1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravitation homework problem

  1. Jan 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Distance of Earth from Sun = 1.50 x 10^11m
    Period = 365.2 days
    Distance of Mars from Sun = 2.28 x 10^11km
    Period of Mars in Earth Years?


    2. Relevant equations
    (Ta/Tb)^2 = (Ra/Rb)^3


    3. The attempt at a solution

    (365.2 days/Tm)^2 =(1.50 x 10^11m/2.28x10^14m)^3
    =(365.2/Tm)^2 = 2.85 x 10^74

    Tm^2= (365.2)^2 (2.85 x 10^74)
    = square root of 3.04 x 10^79
    Tm = 5.5 x 10^39
     
    Last edited: Jan 5, 2009
  2. jcsd
  3. Jan 5, 2009 #2
    any ideas?
     
  4. Jan 5, 2009 #3
    i just put up my work, i know that its wrong because my physics teacher said so, but i'm not sure what my major error is
     
  5. Jan 5, 2009 #4
    Your approch is right but check your math. Note: you made 3 errors.
     
  6. Jan 5, 2009 #5

    jgens

    User Avatar
    Gold Member

    Several errors. First you may want to rewrite your equations (and double check your calculations). Second, you need to convert your time into years.

    Start here (t/1 yr)^2 = (2.28 x 10^11/1.50 x 10^11)^3.
     
  7. Jan 5, 2009 #6
    okay, i tried this way out

    (t/365.2days)^2 = (2.28x10^11/1.50x10^11)^3
    (T/365.2) = square root of 3.5
    T^2/133371 = 1.87
    multiply 133371 to both sides
    t^2 = the square root of 1.87 X 133371
    T = 499
     
  8. Jan 5, 2009 #7

    jgens

    User Avatar
    Gold Member

    Again, you need to check your calculations (and convert your time into years!). When you take the sqrt of 3.5 you have (t/365.2) = 1.87, not (t/365.2)^2 = 1.87. Be careful with the algebra.
     
  9. Jan 5, 2009 #8
    Dm= 2.28 x 10^8m
    De= 1.50 x 10^11m
    Te= 365.2 days or 1 year
    Tm= ?

    (T/1 yr)^2 = (2.28 x 10^8m/1.50 x 10^11m)^3
    (T^2/1yr) = 3.5 x 10^57)
    T^2 = sqrt of 3.5 x 10^57
    = 5.9 x 10^28
     
  10. Jan 5, 2009 #9

    jgens

    User Avatar
    Gold Member

    Again, incorrect. Be careful with calculations and what numbers you substitute into the equation. Your first equation should read:

    t^2 = (2.28 x 10^11/1.50 x 10^11)^3
     
  11. Jan 5, 2009 #10
    t^2 = (2.28 x 10^11/1.50 x 10^11)^3
    t^2 = sqrt of 3.5
    t = 1.87 years or about 2 years
     
  12. Jan 5, 2009 #11

    jgens

    User Avatar
    Gold Member

    Yes. Given your data, I would use 1.87 years (3 sig. figs.).
     
  13. Jan 5, 2009 #12
    thank you! i appreciate your patience
     
  14. Jan 5, 2009 #13

    jgens

    User Avatar
    Gold Member

    You're welcome! :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gravitation homework problem
Loading...