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Mashy Group Title

Why does a planet follow an elliptical orbit and not circular?? I know keplers laws and i know the laws of gravitation.. but I wanna know if there is some reason why the planets follow an ellipse and not a perfect circle

  • one year ago
  • one year ago

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  1. Mashy Group Title
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    I think when you do the derivation using mathematics for a central force which varies 1 over r square... the result turns out to be an ellipse.. is there any intuitive way to come to this conclusion qualitatively?

    • one year ago
  2. Shadowys Group Title
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    note: physics isn't intuitive. example: try to use intuition to understand special relativity. if you want the derivation for the ellipse form, you might want to try Exercise 11.9.32 from "Calculus" by James Stewart. nothing about forces though.

    • one year ago
  3. Mashy Group Title
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    physics isn't intuitive? :P.. what is wrong with you!!.. special relativity is not intuitive TO US cause.. we are not travelling at sub luminal speeds... it was intuitive to EINSTIEN.. if it wasn't.. we wouldn't have the theory in the first place.... Infact.. all the theories are intially laid down by making use of intuitions!!.. sometimes mathematics helps to derive stuff which are totally non intuitive... but at certain level intuition can always be made!!.. PHYSICS IS INTUITIVE mate!

    • one year ago
  4. Shadowys Group Title
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    lol i'm meaning intuitive as in straightforward.

    • one year ago
  5. Shadowys Group Title
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    the above question "is there any intuitive way to come to this conclusion qualitatively", if you count mathematics as intuition (like 1+1=2, duh, thing),then yes...

    • one year ago
  6. Mashy Group Title
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    lol obviously need not be straightforward.. but we can always generate some sort of intuition.. but i agree.. in this case. its not all that straight forward :P

    • one year ago
  7. Mashy Group Title
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    but i think.. its better to understand why the planets DO NOT REVOLVE in a circular orbit.. and thats because.. the initial speed does not satisfy the condition!

    • one year ago
  8. Shadowys Group Title
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    if you want, i could give you some equations where you do a step-by-step derivation and develop your own intuition? technically....the initial speed doesn't really matter? lol how did you arrive to that conclusion?

    • one year ago
  9. Mashy Group Title
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    the initial speed matters.. if mv^2/r is not equal to the force of gravity.. then the object will not give a perfect circle..!!!

    • one year ago
  10. Mashy Group Title
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    "Uniform circular motion Every central force can produce uniform circular motion, provided that the initial radius r and speed v satisfy the equation for the centripetal force If this equation is satisfied at the initial moments, it will be satisfied at all later times; the particle will continue to move in a circle of radius r at speed v forever." form wiki :P http://en.wikipedia.org/wiki/Classical_central-force_problem

    • one year ago
  11. Shadowys Group Title
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    but it's always equal to \(F_G\). if initial speed just affect the altitude of orbit.

    • one year ago
  12. Mashy Group Title
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    NOOOOO!!!.. initial speed decides whether it ll be a circular orbit or not!

    • one year ago
  13. Mashy Group Title
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    otherwise the velocity will not be perpendicular to the force!!.. it ll have some tangential component

    • one year ago
  14. Mashy Group Title
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    i mean.. NON TANGENTIAL component :P

    • one year ago
  15. Shadowys Group Title
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    ah. I saw it. though that might be why most planets are just almost circles.

    • one year ago
  16. Shadowys Group Title
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    *orbits

    • one year ago
  17. Mashy Group Title
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    almost circles cause their eccentricities are too small!

    • one year ago
  18. Mashy Group Title
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    so the deviation from the required speed must be have been very small!!

    • one year ago
  19. Shadowys Group Title
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    hmm,...which brings us to the problem....why? lol

    • one year ago
  20. Shadowys Group Title
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    and how?!

    • one year ago
  21. Mashy Group Title
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    why they are not doing CIRCULAR ORBITTING :P..

    • one year ago
  22. Mashy Group Title
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    wait.. what why??? that depends upon the initial conditions.. when the planet was formed... why should the exact speed be matched!!?

    • one year ago
  23. Shadowys Group Title
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    so basically those planets are just...lucky that they managed to do so? lol :P

    • one year ago
  24. Mashy Group Title
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    i dunno. .. you tell me :P..

