0
$\begingroup$

The equation $$V_{rel} \frac{dm}{dt} - mg = ma $$

  1. what is the meaning of m in the given equation is it the mass of the rocket and the point with which we are concerned or is it the initial given mass?

  2. why is there a need of V relative in the formula?

$\endgroup$
2
  • 2
    $\begingroup$ If m is the initial mass, would dm/dt make sense? $\endgroup$ Mar 24, 2019 at 23:46
  • $\begingroup$ This can maybe help $\endgroup$
    – rdbisme
    Mar 25, 2019 at 14:37

1 Answer 1

0
$\begingroup$

You might notice that Newton's second law has a form that rate of change of momentum of a system is what external force is as we might assert that internal forced to an system of particles can't change the momentum of the system now as we can se that by simple product rule we have the following result derived. Now for the significance of V and m let's start by saying m as the mass of rocket as a function of time as we can notice that as rocket propels up in the space it uses the thrust due to released gases. Now let's look at momentum of system it can be written as $(M-\Delta m) v_1 + (\Delta m) (-v_2) = 0$ which would imply that $M = \Delta m ( v_1 + v_2)$ now from kinematics point of view the velocity term can be represented as $V_{\rm rel}$ and differentiating the same would yield what you need.

$\endgroup$
1
  • $\begingroup$ Please don't post formulas as pictures or plain text, but use MathJax instead. MathJax is easy for people on all devices to read, and can show up clearer on different screen sizes and resolutions. I've edited it here as an example. Look at this Math SE meta post for a quick tutorial. $\endgroup$
    – user191954
    Mar 28, 2019 at 7:56

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.