On Conditions of Rocket Lift-off

I’ve encountered with a post in a forum asking that what the conditions would change if a rocket is lift off on the moon rather than on Earth. It’s an intriguing question and therefore I spent some time attempting to solve the problem. And now the solution which I believe is correct is shown as follows.

Let$F$denotes the gravity, while$M$and$\mu$

respectively denote the mass of the rocket itself and the one of fuel at a moment. After a moment of$\mathrm{d}t$, the combustion engine exhausts$\mathrm{d}\mu$mass of the fuel.

To simplify the problem, I ignore the details how a rocket lift off from the origin stationary status otherwise the force from the ground is likely to complicate the scenario, i.e. somehow the rocket now has already left the ground but I need to consider the conditions of the force that can drive the rocket up. Let the rocket’s velocity right at this moment be$\vec{u_{0}}$, and the gush-out fuel’s velocity with respect to the rocket’s be, supposingly, a constant$\vec{u_{0}}$.

Then according to momentum law, we can yield the following equation:

$\vec{F}{\mathrm{d}t}=(M+\mu-{\mathrm{d}\mu})\vec{u}+{\mathrm{d}\mu}(\vec{u}+\vec{v}_{rel})-(M+\mu)\vec{u}_0$

Now supposing that the rocket’s motion is parallel to the gravity, so all of the vectors can be introduced into scalar quantity easily. Let the gravitation to be a negative value, hereby yielding

$-F{\mathrm{d}t}=(M+\mu-{\mathrm{d}\mu})u+{\mathrm{d}\mu}(u-v_{rel})-(M+\mu)u_0$

Step further, we have

$-F{\mathrm{d}t}=(M+\mu){\mathrm{d}u}-{\mathrm{d}\mu}(u_0+v_{rel})$

or

$F+(M+\mu)\dot{u}=\dot{\mu}(u_0+v_{rel})$

Introducing gravitational acceleration, hence the gravity becomes$F=(M+\mu)g$, which then substitutes the equation above, yielding the final equation as following:

$\huge g=\frac{\dot{\mu}(u_0+v_{rel})}{(M+\mu)}$

Now we can draw conclusions from this equation that compared to lift-off on the earth as $g$ is smaller, there’s no need for the rocket covering so swiftly as on Earth, the gushing fuel’s speed with respect to the rocket itself doesn’t have to be so fast as well, and furthermore, the mass of fuel exhausted per unit time is unnecessarily so larger as on Earth.

Despite that there’s a omission of the detailed discussion about how a rocket can leave the ground, I think these conclusions are quite reasonable.