    • one year ago
  25. Shadowys Group Title
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    lol let's leave that to the NASA boys

    • one year ago
  26. Mashy Group Title
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    http://www.physicsforums.com/showthread.php?t=297095 this says.. drag forces must have reduced the velocity somehow.. !!

    • one year ago
  27. Mashy Group Title
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    i can't i am a teacher.. i need answers :P

    • one year ago
  28. UnkleRhaukus Group Title
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    |dw:1354694312598:dw|

    • one year ago
  29. UnkleRhaukus Group Title
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    |dw:1354694460654:dw|

    • one year ago
  30. Mashy Group Title
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    uncle what are you trying to suggest? :P

    • one year ago
  31. UnkleRhaukus Group Title
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    difference in energy

    • one year ago
  32. Mashy Group Title
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    i didn't get it :-/

    • one year ago
  33. MuH4hA Group Title
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    • one year ago
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  34. MuH4hA Group Title
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    Hmm - doesn't show the image for some reason. (I'm going to do a feature request, I guess ^_^) Anywhoo... as you can see in my ahm.. masterpiece ^^ The "effective potential Energy" is a sum of the (green) potential gravitational energy plus the centrifugal one. Now there are various cases - depending on the initial conditions. If the resulting energy is negative but there is radial-kinetic-energy, then that's an ellipse. The blue line I shows that case. r1 and r2 are the minimal and the maximal distance of the planet to the sun. If you do not have radial-kinetic-energy, you are going to have a circle. This is the minimal effetive energy you can possibly have - labeled II. The distance is a constant r0. If your energy is positive, then you are going to have a hyperbola. Meaning your celestial body will escape the gravitational field of the sun. (labelled III) Note here, that the point C is the minimal distance to the sun, the body is going to have. Also note that the transistion from elliptical to hyperbola (E = 0) is called a parabola. Please ask, what you don't understand yet, I'll elaborate. \[E_{kin}^{\tan} = \frac{1}{2}\, m\, r^2\, \dot \phi ^2 = \frac{L^2}{2\, m\, r^2}\]\[E_{pot}^{eff} = E_{pot}(r) + \frac{L^2}{2\, m\, r^2} = -G \frac{m\, M}{r} + \frac{L^2}{2\, m\, r^2}\]\[E_{kin}^{rad} = \frac{1}{2}\, m \, \dot r ^2 = E - E_{pot}^{eff}\]\[\frac{d\, E_{pot}^{eff}}{dr}= 0 \rightarrow r_0 =\frac{L^2}{G \, m^2 \, M}\]

    • one year ago
  35. Mashy Group Title
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    ok i need to study more i guess :D.. what is centrifugal energy?? isn't it just its inertia? I always hate the word centrifugal force... cause its not a real force!!

    • one year ago
  36. kr7210 Group Title
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    |dw:1354712941270:dw|

    • one year ago
  37. MuH4hA Group Title
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    Well, why do you hate easy things? There is an acceleration acting on the body - I don't think you'll doubt that. Now just taking F = m*a and naming that the centrifugal force makes things easier. Yeah you can always choose a coordinate system, where you wouldn't need it, but why hate on easy principles? And about it being real or not.. that's pretty much a point-of-view-thingy.. Are you familiar with D'Alembert's priciple? [1] I meant the potential energy of the centrifugal force/movement. Does that name make it clearer? 1. http://en.wikipedia.org/wiki/D%27Alembert%27s_principle

    • one year ago
  38. Mashy Group Title
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    no.. i am not really familiar with that potential energy of centrifugal movement :-/.. its like potential to go straight???..

    • one year ago
  39. MuH4hA Group Title
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    Well hmm.. you split the effective potential in 2 parts. One part is the radial part $$\frac{m}{2} \dot r^2$$ giving you the kinetic energy of the radial movement. The other one is the 'angle-part' $$\frac{m}{2} r^2 \, \dot \phi^2$$that'll describe the energy of the tangential movement with a fixed distance r. This tangential part, we can also express with a constant angular momentum $$\frac{L^2}{2 m r^2}$$ NOW: because this part is only a function of r but not of the angle or the radial velocity, we simply say, it's part of the potential energy (which is also solely dependent on r). This sum, we now call the "effective potential energy" (the red line in my drawing).

    • one year ago
  40. kr7210 Group Title
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    if u dont know then u sud read classical mechanics again :)

    • one year ago
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