I want to know what roughly happens if we hit an orbiting body a little. And how is it possible that most of the objects have seemingly circular orbits around. How do they stabilize? I ask because it seems to be that celestial objects must recede from their orbits. The orbit is simply a distance r from the massive object and it depends on the speed v of the orbiting body.
We know that $F=ma=mv^2/r$ and gravitational force is $mMG/r^2$ so that speed of orbiting must be $v_o^2=MG/r_o$ or $v_o=\sqrt{MG/r_o}$, which implies kinetic energy of $E_{kin} = mv^2/2 = mMG/2r$. Also, the binding energy is $U = mMG/r_o = E_{kin}/2$. That is fine. It says that the kinetic energy of our orbital motion is a half of binding energy. That is, our speed is lower than the escape velocity, which from the $mv_e^2/2 = mMG/r$ is $v_e = \sqrt{2MG/r_o} = \sqrt 2 v_o$, twice the orbiting speed. I guess that we will fall onto the Earth surface if move lower than $v_o = \sqrt{MG/r_o}$ and escape the Earth into infinite space if move $\sqrt 2$ faster than that. My question is basically, what happens if we move at the speed between orbiting speed and escape velocity, what would be the orbit?
Furthermore, I wonder how do objects stabilize at their natural orbits. Suppose we have accelerated our naturally orbiting object at speed $v_o$ k times. This means that new speed is $v_2 = k v_o$ and radius must be reduced since orbiting speed and radius are inversely proportional, $r_2 = MG/v_2^2 = r_o/k^2$. However, obviously, accelerated body will not tend towards the attractor, it will tend to escape it. That is why I wonder.
Satellite, accelerated from its $v_o$ to $v_2$ at height $r_o$ will raise up but loose as getting higher. As computed above, $U_{binding} = 2 E_{kinetic}$ so $E_{total} = {3 \over 2} mv_o^2$ and (here, $v_{on}$ stands for natural speed at orbit n) we have $$E_{total2} = {3\over 2} mv_{o2}^2 = E_{total1} = (U = mv_o^2) + E_{KinAcelerated} =\\ mv_o^2 + {m\over 2} v_2^2 = mv_o^2 + {m \over 2} k^2 v_o^2 = mv_o^2(1+{k^2 \over 2})$$ whence speed at the raised orbit will be $v_{o2} == v_o \sqrt{{2 \over 3} (1+k^2/2)} = v_o \sqrt{(k^2 +2)/3}$
That is, if we accelerate the satellite 1 times, $v_0 \rightarrow v_0 $ then the orbit will not change, $v_{o2} = v_o \sqrt{(1 +2)/3} = v_o$. Voila! Increasing the speed $\sqrt{2}$ times, up to $v_e$, will make $v_{o2} = v_0 \sqrt{4/3}$, which corresponds to the orbit $r_2 = 3 MG / (4 v_o)$. It did not gone into infinity. It was reduced by 4/3 with respect to $r_0$. We declined instead of raising the orbit and speed has accelerated! Again, the theory says that satellite must reduce the height when it is accelerated. But, I do not believe that this is what happens in reality. In reality in should go higher and even higher because despite of loss of the speed, attraction force is also reduced at higher heights. On the other hand, if most of the observed satellites orbit at their natural orbits it means that the formula is right and accelerated satellite will go down rather than raise higher.
I assume that it is not important for the orbiting and escaping speeds if our speed is horizontal or vertical. I guess that absolute value is only important and do not know how to make the computations otherwise.
