woelen
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Another calculus brain teaser a la blogfast25
Consider a very simple rocket.
It has mass M, of which there is a payload with mass M0 and fuel with initial mass MF, so M = MF + M0.
While the rocket burns, it loses fuel, so MF becomes smaller over time. M0 does not change.
The rocket with payload is launched in space, no gravity works on the rocket. The thrust of the rocket is precisely along the direction of the line
along which the rocket moves, so the problem can be regarded as a simple 1-dimensional problem, no rotation and torques need to be taken into account.
The rocket is efficient and uses a good solid fuel/oxidizer mix and as long as the rocket burns, for each kg of mix, 2·107 joules of
energy is converted to mechanical energy, the rest is lost in heat, radiation and whatever other inefficiencies there are in the process. For the sake
of simplicity you may assume that at any moment in time all particles of spent fuel/ox. mix leave the rocket with the same momentaneous velocity
vF, no particles move sideways.
Now suppose the useful payload M0 of the rocket is 10000 kg (capsule plus equipment plus fuel chamber and nozzle and nothing of this is
disconnected, the rocket remains in one piece). The rocket is launched from free space with zero gravity, initially floating with zero velocity in
space. We want the final velocity of the rocket which has spent all its fuel to be equal to 10 km/s. How many kg of fuel/ox. mix are needed to achieve
this? How many kg of fuel/ox. mix are needed to obtain a final velocity of 20 km/s and to 50 km/s?
For analysis you may assume that the fuel/ox. mix burns with a constant rate, but this is not really necessary.
For solving this problem you only need very basic newtonian mechanics and a little high-school calculus. There is an analytical solution to this
problem, no need to resort to numerical algorithms.
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For the more advanced: Try to solve this equation for a relativistic rocket, using general relativity with euclidian space metric (curvature of space
may be neglected). The fuel now is much better, we use an antimatter drive, which annihilates ½MF of matter and ½MF of
antimatter. So, we have a matter/antimatter amount of MF. For each kg of "fuel" assume that 5·1016 J of energy is converted to
useful mechanic energy, the rest is lost as radiation perpendicular to the direction of motion or as unreacted matter.
Now, how much fuel is needed for getting a payload of 10000 kg to 10 km/s? To 100 km/s? To 0.01c? To ½c? To 0.9c?
[Edited on 15-4-16 by woelen]
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aga
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So, you got the SpaceX job ?
OK. So you need help. I will start work on the answers.
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blogfast25
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Wonderful.
I suggest participants U2U woelen with their solution(s), rather than create a spoiler alert. woelen can then point to correct participants and ask
them to present their reasoning in full?
What say you?
[Edited on 11-4-2016 by blogfast25]
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blogfast25
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Dodging your calculus class to play with the rocket boys isn't allowed, young man!
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Darkstar
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He's about to be permanently expelled and his full scholarship revoked at this point. We operate on a three-strike system here at B&D University,
and the little stunt he pulled yesterday where he staggered into class over an hour late sweating like a nun in a cucumber field and so drunk he
couldn't find a Dixie coonskin with an Ohio hooker holding his prick and showing him the way was strike two. In addition to interrupting the lecture
and making a complete spectacle of himself by yelling drunken obscenities at the professor and students, he also managed to knock over the projector
and tear its screen off the wall by trying to hold onto them as he stumbled his way to his seat.
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woelen
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Quote: Originally posted by blogfast25  | Wonderful.
I suggest participants U2U woelen with their solution(s), rather than create a spoiler alert. woelen can then point to correct participants and ask
them to present their reasoning in full?
What say you?
[Edited on 11-4-2016 by blogfast25] |
Is OK to me. Let's wait a day or two.
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blogfast25
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Here's woelen's solution for the non-relativistic rocket problem.
In the absence of external force fields the energy provided by the fuel in an infinitesimal amount of time is completely converted to kinetic energy
KE (note that dM < 0, so the correct signs have to be observed):
$$dE_k=-KdM=-\frac12 v_f^2dM+\frac12 (M+dM)(v+dv)^2-\frac12 Mv^2$$
Where K=2x107 J/kg. The first term on the right represents the KE of the ejected mass dM (ejected at absolute velocity vf), the
last two the change in KE of the rocket itself. I went wrong in neglecting that first term, which leads to an under-estimate of the fuel requirements.
After developing and eliminating higher infinitesimals:
$$dE_k = -KdM = -\frac12 dM(v_f^2-v^2) + Mvdv$$
To eliminate vf, we need to apply conservation of momentum. Before and after ejection of dM:
$$Mv=(M-dM)(v+dv)-v_fdM$$
$$\implies 0 = -dM(v_f-v) + Mdv$$
Re-write the equation for dEk:
$$dE_k = -KdM = -\frac12 dM(v_f-v)(v_f+v) + Mvdv$$
Now call:
$$z=v_f-v \implies v_f+v=z+2v$$
And substitute:
$$-KdM = -\frac12 dMz(z+2v) + Mvdv$$
$$0=-zdM+Mdv$$
With:
$$v'=\frac{dv}{dM}$$
$$\implies z=Mv'$$
Divide the energy equation by dM:
$$-K=-\frac12 z(z+2v)+Mvv'$$
$$-K=-\frac12 Mv'(Mv'+2v)+Mvv'$$
$$-K=-\frac12 M^2v'^2-Mvv'+Mvv'$$
$$K=\frac12 Mv'^2$$
As v'2 has a positive and negative root, we need to choose the correct one. With dv > 0 and dM < 0 we need the negative root:
$$v'=-\frac{\sqrt{2K}}{M}$$
Now integrate between the boundary conditions:
$$\int_0^vdv=-\sqrt{2K}\int_{M_0+M_f}^{M_0}\frac{dM}{M}$$
$$v=\sqrt{2K}\ln \Big(\frac{M_0+M_f}{M_0}\Big)$$
$$\large{M_F=M_0(e^{\frac{v}{\sqrt{2K}}}-1)}\large$$
For 10 km/s this leads to MF = 10000 * exp(10000/sqrt(4*10⁷)) - 10000 = 10000*exp(1.581) - 10000 = 38600 kg
For 20 km/s this leads to MF = 10000*exp(3.162) - 10000 = 226000 kg.
For 50 km/s I find 2.70 million kg.
This problem is essentially an application of the Ideal Rocket Equation:
https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
Tomorrow: the relativistic rocket problem!
[Edited on 14-4-2016 by blogfast25]
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woelen
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This exercise nicely shows how much fuel is needed for launching rockets and getting them to speeds in the order of a few tens of km/s. For instance,
the space shuttle used appr. 500000 kg of NH4ClO4/Al mix plus 730000 kg of liquid oxygen+cryogenic hydrogen. The payload only is 100000 kg or so, so
for each kg of useful payload they needed appr. 10 kg of fuel. These fuels have a K-value between 107 and 2*107.
The main problem with this kind of technology is that the fuel itself also needs to be accelerated and a lot of energy goes into moving that fuel. If
it were possible to have an external supply of fuel then the amount of needed energy would be more modest, but unfortunately there is no means of
having the fuel/ox. mix available outside the rocket itself.
[Edited on 15-4-16 by woelen]
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phlogiston
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In addition to accelerating the fuel, it also needs to be lifted partially out of the earths gravity well before it gets burned at altitude, further
compounding the problem.
Your example shows nicely why things like laser propelled craft for instance using light sails are more efficient in this respect.
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"If a rocket goes up, who cares where it comes down, that's not my concern said Wernher von Braun" - Tom Lehrer |
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