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Escaping Our Solar System |
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| Jan29-13, 11:08 AM | #18 |
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Escaping Our Solar System
That is not correct. r is not the radius, it is something else. Do you understand how this formula is obtained?
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| Feb1-13, 07:02 PM | #19 |
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According to Wikipedia, r the distance from the center of gravity. I am not really certain of the meaning of that, though.
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| Feb2-13, 11:08 AM | #20 |
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Perhaps we could go over your simplified problem first, voko? Moreover, would it be to much to ask to go over the particulars of the formula with me?
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| Feb2-13, 11:31 AM | #21 |
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I would say we should do the simple problem first. Re-read the beginning of the discussion regarding kinetic/potential energy consideration. Can you use those to derive the simple formula?
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| Feb2-13, 03:32 PM | #22 |
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Well, a space ship launched from earth will only possess potential energy due to gravity and kinetic energy due to motion. I know our system is the earth and the space ship, so do we need to consider the energy of the earth? The escape velocity will allow use to reach a distance, an "infinite distance, where we are far enough from the earth so that the affects of gravity have effectively diminished to zero by the time we reach that point, implying the final gravitational energy will be zero. Furthermore, the escape velocity will also allow us to reach that same point mentioned in the preceding sentence, so that the gravitational force opposing the space ship's motion will reduce it zero, implying kinetic energy is zero. Hence: [itex]U_{ig}+K_i=0[/itex]. Is this correct?
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| Feb2-13, 05:05 PM | #23 |
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Total energy, meaning the sum of kinetic and potential energy, is constant (conservation of energy). What is total energy at the infinity? What is total energy in some vicinity of a gravitating body?
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| Feb3-13, 09:32 AM | #24 |
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At infinity, the total energy is zero, right? As far as a body being within a finite vicinity of a gravitating body, I believe by convention the energy is negative; although, I don't know why we have chose it to be negative, what makes it so much more convenient?
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| Feb3-13, 09:44 AM | #25 |
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The convention is that the POTENTIAL energy at infinity is zero. This does not mean the total energy must be zero at infinity. It could be assumed zero because, physically, we are looking for the minimal escape velocity, and being at infinity with greater than zero kinetic energy obviously requires more energy at the origin.
Anyway, what is the formula for total energy at the vicinity of a gravitating body? |
| Feb3-13, 10:18 AM | #26 |
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I'm not sure what the formula is for that particular case.
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| Feb3-13, 10:19 AM | #27 |
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Formula for kinetic energy?
Formula for potential energy? Their sum? |
| Feb3-13, 10:26 AM | #28 |
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Oh, so: [itex]E_{mech}=U_g + K\implies E_{mech} = -\frac{Gm_1m_2}{r}+1/2mv_{tan}[/itex]
Is that correct? |
| Feb3-13, 10:35 AM | #29 |
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Exactly. Now you know that total energy is conserved, so must be equal to total energy at infinity.
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| Feb3-13, 10:57 AM | #30 |
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Well, I know that gravitational potential energy will fade to zero, resulting in the kinetic energy term having to compensate for this. Will the term have to change somehow?
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| Feb3-13, 11:28 AM | #31 |
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Review #25.
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