Author Topic: High speed Lunar Magnetic Levitation Train  (Read 22468 times)

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #40 on: 08/14/2007 06:33 am »
Quote
pippin - 12/8/2007  8:50 AM

Quote
A_M_Swallow - 12/8/2007  8:49 AM
1. Leaving a reusable motor unit on the Moon and just throwing the payload into space has many cost and efficiency advantages.

2. a. Rotating the cable.  The linear motors form part of the towers and actually lift the cable.  Since the cable weights several thousand tons it will take a lot of starting and stopping.  If the cable rotates permanently at 2.38 km/s then stationary vehicles need attaching to the cable without touching it.  This design is not easily spun off as an inter-village transport system.

3. b. Nuclear power and/or solar power on the ground, motors in the vehicles.  A method of getting the power up to the vehicle is needed.  Power beaming is difficult because within seconds the vehicle will be over the Moon's low horizon.  Large friction forces between the power collection arms and the power cables produce unwanted wear and heat.


Sorry, but I recommend that you read a bit about how maglevs work, for example here: http://www.wikimaglev.org/

The basic idea of a maglev, aside from the levitation, of course, is to have the motor in the track, this way you can adapt power to topology and need. And you don't have to move/accelerate the motor. And this way you can get HIGH acceleration without big losses, so I don't understand why you don't want to use it.

Lots of motors in the track is ruled out by construction cost.  The track is weight that has to be lifted off the Earth to the moon using $billion space craft as trucks.  Consequently to keep costs and construction time under control the mass of the 1209 km long cable has to be minimised.  A track with a linear motor say every 2 metres would need 1209 * 1000 /2 = 604,500 motors.  If each motor weighed 83 kg the motors alone would weigh 604,500 * 83 kg =  50,174 mT.  This would require an extra 50,174 / 23 = 2182 launches making the option financially unenviable.  Producing a working system despite transportation constraints is a high level design requirement.

Details of induction linear motor used for mass estimate.
http://www.h2wtech.com/pdfs/AC%20Induction%20Linear%20Motor.pdf

Since transportation constraints permit only a small number of linear motors, the motors will have to go into either the vehicles or the towers.  Newton's third law makes the system work either way.

The lunar infrastructure to find, mine and refine hundreds of tons of carbon for the track is decades away.  An industry able to manufacture hundreds of thousands of linear motors on the Moon is unlikely to exist in the 21st century.  Consequently such parts have to be made on the Earth.

Moon rock suitable for building the towers can probably be found on the Moon.  The use of cement or large nuts & bolts to fasten the blocks together will have to be considered carefully.  Quarrying equipment and dumper trucks will need taking to the Moon.  

Between the track, tower heads, Maglev vehicles, solar panels, people, consumables and construction equipment the lunar high speed magnetic levitation project will rapidly become the biggest customer for Ares V or Jupiter 232 rockets.

Offline pippin

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Re: High speed Lunar Magnetic Levitation Train
« Reply #41 on: 08/14/2007 07:00 am »
That Motor uses eddy currents if the aluminum structure to be moved to generate the force. See my comments above.
It also uses a gap of 1-1.5 mm between the motor and the moving plate. It has an attractive force so you need to find a way to seperate the two. Especially if they have a relative velocity of 2.3 km/s. Again: see my comments above.
Sorry, no way.
Ah, and btw: Why do all this is it consumes so much Ares V flights? Easier to just move fuel for rockets up there.
Does anybody know, whether concrete works in vacuum? I thought it needs water to bind and start the reaction?

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #42 on: 08/14/2007 08:26 am »
Quote
pippin - 14/8/2007  8:00 AM

That Motor uses eddy currents if the aluminum structure to be moved to generate the force. See my comments above.
It also uses a gap of 1-1.5 mm between the motor and the moving plate. It has an attractive force so you need to find a way to seperate the two. Especially if they have a relative velocity of 2.3 km/s. Again: see my comments above.
Sorry, no way.

You can separate the motor and the moving plate by using 2 stators.  This is used when moving sheets of metal on a conveyor belt.
see
http://www.force.co.uk/page2.html

Alternatively add steel plates to the ribbon and be the first person to use Magnetodynamic Suspension (MDS).
http://en.wikipedia.org/wiki/Maglev_train

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #43 on: 08/14/2007 08:57 am »
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pippin - 14/8/2007  8:00 AM
Ah, and btw: Why do all this is it consumes so much Ares V flights? Easier to just move fuel for rockets up there.

