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Pot. energyby Kamataat
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#1
Nov2905, 12:16 PM

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If h, the height of a body from the ground, approaches infinity, then the body's potential energy approaches 0? I'm assuming a nonconstant value for g.
E_p = mgh. In this equation I get 0 times infinity, which is mathematically indeterminate, but I'm guessing that the physical interpretation takes it to equal 0?  Kamataat 


#2
Nov2905, 12:35 PM

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That equation[itex]E_p = mgh[/itex]is only good for measuring gravitational PE near the surface of the earth. (As you note, g is not constant as the height of a body increases, but that equation ignores that fact.) Also, the implied reference point in that equation takes the surface of the earth as the point where PE = 0.
If you want to go to infinity, you'll have to integrate the work done against gravity which varies with distance. Also, it's more convenient to take infinity as the PE = 0 point, which makes the potential energy negative for subinfinite distances. Just for the record, the gravitational PE for an object of mass m at a distance of r from center of the earth (where r > radius of the earth) is given by: [tex]\frac{G M_{(earth)} m}{r}[/tex] 


#3
Nov2905, 12:40 PM

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h<<R(earth). The formula for larger h is E_p=(mgR^2)/(R+h), which>0 as h>infty. 


#4
Nov2905, 01:00 PM

P: 135

Pot. energy
So if a question asks to find the speed which a body must achieve at ground level to escape the Earth's grav. field, then I can take E_p@infinty = 0? From that E_k=E_p to find the speed?
 Kamataat 


#5
Nov2905, 01:04 PM

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Sounds good. (Mechanical energy is conserved. The total energy at r = infinity that an object must have to just barely escape is zero.)



#6
Nov2905, 01:14 PM

P: 135

Thank you!
 Kamataat 


#7
Dec105, 07:58 PM

P: 342

I don't think what you told is correct. This only means that the Kinetic energy of a body that jumps out of earth's gravitational field increases suddenly. This doesn't mean that 0 kinetic energy is enough for a body to be propelled outside the earth's gravitational field.



#8
Dec105, 08:30 PM

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#9
Dec205, 01:09 PM

P: 342

sorryDoc Al
yes, much better interpretation is that escape velocity depends on gravitational force. as r tends to infinity gravitational force tends to zero. thus that's true. we were confused since we were comparing it with the potential energy. 


#10
Dec205, 09:38 PM

P: 42

ya i must ask what r going to infinity has to do with escape velocity, when one can simply assume its when KE=PE



#11
Dec305, 07:42 PM

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"Going to infinity" is what escaping the earth means. To just barely make it, you'd need zero speed at r = infinity. Since PE = 0 (by definition) at r = infinity, the total energy needed is zero. At the earth's surface, that translates into: KE + PE = 0. Thus KE =  PE. (PE is negative at the earth's surface.)



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