1. PF Insights is off to a great start! Fresh and interesting articles on all things science and math. Here: PF Insights

Satellite period, which equation?

  1. 1. The problem statement, all variables and given/known data

    I'm looking for the period of an orbiting object a certain height from the earth's surface, I am given this height. So I have the total radius of 6,5OO,OOO m, g = 9.81 m/s^2 and the mass of earth = 5.98*10^24 kg

    Please note that for this problem G is another constant than what it usually is.

    2. Relevant equations

    Here is where I am confused I do not know whether to use the T = (2[tex]\pi[/tex]r)/[tex]\sqrt{gr}[/tex]

    or the T[tex]^{2}[/tex] = (4[tex]\pi^{2}r^{3}[/tex])/(GM[tex]_{earth}[/tex])

    where G = 6.67*1O^-7

    3. The attempt at a solution

    K. So when I use the method of going with T = (2[tex]\pi[/tex]r)/[tex]\sqrt{gr}[/tex]

    I get about 5114 seconds for the period.

    when I use T[tex]^{2}[/tex] = (4[tex]\pi^{2}r^{3}[/tex])/(GM[tex]_{earth}[/tex])
    I get 52.13584223 seconds, which doesn't logically seem right but since G is different I don't know.

    does anyone know what the right method is?
     
  2. jcsd
  3. malawi_glenn

    malawi_glenn 4,727
    Science Advisor
    Homework Helper

  4. Dick

    Dick 25,832
    Science Advisor
    Homework Helper

    And your value for G is wrong. Though it's hard to really say until you put units on it.
     
  5. thx alot for that clarification. Still it doesn't make sense that its period would be 52 seconds here's my work.

    T(secs)^2 = (4(pi^2)(6,500,000^3 m))/((6.67*10^-7 N*m^2/kg^2)(5.98*10^24 kg))

    are the units for T in seconds? was it alright that I changed km to m for the radius?
     
  6. I'm using a different value for this problem
     
  7. Dick

    Dick 25,832
    Science Advisor
    Homework Helper

    G=6.67*10^(-11)*N*m^2/kg^2. Note the exponent.
     
  8. Dick

    Dick 25,832
    Science Advisor
    Homework Helper

    Is it an 'alternative universe' problem? Why would you use a different value for G? It's a 'universal constant'.
     
  9. you could say that. But either way I don't think it would have much difference for this equation than the plug and chug. Right now I want to know if I am calculating everything else right. I assume I am.
     
  10. Dick

    Dick 25,832
    Science Advisor
    Homework Helper

    The reason you getting 52 seconds is because you are putting in a value of G that is 10000 times too large. Other than that you are doing fine.
     
  11. thank you
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?