Is g₀ a necessary term in Tsiolkovsky's Rocket Equation?

Question

I learned Tsiolkovsky's Rocket Equation and have been thinking about its gravity constant. Is it necessary to include such a constant in the equation in regions of deep space, where Earth's gravitational influence has been escaped? I've visited many websites, but I wasn't able to reach a conclusion.

Answer 1

In the Tsiolkovsky rocketry equation, the $g_0$ can be confusing at first. The reason why it is there is because of the way ISP is calculated. The Isp can be calculate with the following equation:

$$I_{\mathrm{sp}} = \frac{\text{force}}{\text{mass flow rate} \cdot g_0} = \frac{F}{\dot{m} \cdot g_0}$$

If you place that in Tsiolkovsky rocketry equation, you can see that the $g_0$ is removed:

$$\Delta v = I_{\mathrm{sp}} \cdot g_0 \cdot \ln\left(\frac{m_0}{m_f}\right)$$ $$\Delta v = \frac{F}{\dot{m} \cdot g_0} \cdot g_0 \cdot \ln\left(\frac{m_0}{m_f}\right) $$ $$\Delta v = \frac{F}{\dot{m}} \cdot \ln\left(\frac{m_0}{m_f}\right) $$

Since $g_0$ is simply a reference value tied to Earth's gravity, it cancels out in the Tsiolkovsky rocket equation and does not reflect the local gravitational field. This means that even in deep space, where Earth's gravity no longer applies, the specific impulse $I_{\mathrm{sp}}$ calculated with $g_0$ remains valid.

Answer 2

The US uses, and I'll be blunt, goofy units. The US definition of specific impulse explicitly uses $g_0$ to convert the effective exhaust velocity to units of time. Other places simply use the effective exhaust velocity as the measure of specific impulse.

There are times when specific impulse measured in units of time makes a modicum of sense. Those times are launch from a planet whose surface gravitational acceleration is close to $g_0$. Specific impulse, in units of time, yields a rough estimate of how long a well-fueled set of engines can fire. There are times when measuring mass in units of pounds-mass and force in units of pounds-force make a modicum of sense. Once again, this is launch from a planet whose surface gravitational acceleration is close to $g_0$. A launch vehicle with a total mass of one hundred thousand pounds needs an engine (or set of engines) with a launch force of at least one hundred thousand pounds.

In space, it makes much more sense to use consistent units so we can write $F=ma$ rather than $F=kma$, where $k$ is a non-unitary and non-unitless scale factor that is needed to make Newton's second law work due to the use of inconsistent units. Note that using consistent units does not merely make the scale factor $k$ have a numerical value of one. It is a very strong statement that force is not an independent unit. Force is mass times acceleration, period.

Similarly, in space it makes much more sense to use effective exhaust velocity as the measure of specific impulse rather than effective exhaust velocity divided by $g_0$. When specific impulse is expressed in units of length over time (i.e., effective exhaust velocity), the ideal rocket equation has no dependence on $g_0$.

Answer 3

No and No again! If there was anything I could banish to the great beyond, it's specific impulse. Please for the love of all that is holy, use effective exhaust velocity instead.

Impulse by itself makes sense. You measure it in Newton-seconds, or poundforce-seconds. It's the integral of the force the engine generates vs time. For an ideal constant-thrust engine, it's that constant thrust times the burn time. If you hook up a tank twice as big to the same engine, this doubles the impulse. You can run the total impulse through a close analog of the rocket equation to get delta-v directly from impulse.

But how do we compare a Space Shuttle booster with the E9-6 engine in my model rocket? Sure the booster has a thousand tons of thrust, but it weighs 500 tons by itself. So we divide by the mass of the propellant used to get specific impulse. This makes lots of sense, since we get some kind of efficiency reading. You get more impulse out of one kilogram of hydrogen and oxygen than you do out of one kilogram of modern solid propellant, and a lot more than from one kilogram of black powder.

So if the units of impulse are newton-seconds and the units of mass are kilograms, you get a unit for specific impulse of newton-seconds per kilogram. Nothing wrong with that, makes perfect sense.

