How to use the ideal rocket equation?

You need approximately: 15 minutes

Follow the steps and learn how to use the ideal rocket equation!

1. Understand what the ideal rocket equation is and why it’s useful

   There is no point in calculating something, if you don’t understand what you are doing. Hence, we start with some very brief ideal rocket equation basics. The ideal rocket equation is also known as Tsiolkovsky rocket equation or classical rocket equation.^[1] It describes the maximum achievable change in velocity for a vehicle (rocket) in relation to its mass and its exhaust velocity. And although we are dealing with rocket science, we’ll see that the ideal rocket equation is not overly complicated.
   
   Konstantin Tsiolkovsky^[2] was among the first to derive the ideal rocket equation, hence its alternative name: Tsiolkovsky rocket equation. Konstantin Tsiolkovsky was a brilliant scientist, but we are just not a big fan of naming equations after people who derived them. So we will stick to calling it ideal rocket equation in what follows. Sorry, Konstantin!
   
   The ideal rocket equation is particularly useful to understand the economics of space travel. While we don’t have futuristic (or sci-fi) capabilities such as canceling gravity, we’ll have to use good old-fashioned chemical rockets to reach space. The ideal rocket equation can tell us the amount of fuel we need in relation to the weight of the empty rocket (without fuel) and the payload (humans, satellites, equipment, etc.). Spoiler alert, the amount of rocket fuel we need is outrageous as chemical rockets are hopelessly inefficient! That’s why we calculate the cost to launch something into Earth orbit in thousands of US Dollars per kilogram.^[3]

   

2. Be aware of the limitations of the ideal rocket equation

   The ideal rocket equation makes some assumptions to simplify the calculation. It makes perfect sense to do so, but it means that the equation ignores certain real-world effects. Most notably, the ideal rocket equation ignores the effects of atmospheric drag and gravity. Such forces work against the rocket when it’s trying to fight its way up. There is also the Earth’s rotational speed, which can give a rocket a push in the right direction at launch.^[4] The equation doesn’t directly include these effects. We can still account for them though, as we’ll see.

3. Grab the correct formula and understand delta-v and the other components

   We can write the ideal rocket equation in several (equivalent) forms.^[5][6] We’ll mostly use the following: $$\Delta v=v_{\text{e}}\ln{\frac{m_{initial}}{m_{final}}}$$ On the left side of the equation, \(\Delta v\) (or delta-v) represents the change in velocity of the rocket. Often, \(\Delta v\) is given because there are known (approximate) delta-v values for certain scenarios. For example, to launch a rocket into Low Earth Orbit (LEO), we require a \(\Delta v\) of around 9.4 km/s^[7] or 33,840 km/h! This value already takes into account the fact that the rocket’s propulsion system is fighting against gravity and atmospheric drag. Without such effects the required \(\Delta v\) would be “only” 7.8 km/s.
   
   Hence, we can use the ideal rocket equation for real-world scenarios by increasing the required \(\Delta v\). This offsets the external forces that are not part of the equation. We should mention that the required \(\Delta v\) of around 9.4 km/s is not an exact figure. This significantly depends on the design of the rocket, because the portion of \(\Delta v\) that we lose to gravity and drag varies for different types of rockets. Furthermore, the extent to which a rocket benefits from the Earth’s rotational speed also differs. This depends on the launch site location (at or near the Earth’s equator is better) and the desired trajectory.^[8]
   
   On the right side of the ideal rocket equation, \(v_{\text{e}}\) is the “effective exhaust velocity”. This is closely related to the velocity of the propellant that’s exhausted (at the back of the rocket) from the rocket engine nozzle^[9]. It’s not the same though, as the effective exhaust velocity includes a correction for a pressure difference.^[10][11][12] Also, in reality the effective exhaust velocity changes during flight. However, the ideal rocket equation assumes it to be constant. And that’s good enough for most practical purposes.
   
   Next, we have \(\ln\), which is the natural logarithm function.^[13]
   
   Finally, we have two mass values. \(m_{initial}\) is the mass of the rocket before launch, and thus includes the mass of the fuel. \(m_{final}\) is the mass of the rocket after it has burned all its fuel. Please note that this assumes a single stage rocket. For multistage rockets the situation is somewhat more complex, as we’ll see later on.

4. Get familiar with some alternative forms

   You may see alternative forms of the ideal rocket equation, such as: $$\Delta v=-v_{\text{e}}\ln{\frac{m_{final}}{m_{initial}}}$$This is still the same formula, but we now have a minus sign and the mass terms in the fraction have been swapped. It’s equivalent to the formula in the previous step, because \(ln{\frac{m_{initial}}{m_{final}}}\) equals \(-ln{\frac{m_{final}}{m_{initial}}}\).
   
