The International Space Station (ISS) circles the earth at an altitude of 347 km.
What is the period of the orbit of the ISS expressed in minutes?
G=6.67x10^-11 N * m^2 /kg^2
T^2/R^3 = (4Pi^2)/(GM)
So: T^2= Sqrt(((4Pi^2)/(GM))*(r^3))
The Attempt at a Solution
Alright, so I've been plugging in these numbers for awhile, and I keep getting the wrong answer: 1.1 seconds. The right answer is 91.3 min.
T is in seconds when initially calculated, right? I'm still doing something wrong, but I'm hoping to just double check on that.