Would someone mind calculating this (total velocity at burn out) for me?

[EndeavourCmdr]

Assume an initial mass of 7,500,000 kgs where 6,500,000 kgs is fuel.
Assume a constant force of 4,000,000 N
This creates an initial acceleration of 0.53 m/s^2

Assume a total burn time of 1,300,000 seconds.
Assume that mass decreases at a rate of 5 kg/s

What is the total velocity at burn out (after 1,300,000 seconds) NOT taking into effect any relativistic effects (which may be negligible anyhow)?

Thanks.

 

[Urwumpe]

Rocket equation.

[math]\frac{400 kN}{50 \frac{kg}{s}} \cdot \ln{\frac{7.5}{1.0} \frac{10^6}{10^6} \frac{kg}{kg}} = 8000 \frac{Ns}{kg} \ln{7.5} = 16119.22 \frac{m}{s}[/math]

 

[EndeavourCmdr]

Where did you get the tool to calculate that?

 

[Artlav]

Why are you dividing 6.5 by 1?
Total initial mass is 7.5.

 

[Urwumpe]

Why are you dividing 6.5 by 1?
Total initial mass is 7.5.

Ah damn, see, that is why you shouldn't do math while dressing up for sports. :lol: Will be corrected.

---------- Post added at 07:12 PM ---------- Previous post was at 07:10 PM ----------

Where did you get the tool to calculate that?

Staples. It is called pocket calculator, you can find various versions of this tool around your places.

The rest is knowing the rocket equation (physics text book), LaTex2 syntax for making Orbiter-Forum display the formula in a nice way (math-tags) and a few reflexes.

But don't do it while changing clothes.

 

[Jarvitä]

And, if you want to take the relativistic effects into account to calculate the final velocity observed from the inert point of reference of the departure point:

[MATH]v_o=\frac{c_0\times v_{subj}}{\sqrt{ c_{0}^{2} + v_{subj}^{2} }}[/MATH]

According to my calculations, the subjective velocity observed shipboard is 16119.224164338118053 m/s, and the velocity observed from the point of departure is 16119.224141037845441 m/s.

 

[EndeavourCmdr]

Okay, this can't be right....

I used the formula you provided, and did a little more research into the type of performance that could be expected from an antimatter propulsion engine.

There has been some research into such engines, as I'm sure you know, but here is a source with some interesting numbers:

http://library.lanl.gov/cgi-bin/getfile?00326047.pdf
A p--NERVA engine based on the most thoroughly tested nuclear rocket, designated
NRX, would have a thrust of 4.4 x 105 N (100,000 lb), a power level of
around 2700 MW, a mass of near 7000 kg, an I of near 1100 s, and a mass flow
of antiprotons of around 13 μg/s.

I used those thrust and flow rate numbers, along with an estimated vehicle mass of 208,656 kgs, which is twice the mass of the space shuttle orbiter at landing, to create my own estimate of speed of an interstellar vehicle (assuming we could produce 2 kilograms of anti-protons).

Which gave me...

440kN
-------- * ln(208658 kg / 208656 kg) = 3.244 * 10^8 m/s = 1.082c ?????
13 μg/s

Really? If we were to NEGATE the effects of relativistic gamma, we would reach 1.082c on only two kilograms of antiprotons? Did I calculate that correct?

 

[Jarvitä]

Yes, antimatter is a ridiculously dense energy storage medium. If you could contain that much of it with that little mass for the rest of the spacecraft, and if you could efficiently channel the products of annihilation into a directional beam, that's what you'd get.

And, at that kind of velocities, it really makes no sense to neglect relativity. With those parameters, the final velocity as observed from the point of departure would be 0.7344c.

 

[EndeavourCmdr]

I got 1.082c for the result. Why did you get a different number?

 

[Jarvitä]

I got 1.082c for the result. Why did you get a different number?

You calculated what the relative velocity between the vessel and the departure point would have been in a universe where Newton got the theory of everything right. To get the velocity, as observed from the departure point, in our universe, you have to use the same formula for departure frame velocity I've provided earlier, [MATH]v_o=\frac{c_0\times v_{subj}}{\sqrt{ c_{0}^{2} + v_{subj}^{2} }}[/MATH]

 

[Urwumpe]

I got 1.082c for the result. Why did you get a different number?

_S_H_E_ corrected for relativity, after getting your result. At such performances, you need to correct for relativity to not believe you have just discovered the warp drive.

 

[EndeavourCmdr]

No that's exactly it, the point was NOT to correct for relativity, and there is a reason for that, but maybe I understand it wrong....

The way i understand is, as velocity increases to near c, three things happen, IE: Time for the traveler slows down, mass increases, and size in the direction of travel decreases, all that the same rate

Assuming that those things are correct, the common argument which is generally accepted that it would be impossible to reach the speed of light, or to even come near to it, because since mass increases, the engine propelling the craft would be less effective. Is this correct? That the "reason" that a craft would not be able to come very close to c is because of its mass increasing, so more energy being required to be able to accelerate it?

