in the classical theory Tsiolkowsky is normally explained as:
$\Delta v=v_{eff}*\text{ln}\frac{m_0}{m_b} $
Then the books say, for staying in LEO one need approx. $\Delta v=7.8$ km/s
But what about the drag resistance the rocket encounters and the potential energy it has to overcome? When I want to include this it should be:
$\Delta v_{requried}=7.8 km/s + t*9.81 m/s^2 + t*\frac{F_{Drag}}{m_0} = v_{eff}*\text{ln}\frac{m_0}{m_b} $
To make it more correct, one needs to integrate this formula since the mass for the aerodynamic drag is changing. Is this formula correct? So this is true for a vertical launch. Otherwise one needs to include some angles.
Is it correct that the term $t*9.81 m/s^2$ stands for the "potential energy". So if the rocket climbs 100 km up, the time t which is needed for this times gravity constant stands for the potential energy?
Thank you for your feedback Best regards
Lucas M.
100 km upthe potential energy should not depend on the time taken to reach there IMHO. Yes, you need integration; but AFAIK, there won't exist an analytic solution for the result of that integration. – AJN Jan 25 '22 at 11:55