We can calculate the minimum delta-v and the corresponding maximum altitude for a given range, d, assuming a spherical earth of circumference 40 000 km and neglecting the earth's rotation and atmosphere. Let θ be half the angle that the projectile is to go around the earth, so in degrees it is 45°×d/10 000 km. The minimum-delta-v trajectory corresponds to an ellipse with one focus at the centre of the earth and the other at the point half-way between the launch point and the destination point (somewhere inside the earth). (This is the orbit that minimizes the semi-major axis, which is equal to the sum of the distances from a point on the orbit to the two foci. Minimizing the semi-major axis minimizes the specific orbital energy and thus the delta-v, which is the speed of launch.) Geometrical arguments lead then to the following (with R being the radius of the earth, about 6370 km):\text{semi-major axis}=(1+\sin\theta)R\text{semi-minor axis}=R\sqrt{2(\sin\theta+\sin^2\theta)}=\sqrt{(R\sin\theta)\text{semi-major axis}}\text{distance of apogee from centre of earth}=(1+\sin\theta+\cos\theta)R/2\text{altitude of apogee above surface}=\left(\frac{\sin\theta}2-\sin^2\frac\theta 2\right)R=\left(\frac{\sin(\theta+\pi/4)}\sqrt{2}-\frac 1 2\right)RNote that the altitude of apogee is maximized (at about 1320 km) for a trajectory going one quarter of the way around the earth (10 000 km). Longer ranges will have lower apogees in the minimal-delta-v solution.\text{specific kinetic energy at launch}=\frac\mu R-\frac\mu\text{major axis}=\frac\mu R\frac{\sin\theta}{1+\sin\theta}\Delta v=\text{speed at launch}=\sqrt{2\frac\mu R\frac{\sin\theta}{1+\sin\theta}}=\sqrt{2gR\frac{\sin\theta}{1+\sin\theta}}

Well, that was the Jim answer. Now let's give the Elon Musk answer, in other words, take it down to the physics:https://en.wikipedia.org/wiki/Sub-orbital_spaceflightQuoteWe can calculate the minimum delta-v and the corresponding maximum altitude for a given range, d, assuming a spherical earth of circumference 40 000 km and neglecting the earth's rotation and atmosphere. Let θ be half the angle that the projectile is to go around the earth, so in degrees it is 45°×d/10 000 km. The minimum-delta-v trajectory corresponds to an ellipse with one focus at the centre of the earth and the other at the point half-way between the launch point and the destination point (somewhere inside the earth). (This is the orbit that minimizes the semi-major axis, which is equal to the sum of the distances from a point on the orbit to the two foci. Minimizing the semi-major axis minimizes the specific orbital energy and thus the delta-v, which is the speed of launch.) Geometrical arguments lead then to the following (with R being the radius of the earth, about 6370 km):\text{semi-major axis}=(1+\sin\theta)R\text{semi-minor axis}=R\sqrt{2(\sin\theta+\sin^2\theta)}=\sqrt{(R\sin\theta)\text{semi-major axis}}\text{distance of apogee from centre of earth}=(1+\sin\theta+\cos\theta)R/2\text{altitude of apogee above surface}=\left(\frac{\sin\theta}2-\sin^2\frac\theta 2\right)R=\left(\frac{\sin(\theta+\pi/4)}\sqrt{2}-\frac 1 2\right)RNote that the altitude of apogee is maximized (at about 1320 km) for a trajectory going one quarter of the way around the earth (10 000 km). Longer ranges will have lower apogees in the minimal-delta-v solution.\text{specific kinetic energy at launch}=\frac\mu R-\frac\mu\text{major axis}=\frac\mu R\frac{\sin\theta}{1+\sin\theta}\Delta v=\text{speed at launch}=\sqrt{2\frac\mu R\frac{\sin\theta}{1+\sin\theta}}=\sqrt{2gR\frac{\sin\theta}{1+\sin\theta}}Anyway, delta-v = sqrt(2*g*R*(sin(theta)/(1+sin(theta)), where theta is the distance in radians and R the radius of Earth. For the distance Florida-Spain, about 1/6 of the Earth's circumference, this gives:http://www.wolframalpha.com/input/?i=sqrt%282*9.81*6378000*%28sin%28pi%2F3%29%2F%281%2Bsin%28pi%2F3%29%29About 7,6 km/s. This is almost the orbital velocity in low Earth orbit, and much closer to the delta-v needed to reach orbit (about 9,4 according to https://en.wikipedia.org/wiki/Low_Earth_orbit) than to the 2,22 km/s that the Falcon 9 first stage currently has at staging according to http://www.spacex.com/news/2015/12/21/background-tonights-launch.

I have heard that the Falcon 9 first stage has a maximum downrange distance of about 400 miles (660 km), so it would need to go about 10 times further to land across the Atlantic. Even the shorter distance across the Gulf from Texas to Florida (which many people have suggested) is not possible.

Quote from: whitelancer64 on 01/20/2016 02:15 PMI have heard that the Falcon 9 first stage has a maximum downrange distance of about 400 miles (660 km), so it would need to go about 10 times further to land across the Atlantic. Even the shorter distance across the Gulf from Texas to Florida (which many people have suggested) is not possible.What about the core of the falcon heavy, Texas to Florida?

So, could it be done?

Quote from: GWH on 01/20/2016 02:27 PMQuote from: whitelancer64 on 01/20/2016 02:15 PMI have heard that the Falcon 9 first stage has a maximum downrange distance of about 400 miles (660 km), so it would need to go about 10 times further to land across the Atlantic. Even the shorter distance across the Gulf from Texas to Florida (which many people have suggested) is not possible.What about the core of the falcon heavy, Texas to Florida?I think that could just about be possible, but the side cores would need to be burned nearly to exhaustion and either caught on two downrange ASDS or expended, and instead of a boost-back burn, they would need to do a boost-forward burn. It's a lot of extra effort that doesn't add any value to the task of launching the customer's payload. It would be much more efficient to just have three downrange ASDS.

Quote from: whitelancer64 on 01/20/2016 02:39 PMI think that could just about be possible, but the side cores would need to be burned nearly to exhaustion and either caught on two downrange ASDS or expended, and instead of a boost-back burn, they would need to do a boost-forward burn. It's a lot of extra effort that doesn't add any value to the task of launching the customer's payload. It would be much more efficient to just have three downrange ASDS.Would be more of a vertical boost after stage separation rather than forward I would think. Advantage being higher probability of landing conditions.Although this tweet from Elon actually states that the center core reenters too FAR downrange as is: https://twitter.com/elonmusk/status/330395232564826112

I think that could just about be possible, but the side cores would need to be burned nearly to exhaustion and either caught on two downrange ASDS or expended, and instead of a boost-back burn, they would need to do a boost-forward burn. It's a lot of extra effort that doesn't add any value to the task of launching the customer's payload. It would be much more efficient to just have three downrange ASDS.