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I didn't initially believe these numbers, but if you look at some real life stats, you are probably right.

Nominal SECO for the last starship mission was at ~8 minutes and it took ~20 minutes from deceleration started (well, from air resistance outweighed the forces of acceleration) to landing. So basically 30 minutes of flight is just the "getting up to speed" and "slowing down" part. Both account for some distance traveled, but still. ~45 minutes is probably a good bet.

Do note however that you may have to go around the world "the wrong way" to get some places due to launch constraints. But living in a world where going around the world "the wrong way" is the easier path is interesting. Imagine that.

by thrownthatway13 hours ago|

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90 minutes is a low earth orbit period.

A suborbital craft won’t be travelling at that speed.

by m4rtink10 hours ago|

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Like, you could do a partial orbit & then drop down over the destination. But it would need much more delta-v & an orbital class heat shield.

It was proposed as nuclear warhead delivery method though: https://en.wikipedia.org/wiki/Fractional_Orbital_Bombardment...

by pfdietz8 hours ago|

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Unless a suborbital trip is nearly at orbital velocity, it will involve a high, arcing trajectory. This will make the deceleration at the end unacceptably (lethally) high for all but short arcs. Some of the Mercury suborbital missions involved deceleration of 15 gees, if I recall correctly.

by phire5 hours ago|

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That was only an issue because they were fired pretty much straight up; They only went 500km down range.

You can also reduce peek deceleration forces by using aerodynamic lift to stretch out the reentry over a longer period.

by pfdietz4 hours ago|

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No, it's an issue for most arcing trajectories. Lift doesn't help much if you're coming in at a steep angle. Reentry from orbit only works well because the entry is almost flat; there even a little lift helps a lot.

by m4rtink5 hours ago|

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If the capsule/rocketplane has some lift & preferably steerable aerosurfaces then you can compensate the purely ballistic deceleration somewhat.

But yeah, if it is going down almost vertically then this will not be enough.

by pfdietz4 hours ago|

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And all but rather short ballistic trajectories (well below orbital speed) will come in at a steep angle.

Unless one has seriously variable aerodynamics, the vehicle will have to swerve to nearly horizontal over a distance of about 1 scale height of the atmosphere, which is about 10 km. The exponentially thinning atmosphere goes from "too thin to matter" to "brick wall" over a short distance.

The acceleration for turning is v^2/r; for v = 5000 m/s and r = 10 km this is 250 g.

Acceleration also limits how rapidly one can reenter from beyond Earth orbit. At > LEO velocity, the vehicle has to use (downward) lift to stay in the atmosphere, and if v is too high the required acceleration is too high.