Say, I'm given the data typical to the Rocket equation in its many variants: $\Delta v, I_{sp}, m_f, m_0, ρ, T$ (thrust) and the likes. And someone asks me "What's the horsepower of the space shuttle main engine?" - Assuming I don't want to cuss them off, or ask you for an answer, how would I go about calculating the power of the engine? Or "how many kilotons TNT worth of energy is spent on getting a Soyuz from launchpad to orbit" - how to calculate the total energy expenditure? And how does even Work fit into that whole image?

  • $\begingroup$ Not sure how they came up with it, but Aerojet Rocketdyne says "the SSME can attain a maximum thrust level (in vacuum) of 512,300 pounds which is equivalent to greater than 12,000,000 horsepower." So there's a data point to check your work when you come up with a formula. $\endgroup$ Commented Feb 7, 2017 at 18:00
  • $\begingroup$ Related: aerospaceweb.org/question/propulsion/q0195.shtml $\endgroup$ Commented Feb 7, 2017 at 18:12
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    $\begingroup$ @OrganicMarble Horsepower is work. Since an engine in a test fixture is fixed (hopefully) the work is moving the exhaust. Power in Watts would be just $\frac{1}{2}\dot mv^2$ where $\dot m$ is in kg/sec and $v$ is exhaust velocity in m/s, or Isp times g. However, it gets tricky when the rocket is moving, thus the real beauty of the question! $\endgroup$
    – uhoh
    Commented Feb 7, 2017 at 18:51
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    $\begingroup$ @OrganicMarble: The article uses " so we can also say that power is equivalent to the force it takes to move an object at a constant speed." - definitely not applicable in void and microgravity. $\endgroup$
    – SF.
    Commented Feb 7, 2017 at 19:02

1 Answer 1


If you're in vacuum, the power represented by the kinetic energy of the thrust is the thrust force times the exhaust velocity (g times Isp) over two:

$$P={g I_{sp}T\over 2}$$

One SSME in vacuum delivers almost seven million horsepower.

(I think they might have forgotten to divide by two on this Aerojet Rocketdyne page, or maybe they were calculating something else.)

The actual work per unit time done on the rocket is different. That is the force of the thrust times the velocity of the rocket. For example, the power applied to the rocket while the SSMEs are running full throttle is zero, while the whole thing is still bolted to the pad. Right before MECO, the maximum velocity while the engines are still running, I get one SSME could do about 24 million horsepower of work per unit time on the orbiter, if at full throttle.

I'm not sure that either horsepower number provides a useful comparison with, say, an automobile engine horsepower, if that's what you're looking for.

The SSME pump ratings on the other hand are directly comparable to an automobile engine's.

  • $\begingroup$ For energy output, just multiply power by time? $\endgroup$
    – SF.
    Commented Feb 8, 2017 at 2:06
  • $\begingroup$ The pump power calculations are indeed useful. You can use standard hydraulic/gas turbine and axial/radial pump power formulae to calculate the power of the turbomachinery devices. In our sim we calculated these powers and from the difference between them accelerated or decelerated the connecting shaft. $\endgroup$ Commented Feb 8, 2017 at 2:22
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    $\begingroup$ @OrganicMarble: I bet the Aerojet Rocketdyne's approach $P=\frac Tt$ is the simplest - torque divided by unit time. What stumped me is how a rocket that hovers (TWR=1.0) expends lots of energy but zero work. It was continuously hammered into our heads that work=energy, and this all goes out through the window here. $\endgroup$
    – SF.
    Commented Feb 8, 2017 at 2:29
  • $\begingroup$ @MarkAdler the kinetic energy of the thrust is calculated in the frame of the rocket (thrust velocity), but the kinetic energy of the rocket is calculated in a rest frame (rocket velocity). Another way to say that is when the rocket hits circa 3 km/sec the thrust is at rest and has no kinetic energy in the frame used to calculate the rocket's kinetic energy). So the two energies can't be added to get "total kinetic energy". $\endgroup$
    – uhoh
    Commented Feb 8, 2017 at 3:14
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    $\begingroup$ @SF. The total kinetic energy put into the thrust is just $mv^2/2$, where $m$ is the total mass of the propellant, and $v$ is the exhaust velocity, $gI_{sp}$. $\endgroup$
    – Mark Adler
    Commented Feb 8, 2017 at 5:12

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