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Total Energy

  1. Jan 30, 2005 #1

    DB

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    Lately there has been alot of talk about this kinda stuff and I am beginning to understand it but I think the best way to fully understand it is if I write it out and ask you guys. Let's say that one mass is orbiting another in space and all we know is the mass of both planets and the gravitational force acting on the smaller planet and we were asked to find the Total energy.

    [tex]F=G(\frac{m_1m_2}{r^2})[/tex]

    [tex]r=\frac{\sqrt{Gm_1m_2}}{F}[/tex]

    Edit: This should be
    [tex]r=\sqrt{\frac{Gm_1m_2}{F}}[/tex]
    ----------

    [tex]E_{total}=E_k+E_p[/tex]

    ----------

    [tex]E_k=\frac{1}{2}mv^2[/tex]

    [tex]v_{planet}=\frac{(2\pi r)^2}{(2\pi\sqrt{\frac{r^3}{GM}
    )^2}}[/tex]

    ----------

    [tex]E_{potential.grav.}=-\frac{Gm_1m_2}{r}[/tex]

    ----------
    Knowing:
    [tex]r=\sqrt{\frac{Gm_1m_2}{F}}[/tex]

    [tex]v_{planet}=\frac{(2\pi r)^2}{(2\pi\sqrt{\frac{r^3}{GM}
    )^2}}[/tex]

    Then:

    [tex]E_{total}=\frac{1}{2}mv^2-\frac{Gm_1m_2}{r}[/tex]

    How'd I do??? :confused: (probably not to well :blushing:)

    Thanks
     
    Last edited: Jan 30, 2005
  2. jcsd
  3. Jan 30, 2005 #2
    When I was in physics, I was confused by this too. The term "total energy" is quite ambiguous, in my opinion. This is because gravitational potential energy is negative, whereas kinetic energy is positive (BTW, kinetic energy is the gravitational potential energy multiplied by -1). In this case, what is meant my the total energy is probably total mechanical energy, which is half of the gravitational potential energy:
    [tex]E_{potential.grav.}=-\frac{Gm_1m_2}{2r}[/tex]
     
  4. Jan 30, 2005 #3
    Sorry, made a little mistake... The equation should be

    [tex]E_{mechanical}=-\frac{Gm_1m_2}{2r}[/tex]
     
  5. Jan 30, 2005 #4

    DB

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    Thanks christinono. From what I know is that as an object moves in the same direction that the force set it in, its potential energy decreases turning into kinetic. So my understanding of why gravitational potential energy is negative is because as an object is accelerated (due to gravity) towards the bigger mass, it is still moving in the direction of the force, but because the force is created from the mass in front of the smaller mass (the oposite direction of the accelerated force) then PE G. becomes negative. So know to get the mechanical energy all I have to do is divide PE G. by 2?
    And does this mean that there is no such thing a total energy between an orbiting mass in space?
     
  6. Jan 30, 2005 #5

    DB

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    o ya, how was my math, did I make any mistakes?
     
  7. Jan 30, 2005 #6
    Yes, the Em is half of Epg.
    As far as why Epg is negative, think of this: A satellite is orbiting around the Earth. It is accelerating toward the center of the Earth. In order to get the satellite to "escape" the Earth's gravitational field, you need to GIVE it energy (called binding energy - additional Ek fot the satellite to escepe the Earth's gravitational field - which is the total mechanical energy multiplied by -1). At that point, the satellite won't have any gravitational potential energy toward Earth. Because you need to GIVE it energy for this, the grav. potential energy is negative. The same concept applies to electrons orbitting around the nucleus.

    Hope I did not confuse you more...
     
  8. Jan 30, 2005 #7

    DB

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    Thanks, and no I'm not to confused. :tongue:

    Does this have anything to do with the concept of centripetal and centrifugal force?
     
  9. Jan 30, 2005 #8
    Well, an object in orbit is accelerating toward the center of the planet (or whatever it is moving around). I'm not sure, though, if it is really considered as centripetal acceleration. You never use centripetal force formulae in orbital motion problems. Usually, you use centripetal motion formulae in banked curves problems, or problems in which a ball is rotating at the end off a string. Truth is, I'm really not sure...
     
  10. Jan 30, 2005 #9

    DB

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    I know I made an algebra mistake here [tex]r=\frac{\sqrt{Gm_1m_2}}{F}[/tex], I'm having a little trouble fixing it....
     
  11. Jan 30, 2005 #10

    DB

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    I found my mistake and I edited my orginal post.
     
  12. Jan 30, 2005 #11
    To correct this previous post:
    Gravitation provides the centripetal force for circular motion when an object is in orbit. (I found this in my Physics 20IB notes) So centripetal force is at play...
     
  13. Jan 30, 2005 #12

    DB

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    Thanks and thanks for looking, apreciate it. :smile:
     
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