# What is the delta-V equivalent to cross the Kármán line in vertical (suborbital) flight?

...or, well, for a typical example - how much delta-V does New Shepard carry?

For orbital launches, there's the ~1km/s estimate to overcome drag and gravitational losses on orbital launches; 8km/s of LEO speed vs around 9km/s worth of fuel on launch - but their flight profile is much different: higher speed means more losses to atmospheric drag, speedy, flattening trajectory diminishes gravitational drag, the need to keep accelerating to orbital speed, higher target altitude etc. For vertical launch, air drag will be dropping faster, MECO will occur much sooner, there will be no savings on gravitational drag whatsoever, speed at Kármán line will be near zero...

I can't even say for sure which way the expenditure changes, because some factors save fuel while others increase its usage. What's the answer?

We can get an best-case estimate by just solving the equations of motion: $$s = \frac{1}{2}(u + v)t$$

Where $$t = \frac{v-u}{a}$$

Since we want $v = 0ms^{-1}$ at $s = 100,000m$, we get:

$$s = \frac{u^{2}}{2a} \Rightarrow u = \pm \sqrt{2as}$$

which means that a single instantaneous impulse giving a velocity of ~1400$ms^{-1}$ would get us to the Kármán line (and no higher). Such a high velocity (Mach 4) in sea level atmosphere would give enormous drag forces and probably destroy the vehicle.

If we take a more reasonable burn time of 110s, a velocity at MECO of 960$ms^{-1}$ and an altitude at MECO of 55km (approximate values for New Shepard), we end up with a further $(110g)ms^{-1} \approx 1,080ms^{-1}$ for gravity losses, giving us a total of ~2030$ms^{-1}$

Drag losses are harder because we don't know the exact aerodynamic properties of the New Shepard, but taking this answer as a baseline, and noting that Falcon 9 flies much lower, much faster than the New Shepard, we can assume the drag losses will be insignificant - certainly less than 100$ms^{-1}$.

Total delta-v: ~2100$ms^{-1}$

Notes:

• Any lateral travel will affect this value (slightly), but New Shepard flies straight up and down.
• Since New Shepard is short and wide, it has quite a low ballistic coefficient, giving higher drag losses.
• New Shepard carries additional fuel for a powered hover-landing, so the initial delta-v budget will be higher.
• New Shepard has a relatively low TWR of ~1.2 - a higher TWR could decrease our gravity losses a lot and slightly increase our drag losses. Of course a higher TWR would require a larger engine, more fuel, more tanks, larger engine, more fuel... something about Tsiolkovsky and tyrants...