It's only rocket science! (II)

Classical Mechanics Level 4

A rocket, dry mass $100\text{ kg}$ needs to stay above a certain point for as long as possible. It has $100\text{ kg}$ of fuel that it can burn and eject at $500\text{ ms}^{-1}$ . Taking $g=9.8\text{ ms}^{-2}$ (and assuming the gravitational field is uniform) how long can the rocket remain above that point for?

Enter your answer in seconds to 2 decimal places.

Details and Assumptions:

The rocket can go above that point but never below it.
The rocket starts on that point. There is no upper limit to the acceleration of the rocket, only the maximum fuel.
Fuel can be burnt at any rate but must be burnt sequentially (i.e. the first unit of fuel burnt must accelerate all the others)

The answer is 70.73.

1 solution

William Whitehouse
Mar 9, 2017

First of all we need to notice that the total impulse on the rocket is fixed: the first unit of fuel burnt will accelerate all of the other units of fuel meaning part of it's Impulse it "wasted". To calculate the "useful" impulse we first formulate an equation for what proportion of Impulse is useful at any point, that is:

$I_{Useful}=\frac{100}{M}*500*\Delta M$ 100/M is the proportion of the impulse going to the rocket and 500 is thrust velocity (or impulse per unit mass).

Converting this into an integral between 200 and 100 (maximum and minimum mass):

$50000*\int^{200}_{100}\frac{1}{M}\delta M = 50000ln(2) \approx 34657kgms^{-1}$

So now we just need to decide how this is best spent. This can change the momentum of the rocket by $346.57ms^{-1}$ so after 346.57/9.8=35.364s the rocket will have to be travelling down (all possible impulse has been cancelled by gravity). In the optimal scenario the rocket will now start falling and take another 35.364s to pass the point again (if all the fuel is burnt and expelled instantly at t=0)

so the answer is 35.36*2=70.728 $\approx$ 70.73

It's only rocket science! (II)

The answer is 70.73.

1 solution

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