   Another form that you may come across is: $$m_{initial}=m_{final}\,e^\frac{\Delta v}{v_{\text{e}}}$$ Here, \(e\) is the famous exponential function, and it appears because it’s the inverse of the \(ln\) function.^[14] After rearranging the equation into this very useful form, you can determine the required amount of rocket propellant (fuel). To be precise, you can find \(m_{initial}\) if you know all quantities on the right side of this equation. And once you know \(m_{initial}\), you can simply calculate the required amount of rocket propellant by taking \(m_{final}-m_{initial}\).

5. Relate effective exhaust velocity to specific impulse

   Specific impulse^[15] is another key concept related to the ideal rocket equation. In fact, specific impulse becomes part of the ideal rocket equation if we write it in yet another alternative form. First, we take the equation from step 3: $$\Delta v=v_{\text{e}}\ln{\frac{m_{initial}}{m_{final}}}$$ Then, we recognize that \(v_{\text{e}}=g_{0}\,I_{\text{sp}}\) , and we end up with: $$\Delta v=g_{0}\,I_{\text{sp}}\ln{\frac{m_{initial}}{m_{final}}}$$Here, \(I_{\text{sp}}\) is specific impulse measured in seconds. This is a measure of effectiveness for rocket propellant. The higher the value for specific impulse, the less propellant (fuel) we need to achieve a given target \(\Delta v\).^[16][17]
   
   Furthermore, \(g_{0}\) is the “gravitational acceleration constant” or “standard gravity”^[18]. This enters the equation because of the definition of \(I_{\text{sp}}\). It is defined as:$$I_{\text{sp}}=\frac{v_{\text{e}}}{g_{0}}$$Now, the gravitational attraction is different on other planets or bodies, say for example the Moon^[19][20]. You wouldn’t be the first to think that you should therefore use a different \(g\) value when considering a rocket launch from the Moon or Mars.^[21][22] However, you would be wrong. The constant \(g_{0}\) is part of the definition of \(I_{\text{sp}}\) for the purpose of unit conversion. More specifically, it’s related to the conversion of units of weight to units of mass (and vice versa)^[23][24]. Thus, \(g_{0}\) is not there to make the equation specific for the situation on Earth. In fact, \(g_{0}\) is the same everywhere (it’s a constant, you know), while the value of the gravitational force varies slightly even on Earth itself depending on location.^[25]
   
   Looking again at \(v_{\text{e}}=g_{0}\,I_{\text{sp}}\) , we see that specific impulse becomes bigger when effective exhaust velocity becomes larger. As mentioned, specific impulse is a measure of rocket propulsion efficiency. And it makes sense that a rocket becomes more efficient when the effective exhaust velocity becomes larger. A larger exhaust velocity simply results in a stronger push in the desired direction.

6. Understand how thrust is relevant to the ideal rocket equation

   While not directly part of the ideal rocket equation, the concept of thrust is highly relevant to it. For rockets, we can express thrust with the following equation: $$F_{\text{thrust}}=v_{\text{e}}\,\dot{m}$$ Here, \(\dot{m}\) is the “mass flow rate”, the amount of mass the rocket loses per unit of time by burning and exhausting fuel. \(F_{\text{thrust}}\) is the amount of force (thrust) that is available to push the rocket in the desired direction. As per the above equation, we can calculate rocket thrust by multiplying the effective exhaust velocity \(v_{\text{e}}\) with the mass flow rate.
   
   Since we know already that \(v_{\text{e}}=g_{0}\,I_{\text{sp}}\), we can write the rocket thrust equation as:$$F_{\text{thrust}}=g_{0}\,I_{\text{sp}}\,\dot{m}$$And, this we can rearrange to: $$I_{\text{sp}}=\frac{F_{\text{thrust}}}{\dot{m}\,g_{0}}$$In this form, it is easy to see that specific impulse is a ratio of thrust to the mass flow of the rocket propellant.^[26] Hence, it’s indeed a measure of rocket engine efficiency. A higher value for \(I_{\text{sp}}\) means that the rocket produces more thrust for a given amount of rocket propellant.

7. Plug in some numbers and consider the results!

   To celebrate your newly acquired knowledge of the ideal rocket equation, let’s do some basic calculations. We’ll limit these examples to just the first stage of a hypothetical rocket. Real-world rockets consist of multiple stages, and thus benefit from dumping empty fuel tanks and accompanying engines along the way instead of dragging them along.^[27] We’ll loosely base our numbers on real-world data^[28], but round them arbitrarily for convenience and simplicity.
   
