text: ## Homework Statement
  
  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
  M(Earth)=5.98*10^24 kg
  
  ## Homework Equations
  
  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.
  
  cepheid
  Staff Emeritus
  Gold Member
  Welcome to PF,
  
  ## Homework Statement
  
  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
  M(Earth)=5.98*10^24 kg
  
  ## Homework Equations
  
  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.
  
  I'm not sure what numbers you're plugging in, but keep in mind that the 'R' in Newton's law of gravitation is the distance between the centres of mass of the two bodies. What is the distance between the ISS and the centre of mass of the Earth? Hint: it is not 347 km.
  
  Ahh, 347000 m + 6.37x10^6 m! Perfect, thanks!