# Why is delta-v the most useful quantity for planning space travel?

Many of the questions and answers on this site make use of the concept of delta-v. Is there an easy to understand the reason why delta-v, the magnitude of the change of the velocity, $$|\mathbf{v}|$$, is so useful for understanding orbital mechanics and planning travel?

My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities. Delta-v doesn't seem to be a good proxy for any of these quantities, since it's not squared like the kinetic energy, but it's also not a vector like the linear and angular momenta.

• I'd wager that it's because that quantity is an increasing value, with respect to time, it can never decrease. You cannot lose delta-v over time, you can only increase your delta-v. Also it's agnostic to the body, unlike angular momentum. For a transfer to Mars, you could say "It will take X change in velocity from LEO to LMO." Where-as what you would say for momentum you'll have to say "I need a momentum/energy increase of X from LEO then a momentum/energy decrease of X from Mars approach to LMO". (Note I actually do not know) – Magic Octopus Urn Mar 25 '19 at 18:10
• Ultimately, we use delta-v to determine the amount of fuel needed to change the trajectory to a desired one using impulse thrusts (sudden change in velocity). But since fuel mass grows exponentially with delta-v, it’s easier to work with delta-v instead of fuel mass directly. It doesn’t matter if you slow down or speed up, the fuel consumed is the same for a given delta-v. Thus, you can accumulate each velocity change over a mission to estimate fuel needed – Paul Mar 25 '19 at 19:23
• The short version: Mass cancels out. – chrylis -on strike- Mar 26 '19 at 1:20
• Spend a few days playing Kerbal Space Program and you should develop a much better understanding of why delta-v is by far the most important factor in spacecraft design and mission planning. – MooseBoys Mar 26 '19 at 5:14
• In addition to other answers and comments, energy, linear momentum, and angular momentum aren't very useful because they depend on mass and in a spacecraft mass is not constant. A workaround could be using change of momentum per mass unit, which is (unsurprisingly) equal to delta-v. – Pere Mar 26 '19 at 22:21

Your orbit is uniquely determined by a current position (three coordinates) and velocity (three more quantities to give magnitude and direction). Going places involves changing your orbit. For instance, from a circular orbit about Earth, enter an elliptical transfer orbit to the moon, then circularize your orbit about the moon. Everything you do in space travel involves changing from one orbit to another orbit, and that is done by changing your velocity.

Heavy spaceships have to change their momentum more than light spaceships, but they both have to change their velocities by the same amount. It can be done with a long, slow acceleration, or a short, fast acceleration. Whatever ship you have, and however you do it, the delta-V is the end result that you must achieve.

Your new orbit definitely does depend on your vector delta-V, but pointing your spaceship is basically a freebie. And you don't get any of your fuel back if you accelerate first in one direction and then in the opposite direction. So, as a characteristic of your spacecraft, it really kind of is a scalar quantity, even if direction does matter when you use it.

• Ahhh... great point. Its agnostic to mass as well. I knew I was missing something. I am glad I didnt answer :). – Magic Octopus Urn Mar 25 '19 at 20:20
• Note that plane-change operations may also be involved depending on where you want to go. And while plane changing does fall under the broad heading of "changing your velocity" its in a perpendicular direction to your orbital plane, as opposed to along your trajectory's path (either forwards or backwards). For instance, the ideal shuttle launch inclination and that of the ISS are quite different, so the easy math of just matching velocities won't be enough delta v. – Draco18s no longer trusts SE Mar 26 '19 at 3:25
• In considering trajectories going between different spheres of influence, it's not uncommon to refer to "(mass) specific energy", which is just energy per unit mass. – Russell Borogove Mar 26 '19 at 16:10
• I think what's missing from this answer is why delta-v is simply additive, i.e., why the relevant cost is linear in the delta-v magnitude (rather than say quadratic, which might seem more natural on grounds of vector algebra and energy). – nanoman Mar 26 '19 at 19:19
• @nanoman Because in space you're not pushing against the ground; the faster you're going in your thrust direction, the slower your propellant in the opposite direction. – wizzwizz4 Mar 26 '19 at 20:11

Delta-v determines the amount of propellant needed.

Suppose a craft with mass $$m$$ and velocity $$\mathbf{v}$$ burns a small mass $$|\Delta m|$$ of propellant and ejects it at relative velocity $$\mathbf{u}$$, so that the craft mass changes by $$\Delta m < 0$$. This occurs over a time $$\Delta t$$ in a local gravitational field $$\mathbf{g}$$. Then the new craft velocity $$\mathbf{v} + \Delta\mathbf{v}$$ is given by $$\text{initial momentum} + \text{change in momentum due to gravity} = \text{final momentum of propellant} + \text{final momentum of craft},$$ $$m\mathbf{v} + (\Delta t)m\mathbf{g} = -\Delta m(\mathbf{v} + \mathbf{u}) + (m + \Delta m)(\mathbf{v} + \Delta\mathbf{v}).$$ Given that the increments are small, this simplifies to $$(\Delta m)\mathbf{u} = m\Delta\mathbf{v} - (\Delta t)m\mathbf{g}.$$ Dividing through by $$(\Delta t)m$$ and passing to derivatives, we have $$\frac{\dot m\mathbf{u}}{m} = \dot{\mathbf{v}} - \mathbf{g}.$$

Taking magnitudes (remembering $$\dot m < 0$$) and integrating over time, we obtain the rocket equation $$|\mathbf{u}|\ln\frac{m_0}{m} = \int dt\,|\dot{\mathbf{v}} - \mathbf{g}|,$$ where $$|\mathbf{u}|$$ is constant since it's a characteristic of the propulsion system. The right-hand side is the general definition of delta-v. We see that it is directly linked to the initial craft mass $$m_0$$, determining the initial amount of propellant needed.

Now, suppose the propellant is utilized in quick burns, during each of which $$\mathbf{u}$$ is constant in direction and $$|\dot{\mathbf{v}}| \gg |\mathbf{g}|$$, separated by coasting intervals during which $$\dot{\mathbf{v}} = \mathbf{g}$$ (i.e., $$\dot m = 0$$). Then delta-v simplifies to $$\int dt\,|\dot{\mathbf{v}} - \mathbf{g}| = |\Delta\mathbf{v}_{\text{burn 1}}| + |\Delta\mathbf{v}_{\text{burn 2}}| + \cdots,$$ hence its name. (In this equation $$\Delta\mathbf{v}$$ is not required to be small.)

• Good answer! I've had a hard time deciding which answer to accept. This one has the mathematical logic that I was looking for, but Greg's has the intuition that I was also looking for. – WaterMolecule Mar 26 '19 at 15:55

My experience in solving physics problems in mechanics has taught me that energy, linear momentum, or angular momentum are usually the most useful quantities.

To put @Greg's answer short: delta-V is a mass-normalized measure to all of the quantities you mention.