Some months ago I was doing research on Tsiolkovsky's rocket equation, and I created a PDF "cheat sheet" (using $\LaTeX$) with the various parameters re-arranged ($m_{\text{i}}$, $m_{\text{f}}$, $I_{\text{sp}}$, $\Delta v$, etc.) to help me better understand the basics of rocket propulsion. Some of the equations I compiled for my document are shown below:

Delta-V: $~~~~\Delta v = v_{\text{exh}} \cdot \ln\left(\dfrac{m_{\text{i}}}{m_{\text{f}}}\right) $

Final mass: $~~~~m_{\text{f}} = m_{\text{i}} \cdot \exp\left(\dfrac{-\Delta v}{v_{\text{exh}}}\right) $

Initial mass: $~~~~m_{\text{i}} = m_{\text{f}} \cdot \exp\left(\dfrac{\Delta v}{v_{\text{exh}}}\right) $

Propellant mass: $~~~~ m_{\text{p}} = m_{\text{i}} - m_{\text{f}}$

Mass fraction of propellant: $~~ \dfrac{m_{\text{p}}}{m_{\text{i}}} = \dfrac{m_{\text{i}} - m_{\text{f}}}{m_{\text{i}}} = 1 - \exp\left(\dfrac{-\Delta v}{v_{\text{exh}}}\right)$

Specific impulse (mass): $~~~~I_{\text{sp}} = 4.55368\sqrt{\dfrac{\text{heat released (kJ/mol)}} {\text{mass of products (kg/mol)}}}$

Specific impulse (thrust): $~~~~I_{\text{sp}} = \dfrac{v_{\text{exh}}}{g_{\oplus}}$

Exhaust velocity: $ ~~~~v_{\text{exh}} = I_{\text{sp}} \cdot g_{\oplus} $

Granted, these equations are hideously redundant, but the stoichiometric equation of mass-based $I_{\text{sp}}$ (units of velocity) had me curious:

What explains the 4.55368 constant?

I realize now that I either (1) found that number somewhere online several months ago when doing my research, or (2) I calculated it somehow. But now (embarrassingly) I cannot remember where, and Google searching earlier today has come up empty.


The below equations are in response to the comments: I claim that the constant value has no units, and I'm defending this claim below. Please note that the radical term reduces to $\dfrac{m}{s}$, which is how the mass-based $I_{sp}$ is measured.

1 Joule is

$$ 1 \text{kg} \cdot \dfrac{\text{m}^{2}}{\text{s}^{2}}$$

From my equation above,

$$I_{sp} \left(\text{as}~\dfrac{\text{m}}{\text{s}}\right) = const \cdot \sqrt{\dfrac{~~\frac{J}{mol}~~}{~~\frac{kg}{mol}~~}}$$

The moles in each denominator (inside the radical) cancel, giving

$$\dfrac{m}{s} = const \cdot \sqrt{\dfrac{J}{kg}} = const \cdot \sqrt{\dfrac{kg \cdot \frac{m^{2}}{s^{2}}}{kg}}$$

Kilograms cancel out, giving

$$ \dfrac{m}{s} = \text{const} \cdot \sqrt{\dfrac{~\dfrac{m^{2}}{s^{2}}~}{1}} = \text{const} \cdot \sqrt{\dfrac{m^{2}}{s^{2}}} = \text{const} \cdot \sqrt{\left( \dfrac{m}{s}\right)^{2}}$$

$$ \dfrac{m}{s} = \text{const} \cdot \dfrac{m}{s} $$

  • $\begingroup$ Are you looking for why the constant is specifically 4.44368 or are you asking why there is a constant there? $\endgroup$ Jan 14, 2017 at 21:54
  • $\begingroup$ Actually, both. $\endgroup$
    – pr1268
    Jan 15, 2017 at 9:45
  • $\begingroup$ The constant is there as a conversion factor and takes in some selection of constants (or things you assume as constant). Just doing rearranging your equation with the constant and the v/g0 one dimensional analysis gives your constant dimensions of $\frac{1}{[L][T]^{-2}}$ which looks like one over an acceleration (perhaps gravity). This doesn't answer your question but I thought it might jog your memory of what constants may have gone in there. $\endgroup$ Jan 15, 2017 at 14:02
  • $\begingroup$ The constant has no dimensions, and the radical term reduces to m/s (velocity). This is a well-documented, but uncommonly-used way of expressing $I_{sp}$ according to Wikipedia. $\endgroup$
    – pr1268
    Jan 15, 2017 at 14:06
  • $\begingroup$ Also, I'm not sure of the accuracy of the value, but when I use that many significant digits, it means I either calculated it carefully, or researched/found it (likely online). The value and the meaning of this constant is what I'm curious about. $\endgroup$
    – pr1268
    Jan 15, 2017 at 14:09

2 Answers 2


Specific impulse is defined through $$F_{thrust}={\dot m}\,g\,I_{sp}$$ with $\dot m$ the mass flow rate in $\mbox{kg/s}$, $g$ gravitational acceleration, $g\approx9.81\mbox{m/s}^2$, and $F_{thrust}$ the thrust of the rocket engine in $\mbox{kg/m s}^2$. Clearly specific impulse thus has units of $\mbox{s}$.

I'm not sure exactly where your factor of $4.55368$ is coming from. Basically the maximum theoretical exhaust speed $v_{exh}$ is equal to $\sqrt{R\,T}$, with $R$ the specific gas constant and $T$ the temperature in the combustion chamber. Obviously its value will depend on the composition of the exhaust gas.

  • $\begingroup$ Well, that makes both of us not sure where the 4.55368 comes from... While I do realize $I_{sp}$ is measured in seconds, the mass-based variant is m/s to eliminate the need for Earth-specific gravity. Thank you for the gas constant- and temperature-based clarification. $\endgroup$
    – pr1268
    Jan 17, 2017 at 17:48

If all the heat released goes into the kinetic energy of the combustion products $$ \frac 12 mv^2 = m\times \hbox{heat per unit mass} $$ you have $$ v=\sqrt 2 \sqrt{\dfrac{\text{heat released (kJ/mol)}} {\text{mass of products (kg/mol)}}} $$ so $I_{sp} = v/ g= \sqrt 2/g\approx 0.144 \sqrt{\ldots} $. Thus the conversion number is neither dimensionless nor is it 4.5.


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