24 (Mass of Earth)
r=610,000

T^2=(39.48/3.99 x 10^14)*2.27 x 10^17
T^2=9.898 x 10^-14*2.27 x 10^17
T^2=22465.81736
T=149.88

I don't know what this means...
but i have another equation for the period of a satelite given by
T=2pi*r/v

mv^2/R+h = GMm/(R+h)^2
v^2/R+h = GM/(R+h)^2
v^2/6.99 x 10^6 = 3.989 x 10^14/4.886 x 10^13
v^2= 8.16 * 6.99 x 10^6
v=7554.3

back to original eqation:
T=2pi*r/v
T=5813.84 seconds
T=96.9 minutes

wikipedia agrees with that number