Let's assume there is no atmosphere and let's assume there is no change in weight due to fuel consumption, will reactive rocket need the same amount of fuel for landing on a planet as for take off?
In theory – I think – you need the same escape velocity to get rocket to orbit as you need to break it to 0 speed after free fall from the orbit, but this changes if the descend is slower than free fall, is that right?
What is the real world(moon) situation in case of lunar module? (extrapolating for fact that the Apollo Lunar Module leaves the descend stage behind)
Best Answer
It is a bit more complicated. You should be able to land without the use of energy but you have to pass your momentum to something - and this passing on of momentum will be the same in both cases. Let us consider a "spherically-symmetric-chicken-in-a-vacuum" model to see this explicitly.
Momentum conservation and propelled mass
The spacecraft of mass $M$ is hovering just a bit above the surface of the planet at zero vertical velocity (it is hovering above the surface to be able to propel fuel downwards). Suddenly, it propels a single pulse of fuel with velocity $v_{\rm f}$, mass $m$ and total momentum $p_{\rm f} = m v_{\rm f}$ downwards. Due to momentum conservation, the spacecraft will start to move upwards with a momentum with equal magnitude. If the pulse is just enough for the spacecraft to escape the planet with velocity $v_{\rm e}$ it must hold that $$p_{\rm f} = m v_{\rm f} = M v_{\rm e}\,,$$ or that the velocity of propulsion of the fuel must be $$v_{\rm f} = \frac{M}{m} v_{\rm e}\,.$$
But your assumptions are that $M \gg m$ which in combination with the previous formula leads to $v_{\rm f} \gg v_{\rm e}$. Even though the model is vastly simplified, the basic conclusions are valid. In more complicated models you will continuously loose mass so that the propulsion will have varying effects but the "before vs. after" equation will look the same on average because momentum conservation always rules.
The very same relation would be valid for a descent. Say you have fallen from rest at infinity up to the surface of the planet and due to energy conservation now have a velocity equal to $v_e$ - but downwards. Momentum conservation! You have to throw the momentum away not to get smashed into pieces, so you propel all your momentum into a single fuel pulse downwards. The momentum of this pulse must be again equal to $p_{\rm f}$ so all the previous relations hold.
Energy balance
It is true you have to propel the same amount of mass at the same speed in both cases, but it is also interesting to see what happens to energy. In the takeoff you have before the pulse $E_0=0$ and after the process a lot more kinetic energy due to the motion $$E_{1} = \frac{p_{\rm f}^2}{2m} + \frac{p_{\rm f}^2}{2M}$$ I.e., the smaller the $m$, the more energy you use in the takeoff. However, in the descent you have before the pulse $$E_0 = \frac{p_{\rm f}^2}{2M}$$ and after the pulse $$E_{1} = \frac{p_{\rm f}^2}{2m}$$ so in principle, it should be possible to save a lot of energy just by an elastic "passing-on" of the momentum. This is actually something we know very well from practice where we have the atmosphere to give the momentum to via a parachute or such.
From this "fuel-pulse" mode you can see that if you are able to pass your momentum to a mass of about your magnitude, in principle you should be able to land without burning a drop of fuel (I.e. without exerting any stored chemical/electric/nuclear energy).
The fact is that even if you just want to use mass and energy stored in your spacecraft, you will not be accelerating it all from zero but from the escape velocity $v_{\rm e}$ to $v_{\rm f}$. You can thus use your fuel to push and propel some dead carriage because your engines will actually propel at their original speed plus $v_{\rm e}$ in the planet rest frame.
CONCLUSION: You have to propel roughly the same mass at roughly the same speed both during take-off and landing, but you do not have to burn so much fuel during landing.
Just a note on the assumption of no mass change for a reaction engine. For a Moon to Earth escape we have $v_{\rm e}\approx 11 000\,\rm m/s$ which is way above the possibilities of a classical bipropellant rocket with maximum $v_f \approx 5000 \, \rm m/s$. Even if you used some of the state of the art ion thrusters, you would not reach more than a few times more than $v_e$. But that is in conflict with the conclusion $v_{\rm f} \gg v_{\rm e}$ from the momentum conservation balance and the assumption $M \gg m$. So the assumption of a small propelled mass $m$ in comparison with the mass of the spacecraft $M$ is not realistic.