Author Topic: Could a Falcon 9 first stage land on firm land by crossing the Atlantic?  (Read 3172 times)

Offline aceshigh

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In another forum, unrelated to SpaceFlight, but where there is a thread about Spaceflight news, somebody made this question and I said it could not be done, because the barge currently is located only about 350 km from the Cape (according to some extrapolation made on Reddit) and the first stage does not reach orbit.

the person answered "Crossing the ocean would be a sub-orbital flight, though. Kind of like an ICBM, just loads more peaceful."

Well, I am no expert, so I told him that I would ask the experts here at NSF instead of being assertive about something that goes beyond my knowledge.

So, could it be done?

I would guess that SpaceX intends in the future to launch from Texas maybe and land on Cape Canaveral, which is a longer hop than the barge landings nowaday but still a loooong way from the huge hop from Cape to the other side of the Atlantic.

My impression is that even being a suborbital flight you would still need to reach higher for less atmospheric drag and faster than the first stage is able to (10 thousand km per hour as far as I know) in other to cross 6000 kilometers to the other side of the Atlantic. (5000 km to the Azores)

Offline Jim

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No.   The first stage doesn't have the energy.  Most ICBM's have 2 or more stages, so that should be an indication.

Offline rocx

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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_spaceflight
Quote
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 45d/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)R

Note 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%29
About 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.
« Last Edit: 01/20/2016 01:04 PM by rocx »
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Offline aceshigh

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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_spaceflight
Quote
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 45d/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)R

Note 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%29
About 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.


thank you for both answers. Jim is always parsimonious with words so I dont take it personally if he sounds a bit rude or too direct. :)

There is value in his directness. Sometimes, it forces other users to think for themselves...
« Last Edit: 01/20/2016 01:06 PM by aceshigh »

Offline tleski

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BTW, there is already a few threads on a similar topic. For example this one:
http://forum.nasaspaceflight.com/index.php?topic=39124.0
and that thread refers also to others discussing related ideas.

Offline whitelancer64

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.
"One bit of advice: it is important to view knowledge as sort of a semantic tree -- make sure you understand the fundamental principles, ie the trunk and big branches, before you get into the leaves/details or there is nothing for them to hang on to." - Elon Musk
"There are lies, damned lies, and launch schedules." - Larry J

Offline GWH

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.

What about the core of the falcon heavy, Texas to Florida?
« Last Edit: 01/20/2016 02:28 PM by GWH »

Offline whitelancer64

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.

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.
"One bit of advice: it is important to view knowledge as sort of a semantic tree -- make sure you understand the fundamental principles, ie the trunk and big branches, before you get into the leaves/details or there is nothing for them to hang on to." - Elon Musk
"There are lies, damned lies, and launch schedules." - Larry J

Offline Garrett

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So, could it be done?
Not if it's delivering a hefty second stage and payload to orbit.

Theoretically, a first stage by itself could reach orbit. Musk mentioned this recently, and a few folks here (well, at least one) have worked out the math and come to the same conclusion. So, for fun say, it could be sent to a low orbit and then deorbit to land in Europe/Africa, but it would be reentering at orbital speeds (Mach 20-ish), whereas the current landings involve much lower speeds (Mach 6-ish). So very unlikely it would survive such a reentry.
« Last Edit: 01/20/2016 02:51 PM by Garrett »
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Offline GWH

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.

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.

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

Offline whitelancer64


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.

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

Really? The East coast of Florida is about 1,000 miles from the Southern tip of Texas.

Maybe with cross-feed the core stage would go too far, but as far as I know, cross-feed has been put on hold. Keep in mind the tweet is from 2013.
"One bit of advice: it is important to view knowledge as sort of a semantic tree -- make sure you understand the fundamental principles, ie the trunk and big branches, before you get into the leaves/details or there is nothing for them to hang on to." - Elon Musk
"There are lies, damned lies, and launch schedules." - Larry J

Offline nadreck

What we have discussed on other threads is landing a centre core in the Keys - my vote is for Kokomo since no one lives there 8)

The highest speed a centre core could be recovered from is probably on the order of 4kms and the longest distance you can get at that speed is about 1800km though the peak altitude to get that range is probably higher than any normal flight profile, more likely the range would be between 1400 and 1600 km the distance from Boca Chica to Key West is about 1600.

Note the centre core can with all cores expended get to about 6000 m/s with a light payload on the 2nd stage. That might take it 4000km while the closest bit of the coast of Africa is at least 7000km away.

It is all well and good to quote those things that made it past your confirmation bias that other people wrote, but this is a discussion board damnit! Let us know what you think! And why!

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