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

Question

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.

Answer 1

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.

Answer 2

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.)

Answer 3

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.

(not addressing vector vs. scalar)