Crashing the lunar lander (science / maths puzzle)

cantab

cantab

I'm sure many people here have at some point played a version of lunar lander. (If not, google it). And I'm also sure some of you have got bored of playing 'properly' and decided to deliberately crash at high speed.

The question then is

How do you obtain the maximum impact speed for a given spacecraft?

As in, when and how fast should you burn the fuel?

Consider the craft to be initially stationary at a certain altitude. (This may be zero.)

For a rocket engine, thrust is directly proportional to rate of fuel consumption, and does not depend on speed. For other sorts of engines (if you wish to extend to say ramjets) things are more complex.

It is up to you whether you want to take into account air resistance (probably negligible for a lunar impactor, probably NOT negligible for a Venusian one), change in gravity with altitude, or any other considerations.
 
I would assume that because thrust (ie force) is directly proportional to fuel consumption, it is best to just use all of your fuel at the beginning (wherever the highest point it), so that by the time you enter the atmosphere and face air resistance, you're already at the highest possible velocity (a greater starting force equates to a greater acceleration, especially since you will be losing the weight of the fuel as you use it) for that height and don't have to try to accelerate against the resistive forces. Unless I am utterly mistaken and just don't know what I'm talking about.

An increase in g is negligible, and even if you do take it into account, it's best to have a higher velocity to start with with the higher gravitational force.

In fact, I would assume that as long as you're able to use all of your fuel before you reach the ground, your ending velocity will be the same no matter what (assuming the thrust is forcing you straight downwards).

Taking AP Physics this year.
 
I would actually think it would be the other way. The energy you lose to air resistance increases as you travel faster through the atmosphere so you should use all your fuel at the last possible point to minimize losses of energy to air resistance.
 
That's true, as the resistive force is somewhat similar to friction (which increases as the normal force increases), but it just makes sense to me that no matter where you use the fuel, you'll have the same velocity at the end.

Think about it - if you are at a higher velocity and thus have higher resistive forces, you will accelerate at a slower rate but will still be traveling at a much faster speed. However, if you start at a lower velocity and use your fuel after you enter the resistive forces, as you accelerate those forces will become greater, but you will have more fuel left to counteract the decrease in acceleration. Either way, at ground zero, v is the same.
 
Think about it from an energy perspective. The fuel provides a set amount of energy and gravity provides a set amount of energy. However, if you accelerate earlier you will have a higher average speed while traveling through the atmosphere and thus incur greater resistive forces so you will lose more energy to these forces.

I'm pretty sure that if we ignore resistive forces then timing becomes irrelevant.
 
I will ask my physics teacher about this tomorrow. I understand the energy perspective but it still seems to me as if the ending velocity would be the same regardless of where you use your fuel.
 
If there is no air resistance (which there shouldn't be on the moon), then just fire your thursters constantly so that you go straight upwards until you run out of fuel. This will allow you to reach your maximum height, at which you have maximum potential energy (PE=mgh). As you fall, all this will be converted into kinetic energy (PE => KE = 1/2mv^2). You will notice that the highest possible PE will yield the highest possible velocity v. Constantly firing your thrusters downward could yield a better result, but the game itself is designed so that you'll hit the ground before you can use all your fuel going downward.
 
It's simple to show that how you burn your fuel DOES matter. Consider burning fuel such that the thrust equals your own weight - your craft stays put, then when you run out you freefall from the initial position. That clearly results in lower impact speed than the method Phantom IV gives.

I didn't clarify before: you can thrust in any direction, and assume rotation is instantaneous and requires no fuel. Impact speed is meant to be perpendicular to the ground, ignoring lateral motion, so you'll probably only want to thrust up or down.

I may work on a fuller analysis myself at some point.
 
I never played so i was not aware it doesn't let you us all of your fuel before you hit the ground.

I agree that rocketing upwards until you are out of fuel and v=0 so that you have maximum potential energy would net you the max possible speed at the bottom, but so would shooting up high enough that you can then face downwards again and thrust, using all of your fuel before you impact. This is because no matter what, you still have the same amount of energy - (mgh max)=(mgh moving + 1/2mv^2)
 
Work backward.

You have two forces working in your favor: gravity and thrust. You have one force working against you: friction.

Assuming we have air resistance, the most efficient use of fuel is to have the smoothest velocity (friction increases with the square of velocity, so you don't want high and low velocity, you want average velocity). However, unused fuel is extra energy you could have added to the collision. Therefore, the best way to use fuel is to use your last bit of fuel at the exact moment of collision, as this has the craft spending the lowest amount of time at the highest velocity, and thus gives friction less time to steal energy.

If you do not have enough distance to rocket straight into the ground, you should rocket upward until the fuel remaining is enough to finish at the exact moment the craft collides. This is a more efficient strategy than the "rocket upward until you run out of fuel" strategy because when you are going up, you then have two forces working against you: friction and gravity. You want to spend the least amount of time going up because you're fighting against a force instead of with it, and your thrust is a constant force that is greater than gravity (so there is no risk of the extra time spent on dropping due to gravity will be stronger than the force from your thrust). So you go straight up to increase your potential energy, but have enough fuel left to constantly accelerate downward at the end, because you get more kinetic energy from thrust than gravity.
 
obi said:
If you do not have enough distance to rocket straight into the ground, you should rocket upward until the fuel remaining is enough to finish at the exact moment the craft collides.
Click to expand...
This is my intuitive expectation, whether or not their is air resistance. Did you carry out a full mathematical analysis, or are you just working from principles and intuition?
 
Well, my solution will work regardless of air resistance. Without any force of friction, there are multiple solutions that should give the same result. The case of there being air resistance just eliminates all other solutions. I didn't conduct any full analysis beyond what I put in my post.

As far as I can see, the two extreme cases here are: You rocket up as far as possible using all of your fuel, and then gravity pulls you back down, or you rocket just high enough up so that the maximum height you reach (rocket up, turn off engines, turn engines back on pointing the opposite direction when the craft begins to fall) gives you just enough time to use up all of your fuel when you crash into the ground.

Turning off the engines until your begin your descent is clearly a superior strategy to rocket up and immediately begin rocketing down, however. A craft that begins falling at height h will not fall with as much force as a craft that begins falling at height h + a. If you rocket up and then immediately begin rocketing down, you will reach your maximum height at a lower point than if you turn off the engines and weight for gravity to halt your momentum (you have gravity + thrust slowing you down in one scenario, and just gravity in the other, so you'll get higher). Therefore, you'll have less potential energy. You will also have used up fuel to do this, so you have energy in the form of fuel, meaning less total thrust available.
 
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