Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Trending News

Promoted
Jim
Lv 7
Jim asked in Science & MathematicsPhysics · 10 years ago

Weightless on Surface of Earth?

Assuming the Earth to be a sphere with Radius = 6,380 km, what would the length of a day have to be in order to become weightless? {a day = period of Earth's rotation}.

Update:

This question relates to the fact that the on the surface the Net Acceleration toward the center of the Earth is *not* equal to g but equals (g - Ac).

5 Answers

Relevance
  • ?
    Lv 7
    10 years ago
    Favorite Answer

    When the centripetal acceleration (Rω²) at the equator equals the free-fall acceleration (g) you would appear to be weightless ..

    Rω² = 9.80 m/s²

    ω² = 9.80 / 6.38^6m .. .. ω² = 1.536^6 .. .. ω = 1.24^-3 rad/s

    Periodic time T = 2π / ω .. .. 2π / 1.24^-3 .. .. .. ►T = 5067.0 s .. (1.40 hr)

  • Anonymous
    10 years ago

    12

  • ?
    8 years ago

    this is simple

    when r (w) squared = g

  • Riaz
    8 years ago

    YOU SEEM VERY VERY GOOD AT PHYSICS AND I NEED YOUR HELP!

    PLEASE ANSWER http://answers.yahoo.com/question/index?qid=201310...

  • Ali
    6 years ago

    when r (w) squared = g

Still have questions? Get your answers by asking now.