Delta V Calculator

Created by Davide Borchia
Reviewed by Anna Szczepanek, PhD and Rijk de Wet
Last updated: Apr 13, 2022

Moving in space is quite different from doing so on the surface of a planet — discover a little bit of spacecraft flight dynamics with our delta-v calculator. How much can you change your speed up there?

Whether studying orbital maneuvers or just playing Kerbal Space Program, you may need to know what delta-v is. Keep on reading to find it out:

• What is delta-v?
• How to find delta-v?
• Some examples and how to use our delta-v calculator!

Space is dark and silent, but also fantastic to navigate! Don your suit, climb in your spaceship, and let us take care of guiding you there.

Ignition sequence start. 3, 2, 1, liftoff! 🚀

What is delta-v?

Delta-v, or $\Delta v$, is a technical way to say "difference in velocity". You can commonly find this quantity in physics, and it has a particular importance when moving in space, between orbits, bodies, or just arbitrary points.

In space, there's no air resistance nor any other source of friction. Once your starship reaches a certain speed, there is nothing to slow it down, and you can just coast until you reach your destination — and when you arrive, you have to brake, or risk overshooting! The concept of distance, then, is not that important: it only matters to humanity because it means we have to wait for our probe to reach the icy moons of Neptune, or because we are getting bored waiting to land on Mars!

When planning your space trip, it's helpful to think in terms of the propellant required to reach your destination more than the distance covered. This quantity translates easily, given the characteristics of the engine in use, to the difference of speed that can be attained while traveling.

What is the difference between exhaust velocity and impulse?

When calculating delta-v, you can use either the effective exhaust velocity $v_e$ or the specific impulse $I_{sp}$ of the engine. Those are different quantities, but they are still closely related. Let's check them out.

The specific impulse $I_{sp}$ of a reaction engine is a quantity defining the efficiency of the generation of thrust. It equals the change in momentum (in physics lingo, impulse) per unit mass of fuel — or using different words, the time for which an engine can generate thrust equal to its mass at $1\ g_0$ of acceleration. We have a whole calculator dedicated to it — go check it out!

The effective exhaust velocity $v_e$is a way to tackle the same problem, but omitting the reference to Earth's gravity ($g_0$). This is a measure of the speed of the gases produced by the engine when they exit the exhaust cone. The term effective is added because, in practice, the value of exhaust velocity varies across the cone's surface. The value used is a good approximation of a uniform velocity.

If we substitute the weight of the propellant with its flow — hence with the exhaust velocity (the two things are similar: more fuel in means more speed out), it is possible to cancel out the factor $g_0$. We use the relation:

$v_e=I_{sp} \cdot g_0$

How to calculate delta-v?

Now we have all the tools to calculate delta-v. We need to introduce the backbone of rocket science, the Tsiolkovsky rocket equation.

In a propulsive system like a rocket, the engine consumes propellant, reducing the mass of the vehicle as it moves. This decrease in mass translates into the desired change of speed following the equation:

$\begin{split} \Delta v & = v_e \cdot \ln\left({\frac{m_0}{m_t}}\right) \\ & =I_{sp}\cdot g_0 \cdot \ln\left({\frac{m_0}{m_t}}\right) \end{split}$

where $m_0$ is the initial mass, while $m_t$ is the mass at the end of the engine's operations. We used the equality between specific impulse and exhaust velocity, see above.

We can derive this equation ourselves — it's not hard (we wanted to say that it's not rocket science, but...). First, define a system where a rocket with initial mass $m+\Delta m$ at $t_0$, traveling with velocity $V$, expels a portion $\Delta m$ of its mass (the propellant) with velocity $v_e$ (the exhaust velocity). We define the momentum at the time $t$ as:

$p_{t_0} = V \cdot (m+\Delta m)$

At the time $t = t_0 + \Delta t$ the value of the momentum changed:

$p_t= (V+\Delta V) \cdot m +V_e \cdot \Delta m$

where $V_e$ is the velocity of the expelled mass from the outside of the rocket's reference system: $V_e=V-v_e$. From the rocket's perspective, the expelled mass travels at $-v_e$.

