newtonian mechanics - 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).

Answer

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 $$F={d mv \over dt}=v{d m \over dt}$$