2
$\begingroup$

If we assume that we have a given launcher that is able to produce a $\Delta v$ of 10km/s at sea level takeoff, and is able to reach an LEO of 200km above sea level, what LEO altitude would that launcher achieve if if the launchpad was located at an altitude X above sea level? Say 15km above sea level?

I stumbled over this function,

\begin{align} \Delta v = \sqrt {2GM_{\oplus} \left( \frac{1}{r_{\oplus}}- \frac{1}{r_{target}} \right)} \end{align}

which is a solution to the underlying work integral with a lot of simplifications

\begin{align} W = \int \vec F \cdot d\vec s \end{align}

Given $\Delta v$ being constant, with all simplifications I do doubt solving the equation

\begin{align} \sqrt {2GM_{\oplus} \left( \frac{1}{r_{\oplus}}- \frac{1}{r_{target}} \right)} = \sqrt {2GM_{\oplus} \left( \frac{1}{r_{\oplus}+x}- \frac{1}{r_{target}} \right)} \end{align}

towards $r_{target}$

will do the job.

$\endgroup$
  • 1
    $\begingroup$ A gain of 1km of elevation at launch should yield slightly more than a gain of 1km of orbital altitude -- gravitational falloff with distance from center of the Earth, reduced air density, increased surface rotation speed all give you increases, but the increases are all going to be very small. I would be very surprised if the result was more than +20km. $\endgroup$ – Russell Borogove Oct 3 '17 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.