# Quickest “flights” across Atlantic and round the world

With the dawning of the private space industry, if someone was to build a rocket-based craft that powers you up and then glides in for a landing, what would be the quickest you could get across the Atlantic? Or half way around the world?

(Assume a typical flight trajectory - e.g. don't go through the earth or travel at ground level + 1 meter!)

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Is there something more to this question than getting us to calculate $t = d/v$ for you? – David Z Feb 17 '11 at 17:09
Follow a geodesic? (great circle). – Gordon Feb 17 '11 at 17:17
I don't think gliding would be a very good plan for getting half way around the world. – Mark Eichenlaub Feb 17 '11 at 17:23
What is the sense of a rocket to this question? Because "flight" means not to leave atmosphere, a rocket drive has no reasonable advance over a jet or ramjet drive. In general, a rocket is good for high acceleration/speed at "sprint" distance, and glide is the domain of sailplanes or U2. Combining both in atmospheric long distance flight is unwise. – Georg Feb 17 '11 at 18:02
Like David, I can also see this question merely as an exercise in $x=ct$. Moreover, I don't know why flights appear in quotes. Don't you believe that they're real flights? I gave you a minus one point. If there is some hidden depth in the question, you may reveal it, and I may change -1 to +1 after your edit haha. – Luboš Motl Feb 17 '11 at 18:58

Oh, I see. I may have understood what Alistair is asking. He wants the rocket to accelerate only at the very beginning of the flight, and then fly by free fall, without any force except for Earth's gravity. For short flights, you would imagine that the trajectory will be a parabola.

However, for significant distances in a spherically symmetric gravitational field of the Earth, the trajectory still has to be an ellipse with the Earth's center as the focus - using Newton's approximation for the law of gravity (Kepler's laws hold, but the Earth is now at the center instead of the Sun).

One could calculate the timing more accurately but it's clear that the optimal trajectory will have a pretty constant altitude - so the speed will be close to the first cosmic velocity - the velocity corresponding to the circular orbit. The optimal trajectory will be an ellipse that is very close to the circle around the Earth, of radius $6378+10 km$ or so.

The first cosmic velocity is $$v = \sqrt{gR}$$ where $g=9.8 ms^{-2}$ and $R=6,378,000 m$. You may see that $v=7.9 km/s$. For 4,000 kilometers between East Coast and Central Europe, you need about 500 seconds which is 10 minutes. Faster than BarsMonster. ;-) It takes about 1 hour and 25 minutes to fly around the Earth (40,000 km) at the orbital speed; the International Space Station is doing it 17 times a day or so.

I don't understand why BarsMonster thinks that "optimal" (for what?) velocity is suborbital. My understanding is that Alistair wants the speed. So the ellipses that are very close to circles - orbital speed - are "optimal" according to my definition of the word. They're faster. Let's hope that you have enough money to pay Richard Branson both expenses and the profit.

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Well, optimal means 'optimal energy/km'. Surely you can go at 50km/s, you will just need extra energy to keep you near earth. at 7.9km/s you will need special actions to ascend/descent. At 5km/s it will be just ballistic trajectory. – BarsMonster Feb 18 '11 at 10:03

Well, while some think it's simple x = ct, speed is a question :-) Optimal trajectory would be suborbital (but still not higher than ~150-200km), for these distances suborbital speed is gonna be ~5km/sec.

So, as a rough estimate, for 15'000km flights you will need 50 minutes as a rough estimation. If we assume that non-prepared humans can handle 2g overload at takeoff/landing, we'll need 5000/19.6 = 4.5*2 = 9 minutes for takeoff and landing.

So, in total it's gonna be about 1 hour. And tonns of fuel ;-)

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Where did you get those "optimal trajectory" and suborbital speed numbers? – spencer nelson Feb 17 '11 at 21:43
Well, this is an estimation, and I should not be off by more than 20%. – BarsMonster Feb 18 '11 at 10:00