Given the absence of external forces, the momentum obeys the law of conservation, and using Newton's second law, it's possible to write:

$F_\text{net} = \lim_{\Delta t\rightarrow 0} \frac{p_t - p_{t_0}}{\Delta t} = 0$

We can substitute and then exchange $\Delta m$ with $-dm$ since we are considering a decrease in mass. Erasing two pairs of equal terms and taking the limit, we have:

$\begin{split} \frac{m\cdot dV+v_e\cdot dm}{dt}&=0\\ -m\frac{dV}{dt}&=v_e\frac{dm}{dt}\\ -\frac{dV}{dt}&=v_e\frac{dm}{m \cdot dt} \end{split}$

Finally, let's integrate. We change the limits of integration for the mass to $m_0$ and $m_t$; thus we can write:

$-\int^{V}_{V+\Delta V} dV=v_e\int^{m_t}_{m_0}\frac{1}{m}dm$

which finally yields the delta-v:

$\Delta V = v_e \cdot \ln{\frac{m_0}{m_t}}$

Here you can obviously substitute the value of $v_e$ with the specific impulse. That's all you need to know on how to calculate the delta-v!

What is a delta-v budget?

Well, you know how to calculate the delta-v, now what? Let's see how to use it.

When you launch a space vehicle, you load a certain amount of fuel with you: this way, you set in advance the amount of thrust you can generate overall, which sets the amount of velocity you can acquire (or lose) during the flight. We call this quantity delta-v budget.

Maneuvers in space are elegant, to say the least. There are no roads and routes — only geometry, gravity, and a lot of math. Freeing yourself from the Earth's gravity well requires a huge delta-v of about $9\ \text{km}/\text{s}$ just to reach low Earth orbit, or LEO), but once you're there, it takes a relatively small amount of fuel to move yourself to higher orbits, or in trajectories to other bodies.

However, refuelling in space is hard — if it's even feasible at all. Engineers need to carefully consider every eventuality before launching a spacecraft on its way. Think of the James Webb Space Telescope. It sits in an orbit around a point called Lagrange point L2, which requires constant corrections to the trajectory to be stable. The telescope has enough fuel for about ten years of operations, equalling a delta-v budget of less than $30\ \text{m}/\text{s}$.

How to use our delta-v calculator

Using our delta-v calculator is extremely easy. First, choose if you want to input the specific impulse or the exhaust speed, then fill the other required fields: the initial and final mass. Be careful to write them in the correct order!

🙋 You can use our delta-v calculator in reverse, too — input the delta-v and find out how much fuel you need, or if you need to change your engine!

Some practical examples of dubious use 😉

You are in orbit around Earth, at the altitude of the International Space Station: you want to reach the Moon. How much fuel do you need to get there?

We plan on using a Hohmann transfer, a set of two burns described by this sequence:

• The departure from a circular orbit;
• The first burn, that accelerates the vehicle from the circular orbit to an elliptic one, at its perigee;
• The coasting phase;
• The second burn to transfer the vehicle on the target circular orbit from the apogee of the elliptic one by changing its speed to match the circular orbit's one;
• The destination: a circular orbit with a larger radius.

🔎 The delta-v of a maneuver depends on the trajectory the vehicle will follow. There are many types of maneuvers, with different values of delta-v. The Hohmann transfer is generally a good choice in terms of efficiency, but it's not necessarily the best one!

The first circular orbit, at 400 km of height over the surface of Earth, requires a speed of about $v_{c_1} =7.67\ \text{km}/\text{s}$ — you can calculate the orbital speeds by using our orbital velocity calculator.

The transfer orbit, with elliptic shape, has two characteristic velocities:

• Velocity at the perigee $(v_\text{elliptic})_p=10.76\ \text{km}/\text{s}$, higher than the one in the circular orbit;
• Velocity at the apogee $(v_\text{elliptic})_a =1.82\ \text{km}/\text{s}$, lower than the one of the target orbit.

