Why does a spacecraft propulsion system either have high thrust or high specific impulse, but can’t have both at the same time?

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I just started learning about space travel. I’ve heard that a spacecraft propulsion system either has high thrust + low specific impulse, or low thrust + high specific impulse.

As far as I know, high thrust means a propulsion system has high mass flow rate, achieving higher acceleration.

High specific impulse means a propulsion system can generate more thrust for given amount of propellant used, achieving higher fuel efficiency.

So if a propulsion system such as ion thruster has higher specific impulse than chemical rocket, why can’t engineers increase the output of ion thruster to increase thrust, achieving both high fuel efficiency and high acceleration to replace chemical rocket?

In: Engineering

12 Answers

Anonymous 0 Comments

Energy. For all of the engineering we put into rockets, there is one thing that is in short supply no matter what we do; energy. A rocket powers itself by throwing something out the back, and it costs energy to do that. The faster you throw it, the less of it you need, *but the more energy it costs to do so*.

So if our rockets all have the same power source (not true but let’s ignore that for a moment) then we must choose – if we throw more stuff out the back then we get a higher thrust without needing more energy. If we conversely wanted a high efficiency, we’d have to take our time and be patient to get enough energy from our limited power source, which is the same as saying we’d have a low thrust (or a high thrust in short bursts but a low average thrust).

It happens that our energy source can be tailored to our engine choice. A chemical rocket uses the stuff it throws out the back as its energy source, which makes a ton of sense. This also gives chemical rockets access to more power than most other rockets – they’re willing to carry huge amounts of fuel *because that fuel is not dead weight but also serving as the stuff that gets thrown out the back*.

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