Orbital escape velocity [closed]

Question

Given a satellite (black) orbiting a planet (green) in a circular motion with $r_o$ being the orbital radius.

1. The orbital the velocity of the satellite is $v_{orbit}=\sqrt{\frac{GM}{r_o}}$, right?

2. If the satellite would have started from the earths surface it would have at least needed the escape speed $v_{escape}\geq\sqrt{\frac{2GM}{r_e}}$ with $r_e$ being the earths radius ($=6371$km), to escape the earth's gravitational field for ever, right?

3. If the satellite is stable in it's circular orbit $r_o$ and I then would permanently increase $v$ to a velocity of the satellite in the range of $v_{orbit} < v < v_{escape\,orbit}$ it would just turn the orbit into an ellipse, right? [as suggested here: what-happens-to-a-satellites-orbit-when-you-speed-it-up]

4. So it would suggest that if the satellite is stable in it's circular orbit $r_o$, the velocity to escape the earths gravitational field from this orbit for ever would be to increase the velocity of the satellite to $v_{escape.orbit}\geq\sqrt{\frac{2GM}{r_o}}$, right? [This answer suggests differently if I understand it correctly: https://physics.stackexchange.com/q/382753]

5. So why is it that when comparing the oposing gravitational force $F_G$ and centrifugal force $F_Z$ with $F_G = F_Z \rightarrow \frac{GmM}{r_o^2}=m\frac{v_{orbit}^2}{r_o}$ a small increase in velocity would lead to a small increase in radius by which $F_G$ would decrease quadratically, but $F_Z$ just linearly. So it would suggest that an infinitally small increase in radius at a stable orbit would lead to an irreversible escape out of the orbit. [Which would suggest the solution from Farcher at the answer above]

So Is there something wrong with my logic or the understanding of the given questions/answers. How would my statements above be true? Is it generally better to think in terms of energy than forces when talking about orbits?

Answer 1

In (5) you say "a small increase in velocity" (whichI'll take to mean the azimuthal speed). Of course there are many ways to increase the velocity; since we're interested in circular orbits, let's say you pick one such that the orbit will remain circular. What you'll find is that if you (infinitesimally) increase the speed, then the satellite will (infinitesimally) increase its radius, as you expect, which will then cause a (infinitesimal) decrease in speed to match the new circular speed at the new orbital radius. This is of course given that we've contrived to keep this object on a circular orbit.

I would agree that thinking in terms of energy is more intuitive: if you go through that reasoning you'll find that a bit of added energy will end up partitioned between increasing the potential energy and increasing the kinetic energy, and you won't see the same misleading hint of an unstable/runaway process. In a spherical potential like the point mass you're considering, (total) energy and angular momentum are conserved along any orbit, which make them very convenient to work with indeed. This approach also allows handling more general (non-circular) orbits much more easily.

Answer 2

Your conclusion in 5.) is incorrect as the velocity decreases when the satellite goes to larger r as it does work against gravity. Assume you have initially the circular orbital velocity $v$ and increase this suddenly by an amount $\Delta v$. This increases the orbital angular momentum from $mrv$ to $mr(v+\Delta v)$. And angular momentum conservation requires that in the apogeum (largest distance) $r'$ of the subsequent elliptic orbit

$$mr'v'=mr(v+\Delta v)$$ so $$v'=(v+\Delta v)\frac{r}{r'}$$

The centrifugal force in the apogeum is therefore

$$F'_Z=\frac{mv'^2}{r'}=m(v+\Delta v)^2\frac{r^2}{r'^3}$$

So the centrifugal force decreases actually faster with $r'$ than the gravitational force, which means that, on the contrary, the satellite will be falling back again in its elliptic orbit to smaller radii rather than escaping.

Answer 3

From the expression: $$v_o = \sqrt{\frac{GM}{r_o}}$$ we can get the stable circular radius of the orbit as a function of $v_o$: $$r_o = \frac{GM}{v_o^2}$$ The effect of a small change of velocity on the stable circular orbit is: $$\delta r_o = \frac{\partial r_o}{\partial v_o}\delta v_o = -\frac{2GM}{v_o^3}\delta v_o$$ The minus sign indicates that for an increment in the velocity, the stable circular orbit is now more internal, what is logical.

Of course it is possible to increase the (instantaneous) velocity and go to a more external orbit. But it will be no more circular.

Answer 4

In a circular orbit, the magnitude of the kinetic energy of an object is half the magnitude of the potential energy (which is negative when measured from infinity). (In an exchange, the fractional change in the K.E. is larger.) If the object is given extra energy, it will move out from the circular orbit. The potential energy goes up (toward zero) and the kinetic energy (and velocity) go down. Eventually, the tangential component of velocity is less than what is required for a circular orbit and the radial component starts to decrease. When the object reaches the far side of an elliptical orbit, the radial component is zero, but still accelerating toward the central body. To escape from a circular orbit, the total energy must be zero.