The issue comes up when you use pounds force and pounds mass. You end up with poundforce-seconds per poundmass. Now someone goes and has the great idea to cancel the poundforce on top with the poundmass on the bottom, and leave "seconds".

This isn't specifically a US unit thing, it's a confusion of force and mass thing. You can do the same thing with SI, you just have to use kilograms-force instead of newtons as your unit of force. Then you get so many kilogramforce-seconds of impulse per kilogrammass of propellant, and have the same temptation to cancel the different kinds of kilograms. We just customarily use Newtons in SI, since Newtons are the one and only SI unit of force. Kilograms-force are an abomination that so far I've only seen SpaceX use, and then they usually talk about tons of force where a ton is 1000kg.

Whenever we try to rationalize specific impulse in seconds, we end up talking about how long you can burn 1 pound of fuel to get 1 pound of thrust. It's a cursed unit.

All of this hides some interesting and important physics, that comes for free just by analyzing the units of specific impulse:

$$\begin{eqnarray} \frac{N\cdot s}{kg}&=&\frac{kg\cdot m}{s^2}\frac{s}{kg} & \text{(expand N in base units)}\\ &=&(kg)(m)(s^{-2})(s)(kg^{-1}) & \text{(separate all terms)}\\ &=&(kg)(kg^{-1})(m)(s^{-2})(s) & \text{(group kg terms and s terms)}\\ &=&(m)(s^{-2})(s) & \text{(cancel kg terms)}\\ &=&\frac{m}{s} & \text{(cancel s terms and get speed!)}\end{eqnarray}$$

And at the end we are left with a speed, and not just any speed. This is the speed you need to throw stuff out the back in order to generate the required force. This is a physically present and important concept. If your engine has an effective exhaust velocity of 4000m/s and is operated in space, the exhaust will be 4km away in one second.

The only reason we call it effective exhaust velocity is that some of the flow is slower because of things like turbulence in the jet or friction against the nozzle, but conversely some of the jet is faster. It really is the "average" exhaust velocity.

Now, if you divide 4000m/s by 9.80665m/s^2, you get the conventional ~408 "seconds" of specific impulse. The reason $g_0$ is there is because that's the conversion factor between kilograms-force and kilograms-mass. Earth's gravity exerts exactly $g_0$ Newtons of force on each kilogram of mass. Conversely if you stick to pure US customary units, the exhaust velocity will be about 12000 feet per second, and you have to divide by $g_0$=about 32 feet per second squared. That $g_0$ in feet per second is the conversion between pounds force and the US customary unit of mass slug which no one actually uses. In either case since this is just a conversion factor, we use the nominal value for the surface of the Earth rather than the actual gravity at any point in the trajectory.

So if someone gives you a specification where the specific impulse is in "seconds" you just divide by the appropriate $g_0$ to get exhaust velocity, and use that whenever you use the rocket equation, just as is explained in other answers.

We only use specific impulse as "seconds" because it is conventional, but conventions can change. With your help, we can stomp out the menace of specific impulse and proclaim the gospel of effective exhaust velocity.

Answer 4

Looking at the translated copy of Tsiolkovsy's works I have bookmarked, it doesn't appear that he ever considers making up silly new units such as "seconds of specific impulse".

All calculations and equations use the natural notion of velocity of the exhaust. See for instance page 83-84 for the derivation of the equation from change in mass that you can find in any physics lecture, resulting in:

$$\frac{v}{v_l} = \ln\left(1 + \frac{M_2}{M_1}\right)$$

Looks slightly unfamiliar with using $M_2$ for propellant mass and $M_1$ for all other mass, instead of dry and total mass, but shuffling those around and using the most common notation nowadays, it's exactly the same as:

$$\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)$$

So while Tsiolkovsky can be blamed for inventing a fair chunck of rocket science, he doesn't appear to be the culprit when it comes to the tradition of throwing around arbitrary $g_0$ conversion factors to confuse students in every new generation.