For simple rockets, using the rocket equation is simple enough.

$$ \Delta v = v_e ln\left(\frac{m_i}{m_f}\right) = I_{sp} \ g \ ln\left(\frac{m_i}{m_f}\right) $$

plug in full and empty weights and Isp, and you're done.

But what to do when a rocket uses boosters? You get 2 Isp values. I expect you can't just add the delta-V of the main stages to the delta-V of the individual boosters.

Is there a way to use the Rocket equation on rockets that use boosters?

  • 1
    $\begingroup$ I'd imagine you'd do a weighted average based on the fuel mass. Just a guess on my part, though. $\endgroup$
    – UIDAlexD
    Commented Feb 9, 2018 at 15:52
  • $\begingroup$ Not at all an answer, but there's some deriving going on here. $\endgroup$
    – uhoh
    Commented Feb 9, 2018 at 16:04

4 Answers 4


For each phase of flight (stage or throttle variation or what have you) you weight the specific impulse (= exhaust velocity) by the propellant mass flow rates (i.e. consumption rates) of the different engines operating.

So if, for example the boosters are consuming 5000kg/s at 240s specific impulse and the core is consuming 1000kg/s at 310s, it works out to

$$ \frac {5000 \frac {kg} {s} * 240 {s} + 1000 \frac {kg} {s} * 310{s} } {5000 \frac {kg} {s} + 1000 \frac {kg} {s} } = 251.66 {s} $$

For 251.66s specific impulse.

  • 3
    $\begingroup$ What does that equation mean? What are the units? $\endgroup$
    – kingledion
    Commented Feb 9, 2018 at 20:32
  • 4
    $\begingroup$ I reckon the units are (kg/s * s) / (kg/s) = s, yielding 251.66s combined Isp. Just a guess. $\endgroup$
    – Brian
    Commented Feb 9, 2018 at 21:28
  • 1
    $\begingroup$ @Brian Good guess. $\endgroup$ Commented Feb 10, 2018 at 1:05

You have to consider each "phase" of flight.

For each phase of flight you need to work out the initial mass, the final mass and the overall specific impulse (total thrust/total propellant flow rate). You can then apply the rocket equation to each phase of the flight and add up the results.

  • $\begingroup$ This is also true, but the problem is that the boosters can have different Isp and dm/dt also. $\endgroup$
    – uhoh
    Commented Feb 9, 2018 at 16:34
  • 1
    $\begingroup$ If you install the Flight Engineer mod for Kerbal Space Program, it will tell you the ISP, the dV, the TWR, and burn time for every stage. That's how I get it to work ;) $\endgroup$
    – Mazura
    Commented Feb 10, 2018 at 0:03

The derivation is based on the rate of change of total momentum


For each engine,


the exhaust velocity times the rate that the propellants' mass is ejected (kg/sec). You calculate that for each engine, then add them up.

So where you would normally have

$$m \frac{dV}{dt} = -v_{ex}\frac{dm}{dt},$$

you would have instead

$$m \frac{dV}{dt} = -\Sigma \left(v_{ex, i}\frac{dm_i}{dt}\right),$$

where $v_{ex,i}$ and $\frac{dm_i}{dt}$ are the exhaust velocity and mass rate of the ith engine. You can then introduce the weighted exhaust velocity:

$$ v_{ex,m} = \frac{\sum v_{ex,i}\frac{dm_i}{dt}}{\sum \frac{dm_i}{dt}} = \frac{\sum v_{ex,i}\frac{dm_i}{dt}}{\frac{dm}{dt}}$$

Hence you have

$$m\frac{dV}{dt} = -v_{ex,m}\frac{dm}{dt}$$

which is the same equation as in the single engine case. Integrating this equation yields the following rocket equation:

$$\Delta V = v_{ex,m} ln\big(\frac{m_i}{m_f}\big)$$

  • $\begingroup$ Oh, that's much nicer, as well as pretty obvious in retrospect. Thank you Community♦ $\endgroup$
    – uhoh
    Commented Feb 10, 2018 at 0:19
  • 1
    $\begingroup$ "exhaust" - such a depressing term. $\endgroup$ Commented Feb 10, 2018 at 2:14
  • $\begingroup$ @OrganicMarble lol, indeed. $\endgroup$
    – uhoh
    Commented Feb 10, 2018 at 3:34

uhoh's answer shows how to derive combined ISP from momentum and rate. Here is the practical application of that, using the list of variables that's usually available:

  • $M_{p} = \text{mass of payload}$
  • $M_{c0} = \text{gross mass of core}$
  • $M_{c1} = \text{dry mass of core}$
  • $M_{b0} = \text{gross mass of boosters}$
  • $M_{b1} = \text{dry mass of boosters}$
  • $F_{c} = \text{thrust of core engines}$
  • $F_{b} = \text{thrust of booster engines (combined)}$
  • $v_{c} = \text{exhaust velocity of core engine}$
  • $v_{b} = \text{exhaust velocity of booster engines}$

When the engines are burning together, we can get the combined average exhaust velocity the same way from trust and flow rate:

$$v_{avg} = \frac{F}{\frac{dm}{dt}} = \frac{F_c + F_b}{\frac{dm_c}{dt}+\ \frac{dm_b}{dt}} = \frac{F_c + F_b}{\frac{F_c}{v_c} + \frac{F_b}{v_b}}$$

The burn will be in two parts. First the core and boosters burning together, and then the boosters will be detached and the core use its remaining propellant.

First, we find the mass flow of the boosters:

$$\frac{dm_b}{dt_b} = \frac{F_b}{v_b}$$

From that, we can find the booster burn time:

$$T_{boosters} = \frac{v_b \left(M_{b0} - M_{b1}\right)}{F_c}$$

From which we can find how much mass the core has used in the same time:

$$M_{core\ partial} = \frac{F_c}{v_c} \cdot T_{boosters}$$

$$M_{core\ partial} = \frac{v_b \left(M_{b0} - M_{b1}\right)}{v_c}$$

We now have the mass ratio and the combined average exhaust velocity for the first burn:

$$\Delta v_1 = v_{avg} \cdot \ln{\left(\frac{M_p + M_{c0} + M_{b0}}{M_p + M_{c0} - M_{core\ partial} + M_{b1}}\right)}$$

And for the second part, just the core alone, with the remaining propellant:

$$\Delta v_2 = v_{c} \cdot \ln{\left(\frac{M_p + M_{c0}- M_{core\ partial}}{M_p + M_{c1}}\right)}$$

Or combined and expanded:

$$\Delta v_{total} = \Delta v_1 + \Delta v_2$$

$$\Delta v_{total} = v_{avg} \cdot \ln{\left(\frac{M_p + M_{c0} + M_{b0}}{M_p + M_{c0} - M_{core\ partial} + M_{b1}}\right)} + v_{c} \cdot \ln{\left(\frac{M_p + M_{c0}- M_{core\ partial}}{M_p + M_{c1}}\right)}$$

$$\Delta v_{total} = \frac{F_c + F_b}{\frac{F_c}{v_c} + \frac{F_b}{v_b}} \cdot \ln{\left(\frac{M_p + M_{c0} + M_{b0}}{M_p + M_{c0} - \frac{v_b \left(M_{b0} - M_{b1}\right)}{v_c} + M_{b1}}\right)} + v_{c} \cdot \ln{\left(\frac{M_p + M_{c0}- \frac{v_b \left(M_{b0} - M_{b1}\right)}{v_c}}{M_p + M_{c1}}\right)}$$


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