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SpaceShip Three could be a point-to-point transportation method between two locations on Earth. The types of routes being seriously considered include:

As of 2008, the SpaceShipThree concept spacecraft will be used for transportation through point-to-point suborbital spaceflight. This service could provide, for example, a two-hour trip from London to Sydney or Melbourne. (Kangaroo Route)

This simple example indicates that the furthest separated locations on Earth are possibilities for such a suborbital flight. I'm wondering about the parameters of such a suborbital flight...

How elliptic would these jumps be? Would they be elliptic at all? It could just be a LEO which will decay in half a turn, but is this the most energy-efficient? Could you save on your delta v budget by making the arc higher?

I believe apogee and perigee are often given in terms of altitude above sea level, so if you almost circularized the orbit (but not entirely) for a flight from London to Sydney, the perigee might still be above sea level (since the orbit decays when it gets into heavy atmosphere.

If such a flight would be almost circularized to LEO, then the claims of Scaled Composites seem to not make any sense. Suborbital flights are a scaled down version of their plans, where they were previously considering orbital flights. How can a suborbital flight to the other side of the planet have a significant energy advantage over reaching orbit? I'm wondering if maybe this is possible by using a more highly elliptical orbit, where the perigee would remain below sea-level.

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A minimum energy ellipse between departure and destination corners of a Lambert space triangle is described on page 65 of the 1993 edition of Prussing and Conway's Orbital Mechanics textbook.

In this particular Lambert Space triangle, both r1 and r2 would be the radius of the earth, 6378 km. The 3 points of the triangle would be earth's center, Sidney and London. θ would be the angle between Sidney and London, about 152 degrees.

minimum energy ellipse

The second focus of this minimum energy ellipse would lie on the center of the chord connecting Sidney and London.

Distance between foci, (2e*a), is r cos(θ/2). The major axis of this ellipse (2a) is r(1 + sin(θ/2)). So it can be seen that eccentricity e is cos(θ/2)/(1+sin(θ/2)).

In this case altitude of perigee would be about -847 km. Altitude of apogee would be 665 km.

Speed at earth's surface would be 7.84 km/s with a flight path angle of 6.86º, nearly horizontal. Following a nearly horizontal path through earth's troposphere at 7.84 km/s is very impractical. Typically trajectory includes a vertical ascent to get above the thick atmosphere before the major horizontal burn is made. If a 100 km vertical ascent is made, why not just go for an orbital flight from London to Sidney? The delta V would be about the same. This suborbital flight makes little sense to me.

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    $\begingroup$ Why waste the additional fuel required to get into orbit, and then more fuel in the opposite direction to de-orbit? That makes little sense. In fact, I'm trying to figure out how the orbital thing would work at all. You would already be past Sydney by the time you needed to do your orbit circularization. So you would need to deorbit before even getting into orbit. If you wasted more fuel and did a brute force circularization shortly after taking off from London, then your most efficient deorbit burn would be half an orbit before the destination, which is on the other side of London! $\endgroup$
    – Mark Adler
    Commented Apr 5, 2014 at 18:35
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    $\begingroup$ It's also less safe. If you get into orbit, and your propulsion system fails, then you stay in orbit for a long time, you run out of oxygen, and everyone dies. $\endgroup$
    – Mark Adler
    Commented Apr 5, 2014 at 18:41
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    $\begingroup$ By the way, I +1'ed your answer for the Lambert solution and diagram. I did a numerical find minimum to get the answer. I like the geometry better. $\endgroup$
    – Mark Adler
    Commented Apr 5, 2014 at 18:47
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    $\begingroup$ @MarkAdler, on thinking about it you're right. You'd want the ship to re-enter well before the perigee which is 360 degrees from launch. And a deceleration to lower perigee would have you re-entering the atmosphere 180 degrees from deceleration. I guess it's better to start this trip with a 7º flight path angle. $\endgroup$
    – HopDavid
    Commented Apr 5, 2014 at 21:23
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Yes, the minimum $\Delta V$ ballistic trajectory between two points on the Earth's surface will have a negative altitude periapsis, which of course is never reached. For London to Sydney, that trajectory has a periapsis about $800\,\mathrm{km}$ below the surface.

That "orbit" is elliptic with an eccentricity of 0.12. The $\Delta V$ at the surface is $7.85\,\mathrm{km/s}$. So you are correct that it would need to get quite close to orbital speed. A more elliptic trajectory, i.e. one that lofts higher, would require more $\Delta V$.

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    $\begingroup$ How much of that would be "flight" and how much "ballistic" in distance? Does this even matter? I'd like to know if there's a chance this Kangaroo route would go low enough over my stretch of the woods (ground track should be close) to maybe try and point some optical kit towards it, take a shot from under its kilt. :) $\endgroup$
    – TildalWave
    Commented Jan 31, 2014 at 7:06
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    $\begingroup$ I simplified the calculation and just computed the ballistic trajectory from 100 km over London to 100 km over Sydney, and ignored how to get to and from 100 km. The ballistic portion is only 41 minutes, so the rest of the two hours must be "flight". Probably a lot of it just climbing to altitude to launch the SpaceShip 3. The maximum altitude was about 700 km over China. $\endgroup$
    – Mark Adler
    Commented Jan 31, 2014 at 15:58
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Perigee (and apogee) can be defined in two ways - by orbital radius around the barycenter, or by distance above surface.

from the Free Online Dictionary

per·i·gee (pĕr′ə-jē)
n.

  1. The point nearest the earth's center in the orbit of the moon or a satellite.
  2. The point in any orbit nearest to the body being orbited.

If the perigee is defined as orbital radius, then no, a negative perigee is impossible; any such perigee has a minimum of 0 (as one's distance from the barycenter begins to climb once one crosses it).

If using the above surface measure, then a negative perigee would be a body impacting orbit, and yes, it could be used for portions of an orbit.

Both modes of measure are in common enough use.

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