# g-forces of suborbital versus orbital passenger travel

https://www.theverge.com/2017/9/29/16385026/elon-musk-spacex-rocket-transportation-point-to-point

“You can’t fly humans on that same kind of orbit,” Brian Weeden, director of program planning for Secure World Foundation, told The Verge. “For one, the acceleration and the G-forces for both the launch and the reentry would kill people. I don’t have it right in front of me, but it’s a lot more than the G-forces on an astronaut we see today going up into space and coming back down, and that’s not inconsiderable.”

This is a surprising claim, but is it disingenuous? Firstly, what's the implied physical mechanism underlying the bolded text? Given that you start out with an orbital-capable rocket in the first place (as Musk's argument goes), how hard would it be to keep g-forces within tolerable range?

• the 30 minute trip was ICBM trajectory, and it is that which has G forces too high for humans. To keep the G forces in tolerable range, but still point to point in 30 minutes you would have a limited range between the two cities
– user20636
Commented Sep 29, 2017 at 23:52
• The angle of entry makes a difference. Something entering with mostly horizontal velocity and just a little vertical velocity will take a longer time to reach the ground. A suborbital hop might enter the atmosphere at a steeper angle and thus have to shed it's velocity over a shorter period of time. Commented Sep 30, 2017 at 16:35
• Since it is an ICMB trajectory, it requires only around 5000m/s to 5500 m/s of delta-v and the launch vehicle would only suffer small amount of its delta-v to gravity during ascension. Still, no one would be able to predict the deceleration until we run an re-entry simulation.
– Raze
Commented Oct 2, 2017 at 4:36

Back-of-the-envelope:

The Shuttle kept ascent acceleration at or below about 3g - tolerable for most people; other human-rated launches have gone higher. Even under the 3g constraint, burn duration to LEO is only about 8 minutes. LEO period is about 90 minutes. The furthest you would need to travel to get from anywhere to anywhere is equivalent to one half orbit (45 min coast). Assume similar descent acceleration as ascent (not more than 3g). This gives us a worst-case trip time of about 8+45+8 minutes or 61 minutes total. The claim is to get from anywhere to anywhere in not more than an hour, which is reasonable. Another claim is that typical trips may be no more than 30 minutes (because the distances involved are considerably less); this would also be reasonable.

• While the shuttle held ascent to 3g, other historical human-rated launchers have gone higher -- Saturn/Apollo peaked at 4g, and I think Soyuz is in between at 3.5g. Commented Sep 30, 2017 at 1:16
• "Assume similar descent acceleration as ascent" how reasonable is this though? Seems like that would have more to do with aerodynamics. Commented Oct 3, 2017 at 13:08
• @AlanSE The deceleration you get is what you design to achieve; aerodynamics don't just happen, they are designed-in. Commented Oct 4, 2017 at 0:49

Suborbital tourism operators expect g-forces of +3g at the start of the ballistic phase and up to +6g during reentry.

Based on this, stating that the forces would kill people seems hyperbolic (for some values of "people").

However, the paper goes on to state

it seems reasonable to suggest that the average passenger should not be subjected to g-loads greater than 3.0 +Gx and 2.0 +Gz, and that the period of exposure to these maximum g-loads should not exceed thirty minutes.

• The reasonable limits on g force for tens-of-minutes are very different from the peaks of a few seconds you'd get on ascent or reentry, though. Commented Sep 30, 2017 at 1:19
• I read that recommendation as an absolute limit on g's. Open to interpretation, though, I guess. Commented Sep 30, 2017 at 1:21

Actually considering an acceleration of 3G you would be less than 8 minutes to accelerate,. On more a more realistic approach and for “engineering” reasons the upper stage has lower accelerations.

So to simplify on what I managed to find and some back of the envelope calculations the passenger would receive some 1,5 G’s on launch up to 3G’s just before the separation. And on the second stage 1G at the start to 2/3 G’s on final burn.

This acceleration are not that high with the higher end at 3G’s, it is however constant acceleration so someone with “bad circulation” or other cardio circulatory deficiency might have trouble getting blood where it should be. That said it is not really constant acceleration given vectoring and time for acceleration the average acceleration should be at about 1,5, 1,7G’s . sqrt((750/480)^2+1^2)*”Earth gravity effect”/”vertical time”=1,6 (e/v) . For information “ev” shouldn’t be more than 1 ( and if anyone tells me this formula is wrong I know I just “simplified this last division”, I welcome you to do the whole calculation though that would take Excel spreadsheets or c++/java, etc. I’m lazy just back of the envelope calculations today.

Still for comparison when you jump some 2 fett high the ground receive some 300kgf if you have an avg height and some 75kg weight. Aceleration of the bady is about 4 g’s and acceleration of the head of the person jumping should be the second highest pasrt of the body so I would guess 5G’s at least. Droping from 5 feet maybe some 6 ~ 8 G’s of deceleration. Both of those are things anyone can do without problem.

Therefore a 3G’s maximum acceleration should be ok.

About the time a trip to the most distant point in the world would take with an acceleration of 8 minutes and considering constant acceleration and equivalent deceleration it would take 45+8min.(45min trip plus 8 minute of acele + 8 minutes decal minus 8minutefrom the trip because the 8min acceleration +8min decel = traveled distance equal to 8min at top speed.

So the acceleration should be ok for someone healthy limited to 3 Gs and they might even trade off a bit of extra delta v for lower acceleration. Travel time at 53 min for the farthest reaches of earth.(Longer range than any commercial airplane by about 45 % from 14000km to 20000km). Travel time for the longest trips from 14000km at 16 hours to 39,5minutes.

Any doubts ask away.(Btw I’m not English native, just learned the language a couple years ago.)

This had still been eating at me, because I felt like it still wasn't a fully specified problem. I think that people either assumed a ballistic trajectory (parabolic/elliptical), or that the trip would start out by achieving full orbital velocity. The options are not constrained to these two possibilities.

I investigated one option where a small amount of constant upward thrust is maintained throughout the duration of the trip. This allows the trip to proceed at a speed less than full orbital velocity... or at least it does for some trips.

https://medium.com/@AlanSE/suborbital-transit-with-hovering-a516dc8bc383

I found that A->B trips that covered less than 0.5 radians of the Earth could benefit from some of this (I call "hovering"). Distances greater than that, you would still need to achieve basically full orbital velocity. As it happens, basically all of the possible trips mentioned by Musk in the talk were further than this minimum range.

I feel like this is still an important consideration, although ultimately it proves to be irrelevant in the vast majority of practical cases. There may be some other options (that I have not thought of) that similarly reduce propellant use (relative to an orbital trip), while still keeping g-forces in the tolerable range.