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With ISp defined as ${v_e}\over{g_0}$, the naive approach would tell me, the maximum possible specific impulse, with propellant moving at speed of light relative to the craft, is $c\over{g_0}$ or 30,570,322s.

Of course, with relativistic physics naive approach is often wrong and I feel like I'm making some oversimplification here... so, if I am, could someone straighten it out? What happens with Specific Impulse when extreme energies are put into the propulsion, accelerating the propellant to relativistic speeds?

SF.
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3 Answers3

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A photon rocket should have a Ve of c, hence ~30 megasecond Isp. The rocket equation would be tricky to apply, of course -- if you have a magic matter-energy reaction that perfectly converts "fuel" into linear-directed photon exhaust, it works fine, but if energy is going elsewhere, the effective exhaust velocity will be different. You'd account for relativistic effects in applying the ∆v using everyone's favorite sigmoid function, the hyperbolic tangent.

what, you don't have a favorite sigmoid function?

Russell Borogove
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    So - I "naively" plug the speed into ISp, but need to go relativistic when using that ISp for anything else? - on a different note: Considering how weird a unit 'megasecond' is, I checked... that's almost a year. ...And I just got a good line for a sci-fi novel. "...and her ISp is measured in weeks." – SF. Jul 11 '16 at 01:27
  • But the limiting case isn't a photon rocket, though. Radiation pressure is an extremely inefficient way to conduct kinetic energy, and you could do better using a particle accelerator firing a proton beam. Your new bottleneck would be e = mc^2. – Wesley Adams Apr 04 '21 at 17:05
  • And if I power the photon beam with a solar panel, I have a true infinite ISP because my spacecraft expends no mass energy. The physical manifestation of ISP is the number of seconds the engine will burn before it exhausts its fuel supply whose weight is the same as the engine's thrust. If the mass does not decrease while the engine burns, this equation has an infinite result. It's going to take awhile to get anywhere though because that thrust is abysmal. – Joshua Mar 28 '24 at 03:27
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    @Joshua A mirror would work better -- skip the lossy conversion, reflecting the light hitting your ship from behind. You could paint the front side black, run coolant through it that routes through radiators mounted behind the mirror as well. – Russell Borogove Mar 28 '24 at 05:27
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No. The number you gave is for a photon drive. However, if you throw particles out the back at relativistic velocity they weigh more than the fuel you drew from the tank. Since there's no limit on how much relativistic mass you can pile on the particle there's no limit on the ISP. Note, however, that there's no way you can boost particles like that using onboard energy.

Loren Pechtel
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  • Wouldn't that just increase the mass flow? Their mass increase comes from equivalent decrease of mass of on-board power supply.due to its energy loss (RTG battery acting as a fuel tank). – SF. Jul 12 '16 at 04:48
  • @SF. Note that I said it would not be possible from onboard energy. If you have to supply the boosting energy you're going to lose more mass than this gains you. The only way to actually pull it off is to obtain the energy from some outside source. – Loren Pechtel Jul 12 '16 at 15:33
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The highest possible specific impulse could be achieved by using a 100% efficient e = mc^2 mass to energy converter. Specific impulse is defined as the number of seconds that an engine can provide 9.8 newtons of force given one kilogram of propellant. Here's a bit of math to find this value:

c ~ 300,000,000 m/s
c^2 ~ 9 * 10^16 m^2/s^2
Plugging in one kilogram, we get 1kg = 9 * 10^16 kg * m^2 / s^2 or 9 * 10^16 joules. But we have to divide this by two, because you're accelerating both your vehicle and your propellant. This gives ~4.5 * 10^16 joules of usable energy.
Divide that by 9.8 watts, and there you have it:

~4.6 * 10^15 or ~4.6 quadrillion seconds is the highest possible specific impulse. Kinda puts our engines to shame, doesn't it?

Wesley Adams
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    Can you write the equation that you are using separately from the numbers? You are "dividing Joules by 9.8". If 9.8 has units of m/s^2 then your results is kilogram meters, not seconds. If 9.8 is just a number, then the result is still Joules. Either way I don't see your calculation producing seconds as units. Writing your equation first will help a lot. Thanks! – uhoh Apr 02 '21 at 00:23
  • 9.8 is the amount of energy needed to produce 9.8 newtons of force per second. The unit is kg m^2 / seconds^3 or watts. The equation is c^2 / watts where c = 300,000,000 and watts = 9.8. – Wesley Adams Apr 02 '21 at 03:01
  • Wait! There was a mistake in the original post! Because you're accelerating both your propellant and your vehicle, you can only use half of your e = mc2 energy for increased vehicle velocity. That means that the maximum theoretical specific impulse is only half of what I originally said. The correct equation is c^2 / 2 * watts. My bad! – Wesley Adams Apr 02 '21 at 05:34
  • I've added an answer to your question there which may help here. – uhoh Apr 02 '21 at 06:24
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    "9.8 is the amount of energy needed to produce 9.8 newtons of force per second. The unit is kg m^2 / seconds^3 or watts." That's not how it works. If a 1 kg weight lies on a table on Earth, the table constantly pushes it with 9.8 newtons of force without spending any energy at all. This answer is wrong because of this sort of misunderstanding. – Litho Apr 02 '21 at 09:23
  • While that may be true within a stagnant (not changing) field, it does not apply when dealing with particle accelerators like the one envisioned here. In a gravitational field, your potential energy + kinetic energy is constant. Your orbit will return to the same periapsis and peak velocity many times. In a particle accelerator, your can gain energy from an outside source. 9.8 watts is the minimum energy it would take to accelerate a 1kg railgun shell down a barrel in space at 1g. Hope this clears up any confusion! – Wesley Adams Apr 02 '21 at 17:44
  • You are mixing up different physical quantities again. To accelerate a 1 kg object at 1 g, you need to act at it with a force of 9.8 newtons. – Litho Apr 02 '21 at 18:14
  • When professional athletes are pedaling on a bike, they apply force to overcome resistance in the air as well as friction in the wheels. Yet that force is measured by trainers in watts! That's because the force can be used to drive an electric generator which would output that amount of electricity. One newton-meter per second is equivalent to a watt. If you want any further explanation, feel free to respond and I'll try to help you understand this topic a bit better. – Wesley Adams Apr 02 '21 at 18:24
  • 1 N m/s is equivalent to 1 W. But 1 N is not equivalent to 1 W. You say that to accelerate 1 kg at 1 g (i.e., to create a force of 1 kg * 9.8 m/s^2 = 9.8 N), one needs to spend 9.8 W of power. Why? How did you get this value? You would have to multiply the force by 1 m/s, but there's nothing special about this speed. The SI choice of units is arbitrary. – Litho Apr 02 '21 at 22:59
  • Here's a thought experiment that might be helpful:

