How to find a condition that ensures that the rocket immediately takes off? For the rocket in a constant gravitational field, how to find a condition that ensures that the rocket immediately takes off?
Update: I apologize. Let's try again. So my question:  For the rocket in a constant gravitational field, find a condition (involving the constants $m_0,μ,g,v_r$) that ensures that the rocket immediately takes off.
$m_0$ is the mass of the rocket at $t=0$;
$μ$ is the mass per unit time of expelled gas;
$v_r$ is the velocity of the expelled gas relative to the rocket.
After I want to show that, in this case, the velocity of the rocket always grows (until the fuel is used up).  
 A: If $m_0$ is initial mass of rocket, $\mu$ the mass of gas ejected per unit time ($dm \over dt$), $g$ acceleration due to gravity and $v_r$ speed of fuel leaving the rocket motor then....
$$Force ~down = m_0g$$
$$Force ~up =  v_r {d m \over dt} = v_r \mu $$
and condition for lift off is 
$$m_0g \lt v_r {d m \over dt} $$
or
$$m_0g \lt  v_r \mu $$
(may need a minus sign on the right hand side if we consider $dm/dt$ to be negative.)
When you think about times after lift off then you need to look at net force up and how the mass of fuel is lost from the rocket.
Note that force can be calculated from the rate of change of momentum. So $F=ma$ is the same as $F=m {dv\over dt}$ and $F={d mv \over dt}$ or $F={dp \over dt}$.
Here if we differentiate momentum ($p$ or $mv$) with respect to time the $v$ is constant, but the $m$ changes with time so  ${d mv \over dt}$ = $v{d m \over dt}$. 
Putting this all together we get 
$$F=ma=m {dv\over dt}={d mv \over dt}$$
if the mass is constant.
But in this case 
$$F={d mv \over dt}=v{d m \over dt}$$
A: The forces acting on the rocket:
$$F_{thrust} = v_r \mu$$
$$F_{gravity} = - m(t) g = - (m_0 - \mu t) g$$
Net force on the rocket:
$$F_{net} = \Sigma F = F_{thrust} + F_{gravity} = v_r \mu - (m_0 - \mu t) g$$
Acceleration of the rocket:
$$a = \frac{F_{net}}{m(t)} = \frac{v_r \mu - (m_0 - \mu t) g}{m_0 - \mu t}$$
$$a = \frac{v_r \mu}{m_0 - \mu t} - g$$
If we examine the equation for $a$, you can see that $a$ monotonically increases with time, until the rocket runs out of fuel (at which point $\mu$ abruptly becomes zero), so we don't need to worry about the rocket losing positive acceleration then regaining it (note - this requires that we assume that the total rocket mass $m_0 - \mu t \ge 0$, which is good, since negative rocket mass would be non-physical).
Here is an interactive Desmos graph showing the function $a(t)$, with sliders for the other parameters.  It is instructive to play with the various parameters to see how it affects the acceleration of the rocket over time.
Condition for acceleration to be positive:
$$a > 0$$
$$\frac{v_r \mu}{m_0 - \mu t} > g$$
Being careful with the inequality, we can solve for t:
$$t > \frac{m_0}{\mu} - \frac{v_r}{g}$$
This is the time after ignition after which the rocket's acceleration becomes positive; when the rocket becomes light enough to lift off.  
Here is a graph showing a situation in which the rocket is not light enough to take off immediately:

If we set t = 0, we get the condition that the rocket lifts off immediately:
$$0 > \frac{m_0}{\mu} - \frac{v_r}{g}$$
$$v_r \frac{\mu}{m_0} > g$$
