# What altitude would an ocean crossing sub-orbital flight need to reach?

So, let's save some travel time and throw a spaceship, rather than an airplane, across the Atlantic or Pacific ocean on a ballistic trajectory. How high would it have to go? How high do the doomsday ICBM's really go? How far would the Virgin Galactic spaceship go?

I've heard that the altitude required is ridiculous, and the numbers I get when I put some numbers in some equations I've found, are ridiculous too. Does ballistic ocean crossing sub-orbital flight really need to put the passengers through the van Allen Belt?

• Depends on the ocean / distance. Do you mean the so-called Kangaroo Route from UK to Australia that Virgin Galactic planned for SpaceShipThree? See HopDavid's answer to Would a point-to-point suborbital spaceflight have a “negative” perigee? For the Kangaroo Route he calculates apogee of 665 km with a point-to-point ballistic flight between Sidney and London. – TildalWave Nov 20 '14 at 1:24
• @TidalWave Yes exactly, and it looks reasonable when crossing the entire planet. But when moving the foci in order to get a regional (ocean crossing) trip, one gets a big bulge over the surface. Sub-orbital flight seems to imply higher altitude than since Apollo. Super-orbital, really. – LocalFluff Nov 20 '14 at 1:40
• Suborbital and Superorbital is not about elevation but about the velocity perpendicular to the Earth radius. If we ignore other attracting bodies in the universe, one can go arbitrarily high without reaching orbit. – gerrit Nov 20 '14 at 15:32
• It doesn't answer the question exactly, but perhaps makes it less critical - there is no practical reason suborbital transport couldn't start off with a 'non-ridiculous' altitude out of atmosphere, coast near that point, and accelerate further to move the entry point around the world without dumping too much energy into the altitude. – Saiboogu Mar 11 '19 at 20:26

The altitudes are not what I would call ridiculous. Though you seem prone to using unjustified superlatives in your questions.

The optimum, i.e. minimum injection velocity sub-orbit was provided by HopDavid in this answer. With some manipulation, you get that the maximum altitude of the optimal trajectory is:

$${r\over 2}\left(\sin{\alpha\over 2}+\cos{\alpha\over 2}-1\right)$$

where $r$ is the radius of the injection and $\alpha$ is the central angle of the traverse. The optimal sub-orbits look like this: The green lines show the central angle traversed, and the solid orange curve is the trajectory. The dotted orange curve completes the rest of the sub-orbit.

The maximum altitude is at a traverse of $\pi/2$, or a fourth of the way around the planet, about the distance between Los Angeles and Moscow (since we're talking about ICBMs). That maximum altitude is:

$${r\over 2}\left(\sqrt{2}-1\right)$$

or about $0.207\,r$. I would consider a reasonable injection altitude to be 50 km, so this gives a maximum ICBM altitude of 1330 km. That would nick the bottom of the inner van Allen belt, which starts at around $0.2\,r$. For most traverse angles you will be lower and should miss the belt. (Though watch out for the South Atlantic Anomaly if you're going from, say, Los Angeles to Port Elizabeth, South Africa.)

If you accept a penalty in the injection velocity, you can lower the peak altitude. For example, to reduce the peak altitude for a $\pi/2$ traverse by a factor of two to 665 km, you would need to increase the injection velocity from 7.17 km/s to 7.29 km/s.

Theoretically, no height is requires at all. Of course, the atmosphere makes that impractical. I would say most flights will be above the atmosphere most of the time, so probably about 30 mile minimum. 100 or more is likely as well.

• If there was no atmosphere, and the earth was flat, one could orbit just above the surface. Ballistic would be anything below that speed. – PearsonArtPhoto Dec 26 '14 at 19:34