# Why is specific impulse equivalent to effective exhaust velocity?

I am very comfortable with specific impulse in seconds. I know how to calculate it, I know what it is, I know what happens to a rocket when you double in or cut it in half. But it feels like the strangest coincidence that specific impulse in units of velocity is equivalent to exhaust velocity.

Here's what I mean:

If we use the interpretation of specific impulse as the thrust per unit of fuel mass flow, we've got units of

$$\frac{N}{\frac{kg}{s}}$$

Split up our Newtons using Newton's Second Law ($$F=ma$$)

$$\frac{kg \cdot \frac{m}{s^2}}{\frac{kg}{s}} => \frac{kg \cdot \frac{m}{s}}{kg} => \frac{m}{s}$$

Ok, great, like magic we have meters per second. Why? The math said so. But somehow this is supposed to be the exhaust velocity? The $$m$$ in $$m/s$$ came from the thrust of the rocket in the math, not from any property of the exhaust itself. The exhaust $$kg$$ and $$s$$ just cancel out some components of thrust.

Is there a good intuition to explain why this is equivalent to exhaust velocity, beyond "because the math says so"?

• I find it interesting that you are "comfortable with specific impulse in seconds". I've always considered it a historical relic, an arbitrary conversion factor thrown in to simplify hand calculations that I've always found to confuse matters. It's immediately obvious what a rocket with an exhaust velocity of 3 km/s implies, how it relates with mass flow rate to thrust and so on. 306 seconds, though? It's a duration, but not one that seems very interesting except for being related to exhaust velocity by $g_0$. – Christopher James Huff Dec 5 '20 at 22:45
• @ChristopherJamesHuff oddly the very early writers like Willy Ley only wrote about exhaust velocity when discussing efficiency; specific impulse didn't start to be the standard measure until later. "In addition to using the exhaust velocity as a yardstick...performance...can also be expressed by the specific impulse" - the only time the term appears in Ley, 1954 – Organic Marble Dec 5 '20 at 23:00
• "The $m$ in $m/s$ came from the thrust of the rocket in the math, not from any property of the exhaust itself." The two critical properties of the rocket exhaust are its mass and its velocity; the thrust is the product of mass flow rate and velocity. – Russell Borogove Dec 5 '20 at 23:05

Rockets produce thrust by ejecting reaction mass at some velocity. The fundamental quantities involved are mass flow rate and exhaust velocity, thrust is the consequence of these.

It's no coincidence that specific impulse in units of velocity equals exhaust velocity, that's what specific impulse is. The exhaust velocity gives you the impulse per unit of reaction mass simply because $$p = m*v$$.

That parameter is known as the Effective Exhaust Velocity because...

Sutton, 4th edition:

When [the exit plane pressure is equal to the ambient pressure], the effective exhaust velocity is equal to the average actual exhaust velocity of the propellant gases...the effective exhaust velocity is usually close in value to the actual exhaust velocity.

When Specific Impulse ($$I_{sp}$$) is expressed in units of time, the rate of fuel usage is not expressed in units of mass per second.

It's in units of weight (under one Standard Gravity, $$g_0$$) per second.

As such, the unit conversion comes from: $$\rm{\frac{N}{\frac{N}{s}}=s}$$

You must multiply the time-valued Specific Impulse by $$g_0$$ or $$9.80665 \rm{m/s^2}$$, to get the Effective Exhaust Velocity($$v_e$$). As such, the actual unit conversion looks like this when going from $$I_{sp}$$ to $$v_e$$:

$$\rm{\frac{N}{\frac{N}{s}}\dot{}\frac{m}{s^2}} = \frac{m}{s}$$

### Given:

You're in a boat on a calm, windless lake and you're several meters from the dock.

It's essential that you get to the dock while staying in the boat, but you've lost your oars1.

There are several massive objects in the boat that you can throw away from the dock in order to accelerate towards it.

### Question:

With all that given, what could be more fundamental and critically important than the velocity with which you can throw those objects?

### Answer: No, velocity is the most fundamental and critically important here.

Forget all those equations; velocity is what's central to the problem. You know how much fuel (or objects) you have, you know the spacecraft (boat) will get lighter after you throw some of them, you want the most force per unit mass thrown, and starting from velocity you can calculate all the other stuff using math.

1or after you pushed off from the dock you realize that you've forgotten to bring the "keys to the oar locks" (humor).

Of course, specific impulse is "a measure of how effectively a rocket uses propellant", but that's quite a vague statement. So is there a physical interpretation of the value of specific impulse? What does it mean if we have a rocket with a specific impulse of 4 km/s? Where do we end up with the 4 km/s?

It turns out, there is. It's an answer to the question:

What speed could we give to a rocket with constant mass, when expelling a mass equal to itself?

So if we had a rocket of 1000 kg, that had a fuel tank of 1000 kg, where mass would magically reappear so that its total mass remained 1000 kg even when expelling fuel, and this rocket had engines with a specific impulse of 4 km/s, we can now conclude that this rocket could reach a maximum speed of 4 km/s.

Now such a rocket is impossible. But we could approximate it by starting with a 1000 kg rocket, and asking what speed it could attain by expelling 1 kg of fuel (where rocket mass is almost constant). Well, that's 1/1000th the fuel, so it can attain 1/1000th of the speed.

And then we can apply conversation of momentum. We expel a mass of 1/1000th the mass of the rocket, which gives our rocket a momentum of $$m_r \cdot (I_{sp}/1000)$$. And the counteracting momentum of the exhaust is $$(m_r/1000) \cdot v_{exh}$$

So there we have it, $$I_{sp}$$ must be equal to $$v_{exh}$$.