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This question goes to a very basic non-understanding of mine that I have had in the back of my mind for ages - I just read the following here:

ion thrusters are capable of propelling a spacecraft up to 90,000 meters per second (over 200,000 miles per hour (mph). To put that into perspective, the space shuttle is capable of a top speed of around 18,000 mph. The tradeoff for this high top speed is low thrust (or low acceleration). Thrust is the force that the thruster applies to the spacecraft. Modern ion thrusters can deliver up to 0.5 Newtons (0.1 pounds) of thrust, which is equivalent to the force you would feel by holding nine U.S. quarters in your hand.

So when it hits the top speed what is the bottle neck? The logical thing to me is that it takes more and more electricity to maintain the 0.1 pounds of thrust, but if this is the case, does this not violate the premise that you cannot tell how fast you are going without something to compare to? In other words, if I turn the engine on and then off again repeatedly, should I expect different results from one time to the next?

I know I'm confused about something very basic here - that's why I'm asking..

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Does an ion thrust engine consume more energy as it speeds up?

The answer to this question is no.


So when it hits the top speed what is the bottle neck?

The bottleneck is that the vehicle runs out of propellant. The problem is described by the rocket equation, $$\frac {\Delta v}{v_e} = \ln\frac{m_{\text{initial}}}{m_{\text{final}}}$$

Where

  • $m_{\text{final}}$ is the final mass of the rocket, the masses of the structures that previously held the propellant, the engines, the power plants, the structure of the rocket itself, and finally, the payload;

  • $m_{\text{initial}}$ is the initial mass of the rocket, the final mass plus the mass of the propellant;

  • $v_e$ is the velocity of the exhaust relative to the vehicle; and

  • $\Delta v$ is the change in velocity that results from using the propellant.

Note the logarithm on the left hand side of the rocket equation. Adding more propellant has an ever decreasing effect on the change in the rocket's velocity. Another way to look at the rocket equation is to look at the proportion of the initial mass that is propellant:

$$\frac{m_{\text{propellant}}}{m_{\text{initial}}} = 1 - \exp\left(-\frac{\Delta v}{v_e}\right)$$

This means that attaining a $\Delta v$ equal to twice the exhaust velocity requires that 86.5% of the initial mass be propellant. This is quite doable. On the other hand, attaining three times the exhaust velocity requires that 95% of the initial mass be propellant. This is almost possible from an engineering perspective. Anything beyond that is not. Single stage rockets have an upper limit on the change in velocity that is somewhere between two to three times the exhaust velocity.

There are ways to overcome the tyranny of the rocket equation. One approach is to use a multi-stage rocket. The math described above pertains to single stage rockets. A single stage rocket using traditional chemical-based techniques cannot achieve orbital velocity from the Earth's surface thanks to that limit of two to three times exhaust velocity. The rocket equation changes a bit for multi-stage rockets. Another approach is to use a better kind of propellant, one with a higher exhaust velocity. That's what makes ion engines so appealing.

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That page is not well written. The 90km/s speed is the exhaust velocity of the engine. It is not the maximum speed of the spacecraft. There is no maximum speed of the spacecraft, short of the speed of light.

They make the mistake again when they say: "While a chemical rocket's top speed is limited by the thermal capability of the rocket nozzle"

ADDED: Chris Drost makes a good point. If the initial mass of fuel is about 99% of the vehicle's total mass, then it can get up to about 4.6 times the exhaust velocity, or about 400km/s, before it runs out of fuel.

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    $\begingroup$ With that said, the exhaust velocity is not completely irrelevant to the maximum speed of the spacecraft, either. If the spacecraft has "payload" $M$ travelling at its final speed, and a mass of "fuel" $f$, then it is well known that its final speed is $v_e\ln[1 + \frac fM]$ where $v_e$ is the exhaust velocity. Going significantly above the exhaust velocity requires exponentially more fuel, not linearly more fuel. $\endgroup$ – CR Drost Sep 29 '15 at 20:11
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    $\begingroup$ 90km/s * 3600 = 324000km/h. 324000 / 1.6 = 202500mph. I see no problem there. 18000mph * 1.6 = 28800km/h. 28800/3600 = 8 m/s. This is interesting. It's double the exhaust velocity of the shuttle main engines en.wikipedia.org/wiki/Space_Shuttle_main_engine but it's also more than earth escape velocity (and we know a shuttle never flew that fast.) I expect what they mean by "top speed" is not the exhaust velocity, but rather the delta V that can be achieved (though 8m/s is not quite enough dV to get to low Earth orbit, due to losses). Only the author knows what they meant. $\endgroup$ – Level River St Sep 30 '15 at 0:55
  • $\begingroup$ Avoid using the back of an envelope when you are using a browser. Instead, type in the search bar: "90km/s in mph=". $\endgroup$ – Dewi Morgan Sep 30 '15 at 7:09
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The limit in either case is when you run out of fuel. By the Tsiolkovsky rocket equation, if all else is the same, the top speed of a rocket is proportional to the exhaust velocity. So, the faster a rocket ejects its (often burning) exhaust, the higher the final speed when the rocket runs out of fuel.

As a metaphor, imagine being in a boat filled with bricks (please ignore friction). To move the boat, you start throwing bricks off of the stern. If you just do a slow underhand, then each brick will only slightly speed up the boat, and when you run out of bricks the boat won't be going very quickly. If you instead throw the bricks as fas as you can, the boat will be moving fairly quickly when you run out. If you set up a sling shot to shoot the bricks, then you'll end up moving quite quickly.

Although ion thrusters don't push very hard, their claim to fame is that their exhaust velocity is incredibly high. This means that, if you wait long enough, you can achieve velocity changes far higher than any chemical rocket.

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  • $\begingroup$ Top speed of a rocket is only proportional to exhaust velocity for constant mass fraction, and for identical velocity on initiation of thrust. $\endgroup$ – WhatRoughBeast Sep 29 '15 at 21:50
  • $\begingroup$ @WhatRoughBeast: nothing is proportional to anything else unless you control other variables. $\endgroup$ – Daniel Griscom Sep 29 '15 at 23:30
  • $\begingroup$ (Shrug) You're the one who used proportional without controlling (or even mentioning) other variables. $\endgroup$ – WhatRoughBeast Sep 29 '15 at 23:41
  • $\begingroup$ His answer does include a example where everything is held constant except exhaust velocity. $\endgroup$ – LovesTha Sep 30 '15 at 4:41
  • $\begingroup$ @WhatRoughBeast: you're right. Editing. $\endgroup$ – Daniel Griscom Oct 1 '15 at 17:21

protected by Qmechanic Sep 30 '15 at 9:48

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