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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
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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Mashy Group TitleBest ResponseYou've already chosen the best response.0
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

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
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

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
lol i'm meaning intuitive as in straightforward.
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
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

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
if you want, i could give you some equations where you do a stepbystep 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

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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

Mashy Group TitleBest ResponseYou've already chosen the best response.0
"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_centralforce_problem
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
but it's always equal to \(F_G\). if initial speed just affect the altitude of orbit.
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
NOOOOO!!!.. initial speed decides whether it ll be a circular orbit or not!
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
otherwise the velocity will not be perpendicular to the force!!.. it ll have some tangential component
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
i mean.. NON TANGENTIAL component :P
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
ah. I saw it. though that might be why most planets are just almost circles.
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
almost circles cause their eccentricities are too small!
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
so the deviation from the required speed must be have been very small!!
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
hmm,...which brings us to the problem....why? lol
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
and how?!
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
why they are not doing CIRCULAR ORBITTING :P..
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
wait.. what why??? that depends upon the initial conditions.. when the planet was formed... why should the exact speed be matched!!?
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
so basically those planets are just...lucky that they managed to do so? lol :P
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
i dunno. .. you tell me :P..
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.0
lol let's leave that to the NASA boys
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
http://www.physicsforums.com/showthread.php?t=297095 this says.. drag forces must have reduced the velocity somehow.. !!
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
i can't i am a teacher.. i need answers :P
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
dw:1354694312598:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
dw:1354694460654:dw
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
uncle what are you trying to suggest? :P
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
difference in energy
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
i didn't get it :/
 one year ago

MuH4hA Group TitleBest ResponseYou've already chosen the best response.1
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 radialkineticenergy, 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 radialkineticenergy, 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

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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

kr7210 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354712941270:dw
 one year ago

MuH4hA Group TitleBest ResponseYou've already chosen the best response.1
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 pointofviewthingy.. 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

Mashy Group TitleBest ResponseYou've already chosen the best response.0
no.. i am not really familiar with that potential energy of centrifugal movement :/.. its like potential to go straight???..
 one year ago

MuH4hA Group TitleBest ResponseYou've already chosen the best response.1
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 'anglepart' $$\frac{m}{2} r^2 \, \dot \phi^2$$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

kr7210 Group TitleBest ResponseYou've already chosen the best response.0
if u dont know then u sud read classical mechanics again :)
 one year ago
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