# 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

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alphysicist
Homework Helper
Hi warmfire540,

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
I believe this number is your error; this r is supposed to be the radius of the orbit, so r is the altitude of 610 km plus the radius of the earth.