What is the equation that relates Delta V of rocket to the specific impulse of the fuel?

Question

I can find the eqation which includes g0 but I can't find the eqation that compines Isp in secods with delta V

Thanks to the mathematicians out there

Answer 1

That is because the specific impulse is not a measure of speed change, but of the quality of the propulsion system.

I presume the equation you found is

F = Isp * (delta m)/(delta t) * g0

Well, F is your thrust, the other equation to use is

F= ma

where m is the mass of the rocket (which changes as the fuel get used), and "a" is the acceleration the rocket has as a result.

Then to get the speed change, you apply the acceleration for the time you have it

v= a t

Of course, in reality, since all those are changing factors, you need to use differential equations; the one I am giving you are only the instantaneous ones.

Answer 2

Well, you have to take such maximum Delta-V values with caution, they are really "guesstimates". Read here about why it is so: "The fourth is the maximum delta-v this technique can give (without staging). For rocket-like propulsion systems this is a function of mass fraction and exhaust velocity. Mass fraction for rocket-like systems is usually limited by propulsion system weight and tankage weight. For a system to achieve this limit, typically the payload may need to be a negligible percentage of the vehicle, and so the practical limit on some systems can be much lower." It is based on current experience and current materials, but doesn't mention for each kind of propulsion system, which assumption had been made.

Answer 3

If your serious?
Bet you bragging rights you will not find the answer here!
Only Neanderthal and Cro-magnon breeds dwell here.
I'll show you how a wheel works!!!

GOOD LORD!! I just read Pottery Boys Answer!
I stand Corrected!!! Now I add one more speicies to our breeds.
Bet you he cooks his meat! Though.

Answer 4

the impulse is equals to the change of momentum.

so F∆t = M∆v{rocket} + v∆m{fuel}

subtracting mg we get

F - mg = M∆v{rocket} / ∆t + v∆m / ∆t