5
$\begingroup$

As we know the rocket equation,

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

So do $I_{sp}$ and the mass ratio have an inverse relation, or is it that $I_{sp}$ is inversely related to natural log of the mass ratio?

Does it mean that to have logarithmic relation implies in the short interval (early phase) the difference is very large?

$\endgroup$
1
  • $\begingroup$ What do you mean by "in the short interval (early phase)"? There is no reference to time in the equation. Do you mean in the first part of a launch when the payload will be a very small fraction compared to the propellant+rocket structure? Also - what do you mean by difference? Difference between who? (I'm trying to clarify your question, although Russell Borogrove have probably already provided a good answer) $\endgroup$
    – BlueCoder
    Commented Nov 13, 2018 at 11:07

1 Answer 1

14
$\begingroup$

$I_{sp}$ is inversely related to log of mass ratio if delta v is held constant, yes, but that's not how the rocket equation is usually applied.

The way the rocket equation is usually applied is that you have a delta-v requirement given by a particular mission -- for example, the 4100 m/s needed to get from low Earth orbit to lunar orbit. Your $v_e$ will be constrained by your available choices of rocket engine, and $m_f$ will be constrained by the payload you're trying to move.

Rearranging the equation to solve for $m_i$ is left as an algebraic exercise.

$\endgroup$

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.