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!

Centripetal Force

  1. Apr 15, 2007 #1
    When doing problems involving centripetal force, using the equation:

    F = mv^2/r ,

    is r the distance between the centers of the two masses or is it the distance between the surfaces?
  2. jcsd
  3. Apr 15, 2007 #2
    If I am not mistaken, it refers to the distance between the centers of mass of the two objects.
  4. Apr 15, 2007 #3
    It is between the centers of the objects.
  5. Apr 15, 2007 #4
    Okay, but here's my problem. I'm given the radius of the earth (one of the masses), but not the radius of the doghouse (the other mass). I AM given the distance between the two objects. So is radius of the doghouse so small that it is insignificant?

    If so, do I just add the radius of the earth and the distance between the doghouse and the earth for the value of 'r'?
  6. Apr 15, 2007 #5
    Can you give me the full text for the problem? But I will think that it is.
  7. Apr 15, 2007 #6
    It's for a lab so I can't write it exactly as it is or it will make no sense. I'll try my best:

    A doghouse has a mass of 150 kg. The mass of the earth is 5.98 x 10^24 kg. The radius of the earth is 6.28 x 10^8 m. What is the kinetic energy of the doghouse in orbit at 500 km above the earth?

    I'm assuming you do Fg=Fc (force of gravity = centripital force) because the object is in orbit. So mg=mv^2/r.

    Solve for the speed, and then use that in the equation Ek = 0.5mv^2

    Once again, my only problem is that I don't know whether to include the radius of the earth in 'r', especially because the radius of the doghouse isn't given.
  8. Apr 15, 2007 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    A doghouse in orbit above the earth? Sounds a bit weird! Anyway, like you suggest, the radius of the doghouse won't matter as it is neglibile compared to the other two distances. You will need to include the radius of the earth though!
  9. Apr 15, 2007 #8
    just as you do it, and yes the radius of the doghouse is insignificant.
  10. Apr 15, 2007 #9
    Alright, thanks a lot guys.
  11. Apr 15, 2007 #10
    no prob. someday it will be your turn to help me.
  12. Apr 15, 2007 #11
    Oh, one more thing. The next question is asking:

    A dog rocketed over to the moon and landed there. What escape velocity (from the surface of the moon) did the dog need for his return trip to Earth?

    Do you just use the equation for escape velocity or is there some trick to this question? Because isn't escape velocity the minimum speed required to escape the gravitational field, but doesn't the dog also have to return to earth?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Centripetal Force
  1. Centripetal forces (Replies: 17)

  2. Centripetal force (Replies: 6)