When reading any literature regarding space propulsion, I keep getting about these terms deltaV and ISP or specific impulse. I know specific impulse is supposed to be the velocity the propellant mass leaves the exhaust, but I'm not sure what deltaV means.

For instance, I've read that from Earth low-orbit to Mars orbit we require a deltaV of 6.1, what units are those? Also, I would expect ISP to be in units of $\frac{meter}{second}$, but most texts use units of second.


Delta-V is effectively just a change of speed so it has dimensions of $LT^{-1}$ e.g. metres per second.

Suppose you have a rocket with mass $m$ that can generate a thrust (i.e. a force) $F$, then by Newton's first law the acceleration is simply:

$$ a = \frac{F}{m} $$

Acceleration is $dv/dt$, so you get the change in velocity simply by integrating the acceleration from the time you start firing your rocket $t_0$ to the time you turn off the booster $t_1$. If you're accelerating in a straight line and if you can ignore the change in mass of the rocket then this gives the equation we all learned in school physics lessons:

$$ \Delta v = a\Delta t = \frac{F}{m} (t_1 - t_0)$$

where $\Delta v$ is simply the change in velocity. For a real rocket the calculation is more complicated for two reasons. Firstly the acceleration usually isn't in a straight line, and secondly the mass of the propellant changes as it burns so $m$ is a function of time not a constant. That's why the Wikipedia article defines it as an integral. The rsulting $\Delta v$ is not a vector because it doesn't have a defined direction, but it's still a speed so you can work out the change in momentum and kinetic energy just by plugging the speed change into the usual equations.

$\Delta v$ is usually given in kilometers per second. According to the Wikipedia article you mention the $\Delta v$ for Mars is 10.7 km/sec rather than the value of 6.2 you mention.

The $\Delta v$ is useful because it tells you how much work you need to get to your target i.e. how powerful your thruster needs to be and how much fuel you need to carry.

The reason that ISP is in units of per second is because it's quoted using a dimensionless unit i.e. the weight of 1kg divided by the weight of 1kg at the earth's surface.

Why deltaV is not a vector?

Take a very simple example where you're moving the rocket between two stationary objects. You start by accelerating the rocket away from the start point, and there is a $\Delta v$ associated with this. As you approach the destination you fire the rockets again to slow down, and again there is a $\Delta v$ associated with the decelleration. If you're a planetary scientist you add the two $\Delta v$s because you want to know how much fuel is required.

However if you treat $\Delta v$ as a vector you find that the initial $\Delta v$ is positive and the final one is negative, so when you do a vector sum of the two you get $\Delta v = 0$. This is unhelpful as it does not mean you've used zero fuel!

Can you elaborate on ISP a bit?

Impulse is force times time, and it's equal to the change of momentum. Specific impulse is impulse per unit mass of fuel i.e. it's the momentum change you get for every unit mass of fuel burned. So a high ISP means you're very fuel efficient i.e. every kg of fuel burned gives you a large change in momentum and therefore velocity.

Suppose you have a specific impulse $I_{sp}$ given in units of seconds, then if you burn a mass $m$ of fuel the impulse, i.e. change of momentum is given by:

$$ I = I_{sp} \space m \space g $$

where $g$ is the acceleration due to gravity at the Earth's surface i.e. 9.81 m/sec.

  • $\begingroup$ thanks John. two things that i'm not clear: why deltaV is not a vector? if i integrate a vector along a line integral, it is still a vector, right?. the second thing is that i'm still not clear how to convert those ISP in seconds to something physically meaningful. can you elaborate on that a bit? $\endgroup$
    – user56771
    Aug 31 '12 at 13:46
  • $\begingroup$ Is this applicable to a system in a vacuum? If there is no gravity, does the g constant remain the same? $\endgroup$
    – Jack
    Apr 25 '18 at 16:55

Specific impulse and Δv describe different quantities.

Specific impulse describes the efficiency of rocket engines by telling you how much thrust (force) you are going to obtain by burning particular type of fuel at a particular rate:

\begin{equation} F_{thrust} = I_{sp} g \frac{dm}{dt} \end{equation}

Δv on the other hand describes the required change in velocity for orbital transfer to occur. Let us assume you use chemical propulsion where change in spacecraft velocity happens during very short time (compared to orbital periods) and therefore can be assumed to be instantaneous. Imagine you want to change from orbit A to orbit B at some point where they intersect. You would need to find the velocity vA on orbit A at the point of intersection and velocity vB on orbit B at the same point. Then, in order for a spacecraft to change from orbit A to B, it would have to change its velocity by

\begin{equation} \Delta v = v_B - v_A \end{equation}

vA and vB are usually calculated in practice by solving Lambert's problem. Also, see Hohmann transfer for a good example of how Δv is used and calculated.

Knowing Δv and Isp allows you to compute the amount of fuel that needs to be expended in order for a given orbital change to occur. See Tsiolkovsky rocket equation.

Δv is expressed in the same units as velocity, often in km/s. Specific impulse is expressed in seconds since by convention it is basically the exhaust velocity divided by g.

  • $\begingroup$ thanks Adam. what is $g$ in that equation? looks like a typo. $\endgroup$
    – user56771
    Aug 31 '12 at 13:48
  • $\begingroup$ g is Earth's gravity at the surface. It is present in the equations defining specific impulse simply to express the efficiency per weight of the fuel on Earth's surface. $\endgroup$ Aug 31 '12 at 14:51

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