Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Since the earth is spinning, wouldn't centrifugal force make us lighter?

  1. Feb 12, 2008 #1
    so here is my questions
    since the earth is spinning, wouldn't centrifugal force make us lighter?
    if we were not spinning would we be heaver?
    how come i do not have gravity?
    what is gravity
     
  2. jcsd
  3. Feb 12, 2008 #2

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, it does. You are lightest on the equator, no change at the poles.
    You do.
    In 25 words or less?
     
  4. Feb 12, 2008 #3

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Even more so because the centrifugal force makes the Earth bulge out at the equator so you are further form the centre and so gravity is also less.
     
  5. Feb 12, 2008 #4
    how do i have gravity?
    sure make it 26 since your such a sweetheart
     
  6. Feb 12, 2008 #5
    anything with mass has gravity. Your mass is so small that its negligible. Which is why things aren't orbiting you.

    Gravity=(G * m1 * m2) / (d^2)

    G= 6.67300 × 10^-11
    m1=your mass in kg.
    m2=mass of other thing(such as the earth) in kg.
    d=distance in metres
     
  7. Feb 12, 2008 #6
    Everything with mass also has gravity.
     
  8. Feb 12, 2008 #7

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Everything with mass has gravity - you attract the Earth slightly, just as the Earth attracts you.
     
  9. Feb 12, 2008 #8
    You and the Earth, sitting in a tree…
     
  10. Feb 12, 2008 #9
    Just incase he didnt get it in 3 posts:

    "Everything with mass also has gravity"

    P.S. - Everything with mass also has gravity.

    Side: Everything with mass also has gravity

    Note: Everything with mass also has gravity

    ref:

    [1] "Everything with mass also has gravity", CaptainQuasar, et al.

    (Sorry, I just thought it was funny that the same thing was reposted 3 times in a row; just incase he didnt get it the first two times.)

    However, "sure make it 26 since your such a sweetheart" is priceless :rofl:
     
    Last edited: Feb 12, 2008
  11. Feb 13, 2008 #10
    so mass has gravity you say?


    anyway, how can you prove that i have gravity?
     
  12. Feb 13, 2008 #11
    Becuase your attracted to me.
     
  13. Feb 13, 2008 #12

    russ_watters

    User Avatar

    Staff: Mentor

    Stand on a scale.
     
  14. Feb 13, 2008 #13
    Looks like its been answered, time to close it.
     
  15. Feb 13, 2008 #14
    I remember doing a calculation about your first point once. Assuming several idealized things, I calculated that the difference between the force we feel at the poles compared to the equator is about 0.02m/s^2.
     
  16. Feb 13, 2008 #15
    If you could stand on the outside edge of Jupiter at the equator, what would the difference be? doesn't Jupiter have something like an 8 hour day?
     
  17. Feb 13, 2008 #16
    Everything would be different in that case because not only is Jupiter spinning faster but it's more massive and larger.

    There would be relative difference between what you'd weigh at Jupiter's equator versus its poles as there is a relative difference between what you'd weigh at Earth's equator versus its poles. But because you're swinging through a tighter curve on Earth, even though it would involve slower motion than on Jupiter, I think you'd have to do all of the calculations to figure out on which planet the relative difference would be greater.
     
  18. Feb 13, 2008 #17

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    You're off by a bit. Centrifugal acceleration is [itex]r\omega^2[/itex], where [itex]r[/itex] is the distance from the axis of rotation. At the equator, [itex]r=6378\,\text{km}[/itex], and at the poles, [itex]r=0[/itex]. Using the fact that the Earth makes one revolution in one sidereal day, [itex]6378\,\text{km} *(2\pi/\text{sidereal day})^2 = 0.034\,\text{m}/\text{s}^2[/itex].

    Just let the Google calculator does the work for you:
    6378 km * (2 pi radians/sidereal day)^2


    However, this is not the end of the story. A person at the equator is further from the center of the Earth than someone at one of the poles due to the Earth's equatorial bulge. This further reduces the gravitational attraction at the equator. The Earth's equatorial bulge is due to the Earth's rotation about its axis. You need to factor the bulge in as well if you want a full accounting of the effects of the Earth's rotation rate. This raises the effect of rotation on sensed acceleration from 0.034 m/s2 to 0.052 m/s2, or about 0.5% of the gravitational acceleration.
     
  19. Feb 13, 2008 #18
    Sorry when I gave that figure I wasn't intending to be accurate. The 1 sig fig was more or less to emphasize how small of a difference the value is relative to the pull due to gravity.
     
  20. Feb 13, 2008 #19
    If you're within an order of magnitude, you're correct enough. ;)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Since the earth is spinning, wouldn't centrifugal force make us lighter?
Loading...