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Wikipedia defines specific impulse as: $$ I_\mathrm{sp} = \frac{F_\mathrm{thrust}}{\dot{m} \cdot g_0} $$ The $g_0$ is said to be the acceleration at the Earth's surface.

So is it actually the acceleration at Earth's surface, or is it a universal constant that just happens to be equal to that acceleration?

In other words, is the specific impulse of the same rocket different on either of two planets which have different gravitational constants and no atmosphere?

Update: Let me restate the question in yet another way: If I was building a rocket on Mars, would I need to adjust my $I_{sp}$ values (eg. for delta-v calculations to make sense for combining engines)?

Unless I am mistaken, I understand now that the answer is no. However, I don't think that any of the current answers state this clearly enough.

Superbest
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3 Answers3

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The gavitational constant is a constant which is set to give the nominal acceleration due to gravity on the Earth's surface as explained quite well here. It is just that, a constant, there is nothing fundamental and actual gravity varies across the globe.

The reason it is in the specific impulse equation is to convert the thrust/change in mass so that they have the same units. As the units of weight are defined on Earth this is the gravity used for the conversion. If you were to set up a different system then the value of $g_0$ would change.

nivag
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The Wikipedia article you link states:

For all vehicles specific impulse (impulse per unit weight-on-Earth of propellant) in seconds can be defined by the following equation ...

The quantity $\dot{m}g_0$ is the weight flow rate of the propellant when the local gravitational acceleration is $g_0$, so the weight-on-Earth bit of the definition implies that $g_0$ is indeed the gravitational acceleration at the Earth's surface.

John Rennie
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I believe that the reason why specific impulse has the units seconds is to prevent confusion between people who use the metric or imperial system. Because the "efficiency" of a rocket is actually indicated by the effective exhaust velocity: $$ v_e=\frac{F}{\dot{m}} $$ However this has an unit of length, which is different between the metric and imperial system. Therefore by dividing it by a constant, known in both systems, which would lead to unit which is the same in both systems, will lead to less confusion. The constant used is the acceleration due to Earth's gravity at its surface (probably because most people, using specific impulse, will have used it when learning (high)school physics).

fibonatic
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