anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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!

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anonymous
  • anonymous
lol i'm meaning intuitive as in straightforward.
anonymous
  • anonymous
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...
anonymous
  • anonymous
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
anonymous
  • anonymous
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!
anonymous
  • anonymous
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?
anonymous
  • anonymous
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..!!!
anonymous
  • anonymous
"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
anonymous
  • anonymous
but it's always equal to \(F_G\). if initial speed just affect the altitude of orbit.
anonymous
  • anonymous
NOOOOO!!!.. initial speed decides whether it ll be a circular orbit or not!
anonymous
  • anonymous
otherwise the velocity will not be perpendicular to the force!!.. it ll have some tangential component
anonymous
  • anonymous
i mean.. NON TANGENTIAL component :P
anonymous
  • anonymous
ah. I saw it. though that might be why most planets are just almost circles.
anonymous
  • anonymous
*orbits
anonymous
  • anonymous
almost circles cause their eccentricities are too small!
anonymous
  • anonymous
so the deviation from the required speed must be have been very small!!
anonymous
  • anonymous
hmm,...which brings us to the problem....why? lol
anonymous
  • anonymous
and how?!
anonymous
  • anonymous
why they are not doing CIRCULAR ORBITTING :P..
anonymous
  • anonymous
wait.. what why??? that depends upon the initial conditions.. when the planet was formed... why should the exact speed be matched!!?
anonymous
  • anonymous
so basically those planets are just...lucky that they managed to do so? lol :P
anonymous
  • anonymous
i dunno. .. you tell me :P..
anonymous
  • anonymous
lol let's leave that to the NASA boys
anonymous
  • anonymous
http://www.physicsforums.com/showthread.php?t=297095 this says.. drag forces must have reduced the velocity somehow.. !!
anonymous
  • anonymous
i can't i am a teacher.. i need answers :P
UnkleRhaukus
  • UnkleRhaukus
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UnkleRhaukus
  • UnkleRhaukus
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anonymous
  • anonymous
uncle what are you trying to suggest? :P
UnkleRhaukus
  • UnkleRhaukus
difference in energy
anonymous
  • anonymous
i didn't get it :-/
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
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. \[E_{kin}^{\tan} = \frac{1}{2}\, m\, r^2\, \dot \phi ^2 = \frac{L^2}{2\, m\, r^2}\]\[E_{pot}^{eff} = E_{pot}(r) + \frac{L^2}{2\, m\, r^2} = -G \frac{m\, M}{r} + \frac{L^2}{2\, m\, r^2}\]\[E_{kin}^{rad} = \frac{1}{2}\, m \, \dot r ^2 = E - E_{pot}^{eff}\]\[\frac{d\, E_{pot}^{eff}}{dr}= 0 \rightarrow r_0 =\frac{L^2}{G \, m^2 \, M}\]
anonymous
  • anonymous
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!!
anonymous
  • anonymous
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anonymous
  • anonymous
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
anonymous
  • anonymous
no.. i am not really familiar with that potential energy of centrifugal movement :-/.. its like potential to go straight???..
anonymous
  • anonymous
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).
anonymous
  • anonymous
if u dont know then u sud read classical mechanics again :)

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