 

[Urwumpe]

No that's exactly it, the point was NOT to correct for relativity, and there is a reason for that, but maybe I understand it wrong....

The way i understand is, as velocity increases to near c, three things happen, IE: Time for the traveler slows down, mass increases, and size in the direction of travel decreases, all that the same rate

Assuming that those things are correct, the common argument which is generally accepted that it would be impossible to reach the speed of light, or to even come near to it, because since mass increases, the engine propelling the craft would be less effective. Is this correct? That the "reason" that a craft would not be able to come very close to c is because of its mass increasing, so more energy being required to be able to accelerate it?

yes, exactly. relative to your reference - not absolute.

 

[Jarvitä]

yes, exactly. relative to your reference - not absolute.

Redundant disclaimer, there is no absolute frame.

No that's exactly it, the point was NOT to correct for relativity, and there is a reason for that, but maybe I understand it wrong....

Neglecting relativity makes no sense even if you just care about shipboard solutions - you will never experience going faster than c0.

 

[Urwumpe]

Redundant disclaimer, there is no absolute frame.

Has to be said often enough since relativity isn't an easy concept to deal with.

If I put a finger into your nose, I have a finger in your nose and you have a finger in your nose, but I am relatively better off in that situation.

 

[EndeavourCmdr]

Okay, I guess that MIGHT make sense, but I think we are overlooking something.

For these arguments below, I will assume the data I posted above, where the dry mass of the ship is 208,656kgs with 2kgs of fuel, and a force of 440kN.


The arguments of standard understanding would go something like this:

To start, I'm in orbit. Speed is ~0.000027c, and the Relativistic change factor (RTF) is only 1.000000000360219. Hardly enough to notice any difference. Using Acceleration = Force/Mass, I would get an acceleration of ~2.109 m/s^2.

At 0.1c, there would be a RTF of 1.005, so the vessel traveling would be 1.005 * as massive 1.005 * as "short" and 1.005 * slower passage of time, however no change in time, mass or size would be observed by the traveler, because those changes are in their frame of reference. At this speed, the engines are still very much effective as the ship only "weighs" 1.005 times as much as at rest. So, the ship would now have a mass of 209,699.28 kgs. Using the formula for acceleration, Acc = F/M, I would now have the values of 440kN / 209699.28 = Acc = 2.098 m/s^2. A very small change.

At 0.5c, the observed time is now 1.15 times slower, the ship is 1.15 times shorter and time at the point of reference (the traveler) is going 1.15 times slower. Still, not an extremely large difference in mass, however the new formula is 440kN / 239954.4 = Acc = 2.098 m/s^2, which gives an acceleration of 1.837 m/s^2. Starting to be noticable.

At 0.9c, the RTF jumps up to 2.294. Mass of the ship is over twice as much, time is moving half as quicly, and the ship is half it's original length due to the Lorentz Transformation. The new equation would be, 440kN / 478656.864 = Acc = 0.919 m/s^2.

At 0.98c (RTF = 5.025), the ship is now ~5 times as massive, ~1/5th as long and time is moving ~1/5th as fast. Acceleration "should" be 440kN / 1,048,535.84= Acc = 0.419 m/s^2. Getting a lot less effective, and this seems to make sense still, but lets jump up a bit higher.

At 0.9998, the relativistic change factor is 50.0025. The ship is now 1/50th it's initial length. It's looking pretty flat to an outside observer. Time is moving 1/50th as slow. Every second for the traveler is almost a minute to the observer. And lastly, the ship now has a mass of 10,433,324.64 kgs. With these new numbers, acceleration has almost come to a complete halt. 440kN / 10,433,324.64 = Acc = 0.0421 m/s^2.

Looking at these values, it would seem that the generally accepted assumption about acceleration near the speed of light is true, but....

The ENTIRE mass of the ship increases. Anything that is part of the ship, in the ship, or riding on the ship is also experiencing all of the same changes - mass, length and time dilation. Does this not include the fuel?

E = Mc^2? Fine. The total energy output of the engine, E is = to the mass of the fuel, times the speed of light squared... but wait a second, what is the mass of the fuel????

Well, originally, each antiproton has a mass of 1.6726 X10^-27 kg and produces a force of 440,000 Newtons / s at a rate of 13 μg/s. In 13 μg, there are 7.772 * 10^18 antiprotons. The energy then is the mass of the antiproton * c^2 * the number of antiprotons per second. Well, the mass of the proton is 1.6726*10^-27 kg, the speed of light is (2.92*10^8)^2 m/s and so the energy of 1 antiproton is 267 PeV.

