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Potential energy equation?

  1. Oct 26, 2009 #1
    Is the potential energy equation technically invalid as potential energy is defined as mgh but as h changes g changes, so if i launched a projectile at 2000 m/s straight up or 90 degrees, how high would it go. Guess the change in g is negligible at heights below 1000m. So technically the exact potential energy of an object is not mgh.

    What if you launched the projectile at an angle other than 90 degrees, how would you account for the rotation of the earth aka the tangential velocity, not to mention the tangential velocity is different depending on where you are as the radius is different.
  2. jcsd
  3. Oct 26, 2009 #2
    You're right...mgh applies for when changes in g are negligible. There is a more general expression -- Ug = -G*m1*m2/r.
  4. Oct 27, 2009 #3
    The expression for the potential energy doesn't change and is given in the previous post.

    You are right, for the kinetic energy one has to add an extra term [itex]-\frac12 m(\omega\times \vec{r})^2[/itex] due to the rotation of the earth, where r is the position on the earth see from the center.
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