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Energy in the atmosphere.

  1. Jun 20, 2007 #1
    Potential energy: so I guess one form of potential energy would be using mass and gravity and height. Question, in the real world if you take a ball up to 50' and drop it at some point it would reach terminal velocity. Then say you take it up to 2000' and drop it it, it would reach that same veocity and the energy that it would have would be the same when it impacted the ground. If you take it up to say 200,000' and drop it it would go much quicker with out all that pesky air resistance right. Then it would slow down and still hit with the same energy as at 50'. Is that right or am I way off? say though you observed the ball at 100,000' and the ball was dropped from 200,000' the ball could pass you going oh I don't know mach 1. Is there a formula that takes drag into account, would it even be worth it to incorperate a formula thats uses air temp. and pressure? Is that all even needed?
  2. jcsd
  3. Jun 20, 2007 #2
    Yes, there is a height at which a ball reaches terminal velocity and, as a result, dropping the ball from any greater height will always result in it striking the ground with the same energy. Keep in mind that these greater heights correspond to greater initial potential energy but work due to friction drains away this extra potential energy. You can include drag in your formulas, this corresponds to the inclusion of friction when you sum forces. In the real world, friction is usually velocity dependent and is often really messy. As far as air temp. and pressure.....sure, you could also throw in humidity, thermal expansion of the ball at different heights, etc. there all all sorts of things that would be necesssary to get it exactly right. The real issue is whether these extra variable contribute enough to have an appreciable affect on the answer. Thermal expansion is clearly overkill, but the others....
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