THIS IS NOT HOMEWORK.
I'm actually in Physics I atm, and am an avid player of Kerbal Space Program. I hope to do my CS masters thesis several years from now on something space related. But none of that matters.
You're in a rocket, and one moment ago you were orbiting a nice moon (no atmosphere), magically all your horizontal velocity is gone and you are now falling directly towards the moon at the rate of g (which is technically a function of altitude).
Your rocket magically points straight up (with the engine pointing down) for you, at some point you know you want to fire the engine so you can land within some acceptable threshold so your ship doesn't explode on impact killing you.
But, you have no idea how much fuel you have, so you want to use the least amount of fuel possible, and you're smart enough to know that the longer you spend above the planet, the more fuel you're going to waste, so you want to start burning at the last second, leaving just enough time for you to get your downward velocity within that near zero margin so you don't die.
This is known as a "suicide burn."
Problem is, the moment you start burning (let's assume you either burn 100% or 0%), you immediately start increasing the time you spend above the planet, which changes when you should have started.
Let's also assume that your ship has a max thrust, which is greater than g but definitely not infinite.
So to be clear, we're looking for a time to start burning such that you land no harder than x m/s.
My question is, how would I approach such a problem?
I asked my physics professor this question and he said he'd "think on it" so I decided to ask the Internet, thanks!
Answer
I will will assume that you fall perfectly vertical and that the celestial body does not rotate.
If you assume that the mass of the rocket stays constant, then you can find when to start with the suicide burn using time reversal. Namely you "start" st the surface with your desired final velocity, $v_f$, and thrust upwards until your (specific) orbital energy matches that of your current trajectory. You can simulate this as a function of time, however in this case it will be easier to use conservation of energy,
$$ \frac{v_f^2}{2} - \frac{\mu}{R_f} + \frac{Fh}{m} = \frac{v_i^2}{2} - \frac{\mu}{R_i}, $$
where $\mu$ is the gravitational parameter of the celestial body, $R_f$ the final radius (of the surface) relative to the center of mass of the celestial body, $h$ the altitude above the surface at which you have to start the suicide burn, $F$ the amount of thrust the rocket can provide, $m$ the mass of the rocket, $v_i$ the initial velocity of the rocket and $R_i$ the initial radius of the rocket relative to the center of mass of the celestial body.
In order to find the time, $T$, it takes to perform this burn you will have to calculate the integral,
$$ T = \int_0^h \left(v_f^2-\frac{2\mu x}{R_f(R_f+x)}+\frac{2Fx}{m}\right)^{-1/2}dx, $$
however there is no general solution for this, so it will have to be calculated numerically. The used $\Delta v$ can be found with $\Delta v = \frac{F}{m}T$.
If you do want to take into account the variable mass of the rocket then you can also use time reversal and "start" at your final radius and velocity and go back in time, until your specific orbital energy matched that of your initial trajectory. For this you will have to initially guess what your final mass will be, such that at the start of your burn your total mass is equal to $m$.
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