Hover jet

Classical Mechanics Level 3

A jump jet of mass $\SI{400}{\kilogram}$ (without fuel) can hover, stationary, by burning fuel and blasting the waste products downwards at a constant velocity of $\SI[per-mode=symbol]{1000}{\meter \per \second}$ (relative to the jet). It is carrying $\SI{100}{\kilogram}$ of fuel.

How long, in seconds, can the jump jet hover like this before it runs out of the fuel?

Assume that the fuel contains all of the reactants: i.e. no air is being drawn from outside. (A real jump jet would take in air from outside, which would make up a substantial fraction of the exhaust mass.)

Take the acceleration due to gravity to be $\SI[per-mode=symbol]{10}{\meter \per \second \squared}$ .

Give your answer to 1 decimal place.

The answer is 22.3.

2 solutions

Rohit Gupta
Feb 20, 2017

Relevant wiki: Rocket Physics

When the hover jet loses its mass at a rate then it experiences a force called the thrust force. This thrust force equals $F_{\text{thrust}} = - v_{\text{rel}} \times \dfrac{dm}{dt}$ .

Here, $v_{\text{rel}}$ is the speed of the waste exhaust relative to the jet and the negative sign denotes that the mass of the jet is decreasing with time as the fuel burns.

For the hover jet to remain in equilibrium, this thrust force must balance the weight of the jet. Therefore,

$\begin{aligned} mg &= - v_{\text{rel}} \times \dfrac{dm}{dt} \\ \int_0^{{t_F}} {dt} &= - \frac{v_{rel}}{g}\int_{500}^{400} {\frac{1}{m}\,dm} \\ {t_F} &= 100\ln \left( {\frac{5}{4}} \right) = \boxed{22.3} s. \\ \end{aligned}$

Should not the upper limit of integration be 500, lower limit 400?

Joseph Stunzner - 4 years, 3 months ago

No, the mass of the rocket is decreasing. At time $t=0$ , the mass is $500 \text{Kg}$ . Hence, it should be in the lower limit.

Rohit Gupta - 4 years, 3 months ago

It should be ln(4/5) instead of ln(5/4)..... the answer is correct though.

Nikhil Kotra - 1 year, 10 months ago

Andrew Normand
Feb 18, 2017

You can solve this problem using momentum conservation.

If the jet did not have its engines on it would gain $mg \text{ kg ms}^{-1}$ of downwards momentum each second. So, the momentum of the waste being blasted out must match this. Therefore, the mass of the waste coming out is mg/1000 each second (momentum/velocity).

Since g is 10, the rate of mass transfer is m/100, which is $\frac{dm}{dt}$ . This differential equation has a solution of the form $m = Ae^{-kt}$ , where A is 500 kg and k is $\frac{1}{100}$ . If you solve for the when m = 400kg you should get a time of 22.3 seconds.

This problem completely ignores the fact that these are jet engines. The fuel mass will only be a small fraction probably a few percent of the mass flow rate of the exhaust gases. It is completely wrong to assume that the exhaust gases as composed of 100% of the fuel. Whilst this type of calculation would be correct for a hovering moon lander, the question says 'jump jet'. I suggest the question should clarify that the mass fraction of fuel in the exhaust gases is 5% and the answer will extend to 446.3 seconds. The pedants will note that I have neglected the direction in which the air intakes are facing. I'm assuming these are horizontal and that they are relatively so large that air is being taken in at a negligible velocity (this will be far from true - the engines will need around 4 m³/s ).

Ed Sirett - 4 years, 3 months ago

Yeah, that's a good point. I'll probably just change it to assume that the fuel includes all reactants, even though this is not realistic. If I was writing the question now I'd do as you suggest, but it's quite a big change otherwise to a problem that a lot of people have answered already.

Andrew Normand - 4 years, 3 months ago

I agree with Ed Sirett, I suggest mentioning the following points should suffice,

1) The fuel completely converts into gasses and no residue is left

2) The exhaust gasses contains only the gasses produced by burning of the fuel

3) No gasses from outside are sucked.

Rohit Gupta - 4 years, 3 months ago

@Rohit Gupta – Might it be simpler (and more realistic) to change the question to being about a moon lander?

Ed Sirett - 4 years, 3 months ago

I am not sure if I should call it a momentum conservation. It is rather a problem of force equilibrium.

Rohit Gupta - 4 years, 3 months ago

The two ideas are completely equivalent

Andrew Normand - 4 years, 3 months ago

I agree that the ideas are equivalent, it is just that it should not be called as momentum conservation. In fact, if we consider the exhaust and jet as one system then its momentum is not conserved and its center of mass is actually moving down at a constant acceleration $g$ .

Rohit Gupta - 4 years, 3 months ago

Yes, but the fact that the engines are working does not affect the overall momentum of the aircraft/fuel system - which is the principle I used for my solution. You could equally use Newton 2 and Newton 3, and not invoke momentum at all.

Andrew Normand - 4 years, 3 months ago

@Andrew Normand – Yes, the engines are internal to the system and that is why they won't affect the momentum of the overall system.

What you have done is correct. I am just not convinced that this is called the conservation of momentum.

Rohit Gupta - 4 years, 3 months ago

I totally just guessed..I wasn't even sure what value I was suppose to be looking for... Lol 22.3.. now tell me what the chances of getting that right is?

Sean Bouchard - 4 years, 2 months ago

Hover jet

The answer is 22.3.

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