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I can't remember where I heard it but I thought that dividing the exhaust velocity by the gravitational pull of a body, it is possible to find the specific impulse of the rocket engine. However, what about when you are in space around another body with a different amount of pull, suppose Mars. Since the surface gravity is 3.711m/s2, when finding specific impulse that would mean the engine is twice as efficient when around Mars than opposed so around Earth... This can't be right

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The conversion factor isn't the gravitational pull of an arbitrary body, but rather the conversion between pounds as a unit of weight and pounds as a unit of mass -- which is a convention established on Earth, and hence happens to match Earth's gravitational pull.

Some rocket scientists work in feet and pounds, others work in meters and kilograms. Exhaust velocity is likewise measured in feet per second or meters per second, and so can plug directly into the rocket equation.

Specific impulse is most logically measured in velocity units, but there's another physical interpretation of specific impulse: the length of time that the thrust from a given mass of fuel could be used to cancel the weight on Earth of that same fuel. This is a weird and contrived metric, and requires dividing velocity by acceleration to yield time, but the result has a very convenient property: it's in the same units whether you're a metric or non-metric rocket scientist. So there's a long tradition of measuring Isp in seconds even though it has that awkward Earth-centric conversion factor in it.

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$I_{sp}$ in seconds is equal to the amount of time a rocket must be fired to use a quantity of propellant with weight (measured at one standard gravity) equal to its thrust.

The use of strictly earth-bound $g_0$ is the direct result of conversion between units of force (Newtons) and units of mass (Kilograms). Both kilograms and newtons are independent from location, but if you want to compare these apples to oranges, you need some pretty arbitrarily chosen constant that binds them. $g_0$ is one that gives a rather neat physical interpretation of the result, and reasonable numbers corresponding to the range of quantities encountered in reality, so it was chosen here. But it could have been a constant of 1 $m \over s^2$ or anything else, and all that would change would be something along the lines of

$I_{sp}$ in seconds is equal to the amount of time a rocket must be fired to use a mass of propellant such that acting upon this mass with force equal to the thrust of the rocket over period of 1 second accelerates it to 1m/s.

You've got to admit the first definition sounds better :)

But if you're still unhappy, feel free to give $I_{sp}$ in meters per second, as exhaust velocity. Nobody will mind, and with $g_0=9.81 \approx 10$ conversion to get the ballpark value is trivial; AK-47 muzzle velocity is 715 m/s, so its specific impulse is about 70 seconds.

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    $\begingroup$ +1 for the 9.81 = 10 approximation, indispensable tool of back-of-the-envelope engineers. $\endgroup$ – Russell Borogove Mar 2 '17 at 0:44
  • $\begingroup$ Using velocity in place of specific impulse will cause unit errors. Better to just substitute v_e/g for Isp in your equation. $\endgroup$ – Schlusstein Mar 2 '17 at 13:57
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    $\begingroup$ @Schlusstein I may have misunderstood you: in my view Isp in seconds is always converted back to an exhaust velocity when it is actually used. Maybe you meant this too, I read something different betwee your first and second sentences here. $\endgroup$ – Puffin Mar 5 '17 at 16:22
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I prefer this way of looking at the topic, it may help for some people (NB 2nd attempt, edited for clarity). The answers by SF. and Russell Borogove are no more or less correct. In this view however there is no need to talk about "weight".

Specific impulse, as the name suggests is

$ Specific Impulse = \frac{Impulse}{Mass} = \frac {Force \cdot time}{Mass}$

It has the units of velocity, i.e. $ m s^{-1} $ or $ ft s^{-1} $ and so according to one's education its numerical value varies. The next step is to divide velocity by an acceleration to get units of time. Time has the same units in both metric and US standard systems.

The magnitude of this acceleration must be the same in both sets of units, e.g. g0 ( 9.8 $ m s^{-2} $ or 32 $ ft s^{-2} $) though I don't think it especially matters which acceleration is chosen as long as everyone uses the same convention. Isp in seconds is always converted back to an exhaust velocity when it is actually used so, in principle, it doesn't matter which value is used.

Background context

One source of confusion, which is not really part of the problem but may help as context is to consider the nature of the two different systems of units:

  1. In metric the units are

    $ \frac {Newton \cdot second}{kg}$

This is nice and tidy because metric, aka SI, is a consistent set of units where a force of one Newton appplied to a one kilogram mass causes an acceleration of 1 $ m s^{-2} $.

  1. The US standard units are a bit more complicated as the system has diverged into several groups. The customary use of a pound-force, lbf and pound mass, lbm, together is an inconsistent set of units: a force of one lbf applied to one lbm causes an acceleration of 32 $ ft s^{-2} $.

This single difference causes much confusion. It may help to learn that there are variants of the US system that are consistent, and these replace the force or mass terms with the Poundal or Slug which have different magnitudes to the lbf and lbm. If you want to know more about the lb force / lb mass topic try here, https://en.wikipedia.org/wiki/Pound_(force), though it looks like a rather long-winded explanation to me.

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You're right. It's not right.

Isp is exhaust velocity divided by Earth's surface gravity, no matter where you actually are. It's vaguely useful for launch vehicles but the fact that it gets used in deep space is mind-numbing.

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  • $\begingroup$ Oh ok I understand now, thanks for the answer! $\endgroup$ – Jake Blocker Mar 2 '17 at 0:11

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