# What is the constant $g_0$ used for calculating the specific impulse in seconds?

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.

• possible duplicate of No uncertainty for standard gravitational acceleration? – Brandon Enright Feb 12 '14 at 15:30
• @BrandonEnright Thanks for the link, but while it's likely that we are talking about the same $g_0$, the question is about clarification of the rocket equation, unless you would like to assert that $g_0$ has a universal meaning in physics. (Either way, the linked question does not adequately answer my question) – Superbest Feb 15 '14 at 12:38

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.

• So if I was a Martian alien who stole some rocket parts from Earth, complete with their Earth technical manuals, and I decided to build a rocket with them on Mars, I should still use 9.81 to calculate specific impulse and not 3.72? – Superbest Feb 13 '14 at 22:44
• Yes. However I don't think there is any fundamental physics involved. It's just that all the rockets that have been built so far have been built on Earth - well, as far as we know :-) – John Rennie Feb 14 '14 at 6:39
• Thanks, that makes sense - I guess in effect specific impulse of any rocket anywhere is defined relative to an arbitrary rocket on Earth. I proposed an edit to address the question more directly based on what you said, can you please take a look and make sure it's correct? – Superbest Feb 15 '14 at 12:46
• @Superbest: you attempted to edit my answer rather than your question. In your edit you say the 9.81 is an arbitrary constant, but it isn't really. It's chosen to match the acceleration at the Earth's surface so I wouldn't say it was arbitrary.. – John Rennie Feb 15 '14 at 14:15

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).

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.