Discrepancy of the Keplerian speed by a factor when calculated with different methods I needed to derive today the Keplerian speed for a circular motion in a gravitational field. Just for the hell of it I decided to derive it once via the forces and once via the energies and strangely there is a discrepancy by a factor of $\sqrt2$ !? I must have done something wrong, so please let me know. Here are my calculations, first via the forces:
$F_{rad} = F_g$
$\frac{mv^2}{r} = G \frac{Mm}{r^2}$
$\Rightarrow v = \sqrt{\frac{GM}{r}}$
All is in agreement with the Keplerian speed given in text books and wikipeadia. But let's see what happens when we derive the speed $v$ from the energies:
$E_{kin} = E_{g}$
$\frac{1}{2}mv^2 = G \frac{Mm}{r}$
$\Rightarrow v = \sqrt{\frac{2GM}{r}}$
Now unfortunately I have a discrepancy of $v$ from the known formula by $\sqrt2$. It seems that the discrepancy would be resolved if I apply the virial theorem $E_{kin} = \frac{1}{2}E_{pot}$. But I thought the virial theorem is only to be applied to systems with many particles. In the above equations, I am only observing the velocity $v$ of a test particle around a massive central mass with $M>>m$ in a Newtonian limit. So what am I misunderstanding here? Do I really need to use the virial theorem? What is the justification for that? Tnx
 A: The problem is that the for an orbit the kinetic energy is not equal to the potential energy.
To use conservation of energy you have to compare the energy of a system at two different instants in time.
One instant could be "the object is in a circular orbit", where
$$ E_k = \frac{1}{2}mv^2 \quad \mathrm{and} \quad E_p = -\frac{GMm}{r}$$
so
$$E_{tot} = E_k + E_p = \frac{1}{2}mv^2 - \frac{GMm}{r}.$$
We can think of this as the final state of our system.
But $E_{tot}$ is not equal to zero.  You'll need another point of reference, an initial state, where you can determine $E_{tot}$ and any work done to get from your initial state to the final orbit.
Only then can you use conservation of energy to connect things.
$$E_i + W = E_f.$$
From the force perspective you can derive the relation between kinetic and potential energies of a circular orbit by substituting $E_k$ and $E_p$ into the second equation in your question.
Since $2E_k = -E_p$, we can see that the total energy of the orbiting state is negative.  This is because the system is in a bound state.
If you gave the system more energy so $E_{tot}\ge0$, you'd have an unbound state, where the orbiting body can fly off to infinite distance.
In your energy equations you implicitly solved for when $E_{tot} = 0$.
This would occur when $v\rightarrow 0$ and $r\rightarrow \infty$.
Your second solution considers the situation where the particle is a distance $r$ from the central mass and has speed $v$ at the initial point, and at the final point the particle has escaped to infinity:
$$ E_i = E_f $$
$$ \frac{1}{2}mv^2 - \frac{GMm}{r} = 0 $$
You found the escape velocity.
A: The orbital velocity in a circular orbit is $ \sqrt{\frac{G(M+m)}{r}} $.
The escape velocity of an orbit is $ \sqrt{\frac{2G(M+m)}{r}} $.
