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pedrom

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I can understand the mathematics behind the deduction of "expanded" (using Newton's laws) Kepler's First law that shows the orbits can be only parabolic, hyperbolic, linear or ellipses, but a friend (who doesn't know derivatives and differential equations) asked me why, taking in account only the gravity of the sun, the earth will not "fall", I would like to know if it's possible explain that using only concepts like momentum/energy conservation, Newton's laws, etc, without using any complicated mathematical formulas.

And I have another question, the r=k/(1+e*cos(theta+theta0)) equation only shows the positions of a orbiting object at steady state? I mean, if there is a rocket near to the earth with "engine on" it will not follow this position (because the resultant force will not be only the gravity), but, one moment after turning the engine off (assuming the only force acting will be the gravity) it will start following this position equation, or it will take some time to happen?

And I have another question, the r=k/(1+e*cos(theta+theta0)) equation only shows the positions of a orbiting object at steady state? I mean, if there is a rocket near to the earth with "engine on" it will not follow this position (because the resultant force will not be only the gravity), but, one moment after turning the engine off (assuming the only force acting will be the gravity) it will start following this position equation, or it will take some time to happen?

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