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Just concerned with my math

  1. Feb 28, 2005 #1

    DB

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    I'm really just concerned with my math here because I know that I should think of a celestial object having mechanical energy and not total energy, but just using a planet and a star as an example, is this how I would find the Total Energy of a planet orbiting a star, knowing the mass of the planet, the mass of the star and the gravitational force in Newtons exerted on the planet?

    Here' goes:

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

    [tex]E_k=\frac{1}{2}mv^2[/tex]
    [tex]E_{potential.grav.}=-\frac{Gm_1m_2}{r}[/tex]

    We don't know the radius yet so...
    [tex]r=\sqrt{\frac{Gm_1m_2}{F}}[/tex]

    Nor do we know the velocity of the planet so...
    [tex]v_p=\frac{2\pi r^2}{2\pi\sqrt{\frac{r^3}{GM}}}[/tex]
    (M=mass of star)

    So know that we have found both the velocity and radius...

    [tex]E_{Total}=\frac{1}{2}m(\frac{2\pi \sqrt{\frac{Gm_1m_2}{F}}^2}{2\pi\sqrt{\frac{(\sqrt{\frac{Gm_1m_2}{F}})^3}{GM}}})^2-\frac{Gm_1m_2}{\sqrt{\frac{Gm_1m_2}{F}}}[/tex]
    That equation becomes much simpler once you solve for r first.

    [tex]E_{Total}=\frac{1}{2}m(\frac{2\pi r^2}{2\pi\sqrt\frac{r^3}{GM}}})^2-\frac{Gm_1m_2}{r}}[/tex]

    Then back too:
    [tex]E_{Total}=\frac{1}{2}mv^2-\frac{Gm_1m_2}{r}}[/tex]


    I hope I haven't made it too complicated. I think my mistake is with finding the velocity of the planet by dividing distance by time, but I probably have more... :frown:
    Thanks in advance.
     
  2. jcsd
  3. Feb 28, 2005 #2
    Why are you not using the equation GMm/r^2 = mv^2/r to determine v?
     
  4. Feb 28, 2005 #3

    DB

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    Thanks
    [tex]\frac{GMm}{r^2}=\frac{mv^2}{r}[/tex]

    Like so?
    [tex]\frac{GMmr}{r^2m}=v^2[/tex]

    [tex]\sqrt{\frac{GM}{r}}=v[/tex]

    Is that what you mean?
     
  5. Feb 28, 2005 #4
    Thats right DB.
     
  6. Feb 28, 2005 #5

    DB

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    Thanks, one question. Why is GMm/r^2 equal to mv^2/r ???
     
  7. Feb 28, 2005 #6

    cepheid

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    what does each represent?

    gravitational force = centripetal force

    can you tell us why?
     
  8. Mar 1, 2005 #7

    DB

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    Because they are both forces pushing towards a point?
     
  9. Mar 1, 2005 #8

    cepheid

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    The "word equation" is misleading because it might lead you to think that the two are separate forces that just happen to have the same magnitude. NO. In fact, the gravitational force is a centripetal force in this problem. Remember that the term "centripetal" just specifies a specfic type of force that has the property of being "centre-seeking", ie of keeping a particle moving on a curved path. It is not a specific force in nature. I.e. what causes the centripetal force in a specific situation depends on the problem. In this case, the planet "wants" to move in a straight line at a constant speed (Newton's first law). But some mysterious "centre-seeking" force pulls the planet into a rougly circular orbit centred around the point from which that mysterious force originates. What is that centripetal force in this case? I already told you that it is gravity. So by writing:

    GMm/r^2 = mv^2/r, you are basically writing F = ma (in that order). You are NOT equating two separate forces, but just noting that the effect of the gravitational force is centripetal in this case.

    If you were twirling a ball on a string, what force would be the centripetal force in that case?
     
  10. Mar 1, 2005 #9

    DB

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    Would it be your fingers? (as long as the string were at it's full extent?) And in this case would no work be done?
     
  11. Mar 1, 2005 #10
    I don't think so. The explanation you are suggesting is new to me. It is true that in our case while you are not equating "separate" forces but you are still equating two "same" force. There is no restriction in using the word "equation" whether the type of entities are "separate" or "same".

    In your twirling ball example , we would say:
    mv^2/r = T (Tension on the string).
    There is no simple expression (I believe) for the Tension, but nevertheless it does not loose its right to be called an "equation".
     
  12. Mar 1, 2005 #11

    cepheid

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    A bit of misunderstanding here. I have no problem with applying the word: equation here. That's ok. You're right. It is an equation. I was talking about the "word equation" (all one term) which I meant to describe this equation involving words that I wrote:

    gravitational force = centripetal force <--- this is what I was calling a "word equation"

    ...all I was saying is that it might be misinterpreted to mean that those two forces exist independently and have equal magnitudes (as in the case of an "action-reaction" pair) when in reality they are referring to the same quantities, the graviational force acts as a centripetal force here.

    How can you equate two "force?" That is like saying, "he was between the pillar." It makes no sense. You can't have it both ways, i.e agree with me that there is only one force, yet somehow claim that you are equating it to another force. The equation is F = ma. You are equating the magnitude of the graviational force to the mass times the acceleration of the particle, they are equal by Newton's 2nd. But whenever you have a centripetal force acting, it will accelerate a particle by v^2/r, so the particle's mass time acceleration would be mv^2/r, and it is correct to say that mv^2/r is equal to the magnitude of the centripetal force, or that that force is "given by" (meaning that you can calculate its magnitude with): mv^2/r.

    DB, the answer to the question about the string is: the force due to the tension of the string acts as a centripetal force. Think about it...the string is taut, and that's why the ball on the end of it moves in a circle around your hand. As soon as you let go of the string, that tension disappears, and the ball flies off in a straight line.

    Tension is not actually a force, but a property of the string, it is in tension, and that tension is a function of the linear mass density (mass per unit length)...and something else. I can't remember what. But what's making the string taut? Some force on its ends. And if it has a force acting on its ends (say, your hand), then by Newton's 3rd law, it pulls back on your hands with an equal force, so you feel a force action on you from the string's tension. We often call this the "tension force" (that's your "T" here), and it keeps the ball from flying off in a straight line.
     
    Last edited: Mar 1, 2005
  13. Mar 1, 2005 #12
    Mind if I take a shot at clarifying?

    If the sun were not there, the force needed to produce the observed motion is the right hand term. The force that the sun's gravity exerts is the left hand term. We know the sun is responsible for the motion therefore the equivalence. That help?
     
    Last edited: Mar 1, 2005
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