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.


1 Answer 1


There are three possible forms of motion for a small object in a $\propto \frac{1}{|r|}$ potential:

  1. hyperbolic motion
  2. parabolic motion
  3. elliptic motion

The first and second possibilities are unbound, so the object comes from infinity and will go back to infinity afterwards.

The third one is bound, which means that the object cannot escape the gravitational pull. This is the case for satellites.

You seem to assume a satellite in a perfectly circular orbit, which is a special case of the elliptic one. In circular orbit the velocity does not change, because the object is equally far away from the center at all times.

If a satellite in circular orbit were to be accelerated in a short burst, it would leave its orbit for an elliptic one.

To get a better grasp on orbital mechanics, you could try playing with a simulator or the famous video game "Kerbal Space Program".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.