That extra speed, enough to sustain a stable orbit, makes all the difference! We need to compute the speed on a circular orbit, to keep things simple. The radial acceleration is equal to the pull of gravity:

$$ m\frac{\Delta v}{r_f} = \frac{GMm}{r_f^2},$$

which results in the ultra-classic

$$ \Delta v = \sqrt{\frac{GM}{r_f}}.$$

That extra speed, enough to sustain a stable orbit, makes all the difference! We need to compute the speed on a circular orbit, to keep things simple. The radial acceleration is equal to the pull of gravity:

$$ m\frac{\Delta v}{r_f} = \frac{GMm}{r_f^2},$$

which results in the ultra-classic

$$ \Delta v = \sqrt{\frac{GM}{r_f}}.$$

But now you deal with a rocket here and there is an essential point your forgot: the rocket will loose mass as it burns fuel! Let's then analyse the scheme you developed in your question and in your comments to this answer of mine. You want to do two steps.

• In step 1, your rocket, fully loaded with a mass $m_0$, starts at radius $r_o$ with a speed of 0. It then reaches radius $r_f$ with some speed $v_1$, and its mass has been reduced to $m_1 < m_0$.

• In step 2, your rockets then reaches the same altitude $r_f$ but at the orbital speed $v_2$ I wrote above, with a further reduced mass $m_2 < m_1$.

Let us compute the energy balance for each step. We need to take into account the energy produced by the rocket engine. Let's say that burning a mass $m$ of fuel produces an energy $km$ for some constant $k$. The value of $k$ could be computed from the enthalpy of the combustion reaction.

In step 1,

$$ k (m_0 - m_1) = \left(\frac{1}{2}m_1 v_1^2 - \frac{GMm_1}{r_f}\right) - \left(0 - \frac{GMm_0}{r_0}\right)$$

In step 2,

$$k(m_1 - m_2) = \left(\frac{1}{2}m_2 v_2^2 - \frac{GMm_2}{r_f}\right) - \left(\frac{1}{2}m_1 v_1^2 - \frac{GMm_1}{r_f}\right)$$

By summing these two equations

$$k(m_0 - m_2) = \left(\frac{1}{2}m_2 v_2^2 - \frac{GMm_2}{r_f}\right)- \left(0 - \frac{GMm_0}{r_0}\right)$$

which is the amount of fuel to burn to do it in one step, let's call it D for "direct", from motionless at radius $r_o$ to orbital speed at $r_f$. No surprise here, we are dealing with conservative forces but I just wanted to make explicit to you.

There is a big problem however. The design of the rocket engine will fix the rate of mass decrease. Since step 1 + 2 will take longer than step D, if you have enough fuel for step D, then that same amount of fuel will be burnt out before you can finish step 1 + 2. So you will have to provision more fuel for step 1 + 2. A quantitative answer would require solving the equations of motion. So clearly, just looking at energy balances is not enough here.