my orbitz problem
Assume we have a sunplannet system in a perfect, stable orbit.
The gravitation force cancel out with the centrifugal force which equates into: Gm1m2/r^2 = m2v^2/r where m1= mass of the sun, m2 = mass of planet simplify the equation above I get: Gm1/r = v^2 This implies that if we move the plannet faster in tangent direction, the distance between 2 bodies must reduce to achieve a stable orbit. The opposite is true. Slowing down the planet will have to move further away from the sun to attain stable orbit. Let's consider kepler's second law and conservation of momentum. Basically, the law said that when the planet is half radius of a reference position, the velocity increase with the factor of 2. But according the the equation above, the velocity should increase with the factor of 4. Who can shed some light into this? 
Re: my orbitz problem
Hi atom888! :smile:
Quote:
Quote:
They don't apply if you start shoving extra energy in! If a planet is in circular orbit with radius R, and you shove it backward a bit, it will go into an elliptical orbit, and its speed will indeed get more as it comes further in. Its speed at perihelion, at radius r say, will be proportional to R/r (and not to the square). But the speed of a planet in circular orbit with radius r will be (R/r)^{2} times the speed of a planet in circular orbit with radius R. :smile: 
Re: my orbitz problem
Quote:
What you're saying is circular orbits obey my equation and an elliptical orbit got nothing to do with this equation, in another world, obey Kepler's 2nd law and conserve momentum? The part make you confused was that suppose we want to change the planet into another stable circular orbit, say futher out. We have to 2 ways to do this: 1) push it outward and decrease its velocity. 2) increase its velocity and decrease its velocity much more when it reach outter orbit. In reality, it's almost impossible to move rotating planets due to it's gyro effect. I believe that's why planets rotates (to protect itself from external force thus changing orbits). Of course that somehow implies our moon is an artificial satellite. Oh anothe thing. Which plannet in our solar system have elliptical orbit? As I record... none. 
Re: my orbitz problem
tinytim is correct.
The gravitation force cancel out with the centrifugal force which equates into: Gm1m2/r^2 = m2v^2/r is a force balance, which maintains that the centripetal force is equal and opposite to the gravitational force. This is a static situation. If one wishes to change an orbit, one must apply a force to work such that the gravitational potential energy changes. Applying a force means the system is not in equilibrium, but rather it is perturbed. Then we perturb the system again to obtain a circular orbit. Planetary orbits are almost circular, but not quite. A circular orbit has an eccentricity, e = 0. Planetary orbits have nonzero eccentricity. Please refer to this reference: http://www.windows.ucar.edu/tour/lin...ets_table.html http://www.windows.ucar.edu/tour/lin....html&edu=high As for extrasolar planets  See Masses and Orbital Characteristics of Extrasolar Planets http://exoplanets.org/ecc.html http://exoplanets.org/planet_table.shtml 
Re: my orbitz problem
Quote:
[tex]v= \sqrt{GM \left ( \frac{2}{r} \frac{1}{a} \right )}[/tex] Where "a" is the semimajor axis of the orbit, or the average orbital distance. Note for a circular orbit a=r and it reduces to the simple form you used. Quote:
Quote:
Quote:
For instance, due to the Earth's elliptical orbit, its orbital velocity varies from 29.5 km/sec to 30.4 km/sec. 
Re: my orbitz problem
Thx for the answers.
Now let's consider an overexaggerated elliptical orbit so that from a far away point of view, it looks almost like linear. The motion one will see then is a bouncing object. How then one descibe the forces in this picture? 
Hi Janus! :smile:
Quote:
:smile: I like that! :smile: Quote:
It'll be onedimensional: r'' = GM/r^{2} so v^{2} = r'^{2} = 2GM(1/r 1/r_{max} ). :smile: 
Re: my orbitz problem
Quote:

Re: my orbitz problem
Quote:
welcome back tinytim, The cool part is we can interprete this picture as a pure elastic collision except the point of collision is not a physical collision. I'm trying to make this a model in microscopic scale. In theory sub atomic particle is so small that this model might suits well. I'll get back more on this. 
All times are GMT 5. The time now is 09:01 PM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums