As noted in this Wikipedia article, delta-v, used in spacecraft flight dynamics, is a measure of the impulse that is needed to perform a maneuver. However, in general physics (a much longer standing branch of study, i.e., with seniority), delta-v is simply a change in velocity. Therefore, as used in the context of space flight dynamics, delta-v is not the same as the physical change in velocity of the vehicle.

For example, for a delta-v (physical) of 7.8 km/s needed to reach low Earth orbit, a delta-v (impulse) of between 9 and 10 km/s is required, the additional 1.5 to 2 km/s being due to gravity losses (the amount of energy spent fighting gravity rather than on gaining horizontal (orbital) speed) and atmospheric drag (the amount of energy spent pushing the air out of the way).

Why is the symbol "delta-v" repurposed in spacecraft flight dynamics with a significantly different meaning than the previously well established one? Wouldn't it be less confusing to use a different symbol to identify the clearly different quantity?

Is this part of why "rocket science" is generally considered to be so hard - because rocket scientists use previously standardized symbols for significantly different meanings?

  • 4
    $\begingroup$ Both usages of delta-v mean exactly the same thing, a change in velocity. In the context to spacecraft dynamics, the delta-v to perform the maneuver is the change in velocity needed to move from one orbit to another. What is interesting is that delta-v is a good 'fuel' gauge for spacecraft. Think of this as a side effect of rocket engines not another meaning of delta-v $\endgroup$
    – tl8
    Jul 31, 2017 at 6:40
  • 1
    $\begingroup$ No, they do not mean the same thing: If they did, the two numbers I cited (physical and impulse "delta-v" values, from the gravity losses article) for getting to orbit from the ground would be the same. Also, on the Wikipedia page, it explicitly states "As used in this context, it is not the same as the physical change in velocity of the vehicle." $\endgroup$ Jul 31, 2017 at 6:55
  • 1
    $\begingroup$ Delta-v is the magnitude of the change in physical velocity $\endgroup$
    – Rory Alsop
    Jul 31, 2017 at 8:47

1 Answer 1


There's no conflict here. Because delta-v is a scalar figure, the result of applying velocity changes in different directions to a single object in 3 dimensional space over a period of time doesn't necessarily add up linearly.

If a spacecraft applies a delta v of 100 m/s in one direction, then another maneuver of 100 m/s in a perpendicular direction, it has applied 200 m/s of delta v for a net change in velocity of 141 m/s. If the second maneuver is in the opposite direction, 200 m/s of delta v is applied for a net delta v of zero!

Likewise, for an ascent from Earth's surface to orbit, some 9400 m/s is applied, in various directions and with some modification from gravity and drag loss, for a net of ~7800 m/s.

If you have a car whose fuel tank provides 500 km of linear range and you start driving on real world roads with bends and turns and keep going until the tank is empty, then you measure on a map that you're only 400 km from your origin, is your car's range 400 km or 500 km? Is it inappropriate to use the term "range" for both figures?

I do object to Wikipedia's use of the term impulse, though, since that technically means "force times time", incompatible with velocity, and I'd strike that word from your question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.