Gravitation Potential Energy Problem of WHY?

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Gravitation Potential Energy Problem of WHY???

Homework Statement


Energy is required to move a 1540 kg mass from the Earth’s surface to an altitude 3.12 times the Earth’s radius RE. What amount of energy is required to accomplish this move? The acceleration of gravity is 9.8 m/[itex]s^{2}[/itex].

Homework Equations


ΔU = Gm[itex]M_{E}[/itex]/([itex]\frac{1}{R}[/itex] - [itex]\frac{1}{4.12R}[/itex])

The Attempt at a Solution


I know the equation and you just plug it in. But I am having problem of understanding the equation itself. I don't want to just plug numbers in unless I know what I'm doing.
So what I'm confused about is... why do I need the potential energy of the earth? Why can I not just use U = Gm[itex]M_{E}[/itex]/4.12R??
 

Answers and Replies

  • #2
Delphi51
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Isn't the formula for the energy of the mass m due to its position in the Earth's gravity field E = -GmM/r ? This potential energy is negative because it is measured with respect to a mass infinitely far away from Earth, which is said to have zero potential energy. As it falls toward the Earth, it loses potential energy, so it then has negative PE.
So the correct way to think about this is
(Energy at 4.12 R) - (Energy at 1 R)
= -GmM/(4.12R) - (-GmM/R)
 
  • #3
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Yes I forgot the negative sigh.
But still I don't get the answer.
From this equation,
= -GmM/(4.12R) - (-GmM/R)
I feel like only -GmM/(4.12R) is needed for this problem, but why am I subtracting (or adding if you look at the signs) the potential energy of the earth?
 
  • #4
D H
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Why can I not just use U = Gm[itex]M_{E}[/itex]/4.12R??
Because you are asked for the energy to move from the Earth's surface to 3.12 earth radii above the surface of the Earth.

Certainly less energy would be required were that initial position 3.11 earth radii above the Earth's surface instead of right on the Earth's surface. Right off, this says that you need to account for the initial state as well as the final state.

There's always an arbitrary constant of integration in a potential energy function. That constant vanishes when you compute the change in potential energy. It is almost always the change in potential that is of interest. In some applications it appears that potential is only being calculated at one point (and hence no difference). Dig deep and you will find that there is another point as well, but the arbitrary constant of integration has been chosen such that the potential is zero at this point.
 
  • #5
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So since the mass was on the earth's surface, not at 3.12 earth radii to begin with, I need to take account of earth's potential energy?
 
  • #6
D H
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Not the Earth's potential energy. You need to account for object's potential energy when it was on the surface of the Earth.
 
  • #7
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yes yes I mean that yes. So it's basically asking for change of potential energy no?
 
  • #8
D H
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Exactly. It's right there in the question: "What amount of energy is required to accomplish this move?"
 
  • #9
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Well I think it's poorly worded. They should have just asked the delta E <.<
Well thank you for your help :D
 
  • #10
Delphi51
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I feel like only -GmM/(4.12R) is needed for this problem, but why am I subtracting (or adding if you look at the signs) the potential energy of the earth?
It may be important for you to really understand this!
-GmM/(4.12R) is the potential energy the mass LOSES when it falls from infinite radius (the place where it has zero PE by definition) to 4.12R.
The question asks for the potential energy GAINED when it is lifted from R to 4.12R.
 
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