    You have two wheels in space, with an electric motor that can spin them up or down. They each have kilogram masses suspended on poles coming out of the wheels. After one second of the motor running, the kilogram masses have been accelerated to 1 m/s. These are let go. Now with 100% efficient regenerative breaking, they recover any energy wasted accelerating the wheels themselves. How many joules did it take to spin those 2 kg up to 1 m/s?

     – Wesley Adams Apr 04 '21 at 04:44
  • (sorry about splitting this, the character limit got in the way) It takes 2 joules, because a joule is defined as the amount of energy needed to accelerate a one kilogram mass to one m/s. These 2 joules were provided over one second, giving a power input of 2W for the duration. You're right that 1N is not equivalent to 1W, but you can use 1W to produce 1N of force using an electric motor like the one envisioned here. In my original post I assumed 100% efficiency in conversion from mass energy to kinetic energy. – Wesley Adams Apr 04 '21 at 04:54
  • In general, energy is thought about in terms of joules and watts, but kinetic energy is best described in terms of newtons and whatnot. But you CAN convert one into the other, and the difference in nomenclature doesn't mean that they're inherently separate and irreconcilable. – Wesley Adams Apr 04 '21 at 04:58
  • First of all, the kinetic energy of 1 kg moving at 1 m/s is 0.5 J, not 1 J, see the formula. Second, OK, let's assume that you have motors which can convert 1 W of power to 1 N of force. Let's keep them running for 10 seconds, though. By the end of this time, the weights are accelerated to 1 N * 10 s / 1 kg = 10 m/s. Their kinetic energy is 2 * 1 kg * (10 m/s)^2 / 2 = 100 J. But your motors have consumed only 2 * 1 W * 10 s = 20 J. Where did the extra energy come from? – Litho Apr 04 '21 at 09:24
  • Alternatively, let the motors run for 1 s, but replace the weights with 0.1 kg ones. They are accelerated to 1 N * 1 s / 0.1 kg = 10 m/s, their kinetic energy is 2 * 0.1 kg * (10 m/s)^2 / 2 = 10 J, but the motors have consumed only 2 J. (By the way, if you want other users to be notified of your comments, you need to put @ + username into them. In my case, @Litho. You get notified of my comments because they are under your answer, but I was not notified of your replies.) – Litho Apr 04 '21 at 09:31
  • Oh! I didn't know that newton's third law was already accounted for in the definition of a joule. By the way, I just learned f = ma in school so I'm sorry about my gaps in knowledge about kinetic energy. Would the following be correct? ke = mv^2 / 2
    ke/m = v^2 / 2
    v / 2 = sqrt(ke/m)
    v = 2 * sqrt(ke/m)
    so plugging in 10J and one kilogram
    v = 2 * sqrt(10J / 1 kg)
    v = 2 * sqrt(10kg m^2 / s^2 / kg)
    v = 2 * sqrt(10m^2 / s^2)
    v = 2 * sqrt (10) m/s
    v ~ 6.3 m/s
    So 10 watts could produce 6.3 newtons.     – Wesley Adams Apr 05 '21 at 19:53
  • @Litho Plugging in 9.8 m/s in the velocity and one kilogram: v = 2 * sqrt(ke / m) 9.8 m/s = 2 * sqrt(j / 1kg) 9.8 m/s = 2 * sqrt(x m^2 / s^2) 4.9 m/s = sqrt(x m^2 / s^2) (4.9 m/s)^2 = x m^2 / s^2 24 m^2 / s^2 = x m^2 / s^2 x ~ 24 So you'd need ~24 joules to accelerate one kilogram by 9.8 m/s. That means that you'd need 24 watts to get 9.8 newtons of force, but 48 when accounting for reaction mass. Using this number, can we divide 9e+16 joules by 48 watts to get a revised estimate for the maximum specific impulse? If not, where did I go wrong? – Wesley Adams Apr 05 '21 at 20:52
  • Wait! Don't answer that! It turns out that the force from a given wattage is dependent on mass flow, so the 1 kg example would generate significantly more force than a proton beam. This boils down to an optimization problem: how much of your kilogram should you convert to energy, and how much should be used for reaction? I'll get back to you on this, and I'll edit my post to reflect the answer I get. – Wesley Adams Apr 05 '21 at 23:05