   Let’s say that at launch we have an initial mass of 600,000 kg. Furthermore, let’s assume that once the first stage completely depletes its fuel, the remaining mass is 250,000 kg. Finally, we’ll assume an effective exhaust velocity of 3000 m/s. What’s the resulting delta-v? To find out, we simply take the ideal rocket equation,$$\Delta v=v_{\text{e}}\ln{\frac{m_{initial}}{m_{final}}}$$and plug in the numbers:$$\class{gpc-math-s}{\begin{align}\Delta v & = 3000\,\text{m/s}\,\ln{\frac{600000\,\text{kg}}{250000\,\text{kg}}}\\[1ex] & \approx 2626\,\text{m/s}\\[1ex] & \approx 2.63\,\text{km/s}\end{align}}$$So, our hypothetical first stage can achieve a delta-v of “just” 2.63 km/s. Remembering that we need a delta-v of around 9.4 km/s to reach Low Earth Orbit (LEO), you may not be overly impressed. However, since it concerns only the first stage of a multistage rocket, the result makes sense.
   
   Now, let’s flip things around and see if our hypothetical first stage could reach LEO if we vastly reduce its initial mass. For starters, we can simply avoid taking stage two and three along for the ride. But even after removing stage two and three, we might have to reduce the initial mass further. We will ignore the design challenge involved with that, and just focus on the numbers. For this, we take one of the alternative formulations of the ideal rocket equation introduced earlier:$$m_{initial}=m_{final}\,e^\frac{\Delta v}{v_{\text{e}}}$$and divide both sides of the equation by \(m_{final}\), to get:$$\frac{m_{initial}}{m_{final}}=e^\frac{\Delta v}{v_{\text{e}}}$$Then we plug in the numbers on the right-hand side, including the approximate delta-v that we require to reach LEO:$$\begin{align}\frac{m_{initial}}{m_{final}} & = e^\frac{9400\,\text{m/s}}{3000\,\text{m/s}}\\[1ex] & \approx 22.95\end{align}$$The outcome implies that we require an initial mass that is almost 23 times the final mass. And in our first example above, the first stage burned through the following amount of rocket fuel:$$\class{gpc-math-s}{600000\,\text{kg}-250000\,\text{kg}=350000\,\text{kg}}$$If we keep the amount of fuel the same, and want to reach LEO, the maximum final mass is thus:$$\frac{350000\,\text{kg}}{22.95-1} \approx 15945\,\text{kg}$$ Let’s double-check this result. For this, we once again take the ideal rocket equation:$$\Delta v=v_{\text{e}}\ln{\frac{m_{initial}}{m_{final}}}$$and plug in the numbers based on the above result:$$\class{gpc-math-s}{\begin{align}\Delta v & = 3000\,\text{m/s}\ln{\frac{350000\,\text{kg}+15945\,\text{kg}}{15945\,\text{kg}}}\\[1ex] & \approx 9400\,\text{m/s}\\[1ex] & \approx 9.4\text{ km/s}\end{align}}$$So, our revised first stage can indeed reach LEO if the final mass is not more than 15,945 kg. As mentioned, we ignored any involved design challenges. Having said that, the numbers for initial and final mass are not that far off from real-world examples. One such example would be the first stage of SpaceX’s Falcon 9.^[29]
   
   Burning 350,000 kg of fuel to reach LEO with a final mass of 15,945 kg is not very efficient though! This is especially true because that final mass of 15,945 kg includes all parts of the rocket that are not fuel. The final mass would thus consist of the rocket’s engines, propellant tanks, its structural system, and so on^[30]. Hence, we would have very little room left to fit in something useful to launch into orbit. In other words, the payload that such a rocket can deliver to LEO would be very limited.
   
   As mentioned before, multistage rockets can drop parts that they no longer need. Hence, they can avoid dragging empty propellant tanks and such along. Multistage rockets are therefore more efficient and can deliver larger payloads. That’s why currently all rockets capable of launching something into Earth orbit are multistage rockets.^[31][32]
   
   If you want to apply the ideal rocket equation to multistage rockets, you can still use the same formula. You would just apply the ideal rocket equation separately to each stage of the flight. And, \(m_{final}\) for stage one would simply be the total mass of the rocket after stage one has burned its fuel, but before stage one itself is dropped. Hence, \(m_{final}\) for stage one would still include the empty propellant tanks (and other parts) of stage one. However, \(m_{initial}\) for stage two would exclude these parts. Therefore, \(m_{final}\) for stage one does not equal \(m_{initial}\) for stage two! Obviously, the same reasoning applies to stage three and so on.