So 7.772 * 10^18 antiprotons * 267 PeV = 3.325 * 10^17 Joules per second.
So, our 440,000 newtons comes from an energy of 7.557*10^11 J/N (Joules per Newton). Okay. So, at 50.0025 times the fuel mass, what is the energy produced by a single antiproton? Well, that's easy.
(50.0025 * 1.6726 X10^-27 kg) * (2.92*10^8)^2 m/s = 1.334*10^19 eV (electron volts).

Lets multiply the quantity of antiprotons (7.772 * 10^18 ) times their energy per proton (1.334*10^19 eV) = 1.661 * 10^19 Joules. Alright! So now we just take the total Joules, and divide by the Joules per Newton to find out our new force in newtons when the antiprotons are 50.0025 times as massive.

(1.661 * 10^19 J) / (7.557*10^11 J) And the answer is 21,980,000 Newtons.

Alright, lets sub in the new force in newtons at 0.9998c in the equation with the mass of the ship at 0.9998 to see what the acceleration would actually be....

Acc = F/M
Acc = 21,980,000 / 10,433,324.64 = 2.1067 m/s^2

WAIT A SEC! Even at 0.9998c, acceleration did not change!?

If I got this all wrong, I would understand, but this has been sitting in the back of my mind for a loooooong time. Everyone seems to forget that the mass of each individual particle also increases as the same relativistic gamma.

Work it out. Show me where I'm wrong.

Thanks.

 

[Fizyk]

And, if you want to take the relativistic effects into account to calculate the final velocity observed from the inert point of reference of the departure point:

[MATH]v_o=\frac{c_0\times v_{subj}}{\sqrt{ c_{0}^{2} + v_{subj}^{2} }}[/MATH]

According to my calculations, the subjective velocity observed shipboard is 16119.224164338118053 m/s, and the velocity observed from the point of departure is 16119.224141037845441 m/s.

I don't think this takes care of relativistic effects properly. The relativistic rocket equation is:

[math]v = c \tanh \left( \frac{I_{sp}}{c} \log \frac{m_0}{m} \right) = c \tanh \frac{v_{Newtonian}}{c}[/math]

I believe the "subjective velocity" would be something different than the velocity calculated using Newtonian equations.

This equation gives 0.794c if the result is 1.082c in the Newtonian case.

 

[Jarvitä]

Interesting, it seems my relativity needs some brushing up.

The subjective velocity is then calculated by multiplying the departure frame velocity by gamma, no?

 

[Fizyk]

Jarvitä: Yeah, looks like this is indeed the case. I'm guessing subjective velocity is the distance measured in the departure frame divided by time measured in the rocket frame.

EndeavourCmdr: Relativity is a bit more complicated than just using the RTF, that's probably the main problem. Let me show you how you can calculate that using relativity.

It's all based on two conservation laws: conservation of energy and conservation of momentum. Let's assume we have a rocket with velocity v and mass M, which exhausts mass [math]d\mu[/math] with velocity w, decreasing its own mass by [math]-dM[/math] (the minus is so that dM is the change in mass; [math]-dM[/math] will not be equal to [math]d\mu[/math], that's why I denote them differently). This chages its velocity by dv in the departure frame, and by dv' in so-called comoving frame (frame travelling at v).

Conservation of energy in the comoving frame gives:
[math]Mc^2 = \frac{d\mu c^2}{\sqrt{1-\frac{w^2}{c^2}}} + \frac{(M+dM)c^2}{\sqrt{1-\frac{dv'^2}{c^2}}} \approx \frac{d\mu c^2}{\sqrt{1-\frac{w^2}{c^2}}} + (M+dM)c^2[/math]
(I assume dM, dv' etc. to be small, so [math]dv'^2[/math] is of higher order and can be neglected).

We get [math]dM = -\frac{d\mu}{\sqrt{1-\frac{w^2}{c^2}}}[/math].

Conservation of momentum gives:
[math]0 = -\frac{d\mu w}{\sqrt{1-\frac{w^2}{c^2}}} + \frac{(M+dM)dv'}{\sqrt{1-\frac{dv'^2}{c^2}}} \approx -\frac{d\mu w}{\sqrt{1-\frac{w^2}{c^2}}} + Mdv'[/math]
(again neglecting terms of higher order)

We get:
[math]dv' = -w \frac{dM}{M}[/math]

Now we need dv. The velocity is changed by dv' in the frame moving with velocity v, so we need relativistic addition of velocities:

[math]v+dv = \frac{v+dv'}{1 + \frac{vdv'}{c^2}}[/math]

So:
[math]dv = \frac{v+dv'-v(1+\frac{vdv'}{c^2})}{1+\frac{vdv'}{c^2}} = \frac{dv'(1-\frac{v^2}{c^2})}{1 + \frac{vdv'}{c^2}} \approx dv'(1-\frac{v^2}{c^2})[/math]

We get a differential equation:
[math]\frac{dv}{1-\frac{v^2}{c^2}} = -w \frac{dM}{M}[/math]
The solution is exactly the equation I posted earlier (assuming [math]I_{sp}=w[/math]).