The final orbit is circular, at a distance of about $40,\!000\ \text{km}$. The corresponding orbital velocity is about $1\ \text{km}/\text{s}$.

It is straightforward to calculate the required delta-v in both burn:

• $\Delta v_1 = 3.09\ \text{km}/\text{s}$
• $\Delta v_2 = 0.82\ \text{km}/\text{s}$

Which sum to a total delta-v of $\Delta v= \Delta v_1 + \Delta v_2 = 3.91\ \text{km}/\text{s}$.

Let's now take the parameters of a classic rocket engine, the third stadium of the Saturn V, which moved the Apollo spacecrafts on the lunar transfer orbit in real life. We are going to use it for both burns.

The engine has a specific impulse of $421\ \text{s}$ in the vacuum (where it will be used). Then we consider the dry mass of the Apollo craft, assuming we are going to reach the Moon tapping into the fuel reserve. The value is $11,\!900\ \text{kg}$.

Input all of these values in the calculator, with the final mass of the spacecraft as $m_t$. The resulting initial mass $m_0$ is about $31,\!000\ \text{kg}$, almost three times the mass of the dry capsule. This is a lot of fuel!

🔎 The vast majority of our delta-v budget is usually spent on escaping the gravity attraction of Earth. As we explained, the $\Delta v$ required to reach a low Earth orbit is about $9\ \text{km}/\text{s}$ — compare it with the $4\ \text{km}/\text{s}$ needed to reach the Moon! Once we are far enough from this possessive planet, the amount of fuel required is never excessive!

Let's go the other way. Let's assume you are aboard a rocket equipped with a NERVA engine, a colossal nuclear-powered engine studied in the '70s. The specific impulse of that beast was $I_{sp}=841\ \text{s}$. Your spacecraft weights $40\ \text{t}$, and you loaded a mere $20\ \text{t}$ of fuel — and you are going to use it all. We suppose that it will be enough for quite a journey with that engine!

We calculate the delta-v with the rocket equation:

\begin{align*} \Delta v & = 841\ \text{s} \cdot 9.81\ \frac{\text{m}}{\text{s}^2} \cdot \ln\left({\frac{60,\!000}{40,\!000}}\right) \\ & = 3.345\ \text{km}/\text{s} \end{align*}

That is more than enough to reach Jupiter. It's such a pity that we didn't pursue this kind of engine.

A final word

We are sure that this calculator will be useful for your travels. Maybe you can try our UFO calculator too, or other spaceflight dynamics tools like the orbital period calculator or the escape velocity calculator!

Now, off to explore the final frontier! 👨‍🚀

FAQ

What is the delta-v?

The delta-v is the difference of velocity that a rocket engine can impose on a spacecraft as a function of the specific impulse and the variation in the mass of the vehicle itself. It is a fundamental value in planning a journey in space, where distance (even if astronomical 😉) is less of a problem than mass is.

How do I calculate the delta-v?

The delta-v can be calculated using the rocket equation: you can use either the specific impulse or the effective exhaust speed in the calculations. The formula is:

Δv = Iₛₚ * g₀* ln(m₀/mₜ) = vₑ * ln(m₀/mₜ)

What is the delta-v to reach the Moon?

Departing from a low Earth orbit to reach the surface of the Moon, the required delta-v is about 6 km/s. This value consider the capture from the Moon's gravity and the landing. A simple change of orbit would take slightly less: 4 km/s.

What is the meaning of delta-v?

Delta-v means a change in velocity. The naming comes from the Greek letter Δ and from the early use in calculus of the letter d to express a differential.

How much delta-v is needed to enter Earth's orbit?

The value of delta-v required to achieve Earth's orbit varies with the type of orbit, the location of the launch, and many other conditions, but it generally starts at about 9 km/s. We need to subtract a fraction of velocity that is lost (due to air resistance during the atmospheric portion of the journey) from the achieved orbital speed.

Davide Borchia
I want to calculate
Specific impulse
Specific impulse (Iₛₚ)
sec
Initial mass (m₀)
kg
Final mass (mₜ)
kg
Delta-v (Δv)
m/s
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