Does lunar module need the same amount of fuel for landing and take off? 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)
 A: 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.
A: 
Let's assume there is no change in weight due to fuel consumption.

That's a rather unrealistic assumption. If you get rid of that assumption, landing requires considerably more energy than does ascent.
Some realistic assumptions: The lander


*

*Starts docked to some orbiting spacecraft,

*Separates from the orbiting spacecraft,

*Performs a de-orbit maneuver that puts the lander on a ballistic trajectory toward the Moon's surface,

*Coasts ballistically for some time before starting a braking maneuver,

*Follows an optimal gravity turn during the braking maneuver,

*Transitions to an approach phase where the lander or its occupants look for a good landing spot,

*Transitions to a terminal descent phase where the vehicle cancels horizontal velocity and slowly moves down the last tens of meters to the Moon's surface.

*Spends some time on the lunar surface,

*Launches from the lunar surface and follows an optimal gravity turn during the boost phase,

*Coasts ballistically for some time until it comes close to the orbiting spacecraft's altitude,

*Circularizes its orbit at that altitude,

*Approaches the orbiting spacecraft,

*Docks with the orbiting spacecraft, and

*Finally, ends docked with the same orbiting spacecraft.


These steps are not symmetric about step #8, spending time on the lunar surface. On the way down, separating from the spacecraft (step #2) requires considerably less energy than does approach and docking (steps #12 and #13). Separation is a simple matter of firing some explosive bolts. Approach and docking is a delicate dance that requires expenditure of fuel. The ballistic coasts (steps #4 and 10) cost the same amount, zero. That's what "ballistic" means. So far, landing is cheaper than launch. These are the only steps where landing is cheaper than or as cheap as is launch.
The de-orbit maneuver (step #3) has a significantly greater energy cost than does the comparable maneuver after launch (step #11). While the delta-V costs of these maneuvers are more or less the same, the landing vehicle is full of fuel while the launched vehicle is almost empty of fuel.
The gravity turns (steps #5 and #9) have similar delta-V costs, but these are not quite symmetric due to the changes in mass. This means the delta-V cost of step #9 (ascent) is slightly less than the delta-V cost of step #5 (descent). Translating that delta-V to energy means that step #5 costs more energy than step #9.
If you look at the above, you'll see that I omitted steps #6 and #7. These are rather expensive steps, and there is no counterpart to these steps in the launch phase. These two steps could be done away with if there was a beacon on the surface of the Moon that said "LAND HERE!" With that capability (which doesn't exist yet), tiny corrections during the braking phase could do away with those two steps. Without that capability (which is what the Apollo landers had to deal with), those rather expensive steps meant that landing required an additional ~300 meters/second in terms of delta-V than did the launch phase.
The Apollo program reduced the energy cost even more by making the descent stage and ascent stage into separate vehicles. The descent stage had a delta-V capability of 2500 m/s and had to carry 8200 kg of fuel to accomplish this. The ascent stage had a delta-V capability of 2220 m/s and had to carry only 2353 kg of fuel to accomplish this.
A: Alternative - just because it's fun to think about these things.
Moving away a bit from the traditional rocket science, in principle you could land with very little fuel and a good set of wheels/brakes: apply a small fuel burn to change your orbit just enough to make a glancing pass at the surface of the moon, then apply the brakes as you touch down. You dissipate the energy as heat with very little fuel burn (passing the momentum to the moon instead of to the burnt fuel). But you can't easily regain momentum without using a source of energy (regenerative braking?). This would make the take-off much more costly.
Note that the orbital velocity at the surface of the moon required for this maneuver is approximately 1500 m/s - as I said you would need a very fine set of wheels / suspension...
A: Landing is gravity-assisted, so requires less energy. A spacecraft on the moon has used up fuel resulting in its mass being less. If it is significantly less, then it takes less energy to lift a less massive craft back into orbit, than landing a heavier craft.
Escape velocity applies only to non-powered projectiles, such as shooting a canon ball. There is no reason why you could not lift a craft at a constant velocity of 1m/s, if you apply a constant force.
A: Now it occurred to me that the escape velocity/breaking may be symmetrical - that means you need the same amount of energy to counter the gravity on the way up as you need on the way down - but what also matters is how long you stay "hoovering" in the gravitation filed. This is what consumes fuel no matter what way you go and in reality both descend and ascend are slower than free fall(/escape speed). So your total fuel consumption will consist of fuel for generating the escape/breaking velocity plus fuel to compensate for the fact that you are not breaking or accelerating instantly.
I think that in the case of a rocket we can assume that the take off is much slower than the descent, which means you need more fuel for the take off.
Am I right?