By using a light weight track I believe the Lunar Levitation Launcher (LLL) can be constructed for less than 70 Jupiter 232 launches.  Over a 10 year period one lunar launch a month is 10 * 12 = 120 launches, giving the LLL a payback period of less than a decade.  Two decades is 240 launches.
70 * 23 mT = 1610 mT

The proposed Lunar Surface Access Module (LSAM) can take and return 4 people to the moon.  If they stay there for 6 months a village of 4 * 6 = 24 people can be supported at 1 launch a month.
http://en.wikipedia.org/wiki/Lunar_lander

If the rocket fuel is coming from the Earth mining the moon to say build space stations is not worth while, it will be cheaper to lift the materials from the Earth instead of the cargo lifting fuel.  A reusable launcher that does not burn propellant can lift materials from the moon to space.

The machine can do approximately 2 launches a day.
Number of launches = hours in a day / (acceleration time + deceleration time) = 24 / ( 5.8 + 2.9) = 2.76

The extra launch windows can be used for cargo such as regolith to act as a radiation shield or sheet metal.

Offline khallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #44 on: 08/14/2007 12:00 pm »
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A_M_Swallow - 10/8/2007  10:00 PM

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khallow - 10/8/2007  6:02 PM

Keep in mind there's also regular, old-tech train tracks, for when you're not in a big hurry. The thermal flexing might make them high maintenance, but you can ship a lot of cargo and people on cheap aluminum rails. Good, solid 19th century tech on the Moon!

You had better make the rails on the moon they are too heavy to lift from the Earth.
According to this web page good old UK railway track (UIC60) weights 60 kg/m
http://www.railway-technical.com/track.shtml#Track

For a 1210 km track 2 * 60 * 121000 = 145,200,000 kg = 145,000 mT
The sleepers are a lot extra.

To carry a 50 mT vehicle a tether mass of 2.5 kg/m should be possible using 2006 technology.  Any embedded magnetic particles needed are more mass.

Good of you to spot the strength in my proposal. Presumably it'll be a while before the tether can be manufactured on the Moon. So it'll have to be brought from Earth. On the other hand, low tech aluminum rails can be made on site. Just bring a starter factory and you can eventually produce these rails with no further input needed from Earth (aside from customers for the rail).

Incidentally, a lot of these high tech designs as used on Earth depend on having a lot of volume to fund the trains. I don't see that existing on the Moon for some time. A low tech rail solution which can be made on the Moon provides good results for low investment.
Karl Hallowell

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #45 on: 08/14/2007 02:37 pm »
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khallow - 14/8/2007  1:00 PM

Good of you to spot the strength in my proposal. Presumably it'll be a while before the tether can be manufactured on the Moon. So it'll have to be brought from Earth. On the other hand, low tech aluminum rails can be made on site. Just bring a starter factory and you can eventually produce these rails with no further input needed from Earth (aside from customers for the rail).

Incidentally, a lot of these high tech designs as used on Earth depend on having a lot of volume to fund the trains. I don't see that existing on the Moon for some time. A low tech rail solution which can be made on the Moon provides good results for low investment.

The big problem with rail on the moon is that trains operate on the flat, anything over 2 degrees causes problems.  On Earth this is solved by using bridges, embankments and tunnels.  We do not know how to automate the building of these and there is insufficient man power on the Moon.  Now if we can design a train that runs along "rope" bridges with 100 mile spans...

Offline clongton

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Re: High speed Lunar Magnetic Levitation Train
« Reply #46 on: 08/14/2007 02:42 pm »
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A_M_Swallow - 14/8/2007  10:37 AM

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khallow - 14/8/2007  1:00 PM

Good of you to spot the strength in my proposal. Presumably it'll be a while before the tether can be manufactured on the Moon. So it'll have to be brought from Earth. On the other hand, low tech aluminum rails can be made on site. Just bring a starter factory and you can eventually produce these rails with no further input needed from Earth (aside from customers for the rail).

Incidentally, a lot of these high tech designs as used on Earth depend on having a lot of volume to fund the trains. I don't see that existing on the Moon for some time. A low tech rail solution which can be made on the Moon provides good results for low investment.

The big problem with rail on the moon is that trains operate on the flat, anything over 2 degrees causes problems.  On Earth this is solved by using bridges, embankments and tunnels.  We do not know how to automate the building of these and there is insufficient man power on the Moon.  Now if we can design a train that runs along "rope" bridges with 100 mile spans...
Trains on the Moon don't need to be huge. Think San Fransico trolly cars
Chuck - DIRECT co-founder
I started my career on the Saturn-V F-1A engine

Offline renclod

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Re: High speed Lunar Magnetic Levitation Train
« Reply #47 on: 08/14/2007 03:20 pm »
A_M_  what's killing your concept is the 3g limitation and that is because you think manned vehicles.
But people might not need so frequent launches from Luna as lunar-made LOX (LLOX), for example.
I like the circular maglev rail idea, as opposed to linear maglev. I like the gradual buid-up in velocity.
What about the 7km radius rail, 2.4km/s velocity to low energy escape, 82g - a case we are discussing in the neighbour lunar sling thread.
What if we can find a good candidate crater for this ?
And build on the crater's rim ?


Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #48 on: 08/15/2007 03:29 am »
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renclod - 14/8/2007  4:20 PM

A_M_  what's killing your concept is the 3g limitation and that is because you think manned vehicles.
But people might not need so frequent launches from Luna as lunar-made LOX (LLOX), for example.
I like the circular maglev rail idea, as opposed to linear maglev. I like the gradual buid-up in velocity.
What about the 7km radius rail, 2.4km/s velocity to low energy escape, 82g - a case we are discussing in the neighbour lunar sling thread.
What if we can find a good candidate crater for this ?
And build on the crater's rim ?

Interesting idea.
We can make the loop a lot smaller if there are no people on board.
The smaller diameter means that we can beam power to the vehicle using lasers or mirrors permitting lighter solar panels.
The cooling radiators will have to be placed away from the crater side.

When building the first track we probably do not have the man power to smooth the walls of the crater so adjustable struts will be needed.  There are glues that set in ultra-violet light, which the sun provides.  The track will have to be smooth because it no longer approximates a straight line and any bumps will cause vibration.

A launch capacity of 1 to 2 mT of LOX may be sufficient in the first few years because a team of 24 will have difficulty making more.   The same crater can be used for bigger cargoes, the track just needs more metal in it.

Placing motors at 1.5 metre intervals a 7 km radius track needs 2 pi r / d = 2 * pi * 7000 / 1.5 = 29,322 motors.
At 83 kg a motor that is 2,433,726 kg = 2,434 mT.
Too heavy, the motors will have to be on the vehicle (or provide powerful pulses from a small number of places).

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #49 on: 08/15/2007 03:38 am »
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renclod - 14/8/2007  4:20 PM

A_M_  what's killing your concept is the 3g limitation and that is because you think manned vehicles.

Going the other way a 1 g maximum loop has radius r = V^2 / a = 2380 * 2380 / 9.81 = 577,411 m = 577.4 km
and a circumference of 2 pi r = 3628 km

Offline Jim

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Re: High speed Lunar Magnetic Levitation Train
« Reply #50 on: 08/15/2007 12:04 pm »
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A_M_Swallow - 14/8/2007  10:37 AM
Now if we can design a train that runs along "rope" bridges with 100 mile spans...

Spans are too long.

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #51 on: 08/15/2007 05:00 pm »
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Jim - 15/8/2007  1:04 PM

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A_M_Swallow - 14/8/2007  10:37 AM
Now if we can design a train that runs along "rope" bridges with 100 mile spans...

Spans are too long.

The spans are Ok on the Moon using Kevlar.

The spans would be too long using steel or stone on the Earth.


On the Moon spans of several hundred miles are possible using Kevlar and its rivals because Kevlar is over 10 times as strong as steel, much lighter and the Moon's gravity is 1/6 of the Earth's.  This allows enormous structures.

Values from
http://en.wikipedia.org/wiki/Tensile_strength

Longest possible cable length equation from
http://www.brantacan.co.uk/longspan1.htm

The maximum span for a material is a function of the maximum cable length and the tower height.

Lmax = M/(Dg) cos (A)

Where
Lmax = maximum length of cable in m
M = maximum stress in N/m/m.
D = density in kg/m/m/m
g = gravity in m/s/s
A = angle between the cable and the vertical at the tower in radians

Assume that the sag to length ratio is tower Height/Length is 1/10, so that A = 1.2 radians,

Tensile strength of structural steel (M) = 400 MPa = 4 E 8 N/m^2
Density of steel = 7.8 g/cm^3 = 7,800 kg/m^2
Earth gravity = 9.81 m/s/s

L = 4 E 8 / (7,800 * 9.81) * cos(1.20) =  1,894 m = 1.89 km (or 1.2 miles)


Tensile (yield) strength of Kevlar = 3620 MPa = 3.62 E 9 N/m^2
Density of Kevlar = 1.44 g/cm^3 = 1,440 kg/m^2
Moon gravity = 1.622 m/s/s

L = 3.62 E 9 / (1,440 * 1.622) * cos(1.2) = 561,600 m = 561 km (or 348 miles)

348 / 1.2 = 290 times as big.
Welcome to 21st Century construction.


On the Moon spans of several hundred miles are possible using Kevlar and its rivals.  The enormous strength of Kevlar and the weakness of the Moon's gravity are fundamental to my design of Lunar Maglev.

Offline meiza

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Re: High speed Lunar Magnetic Levitation Train
« Reply #52 on: 08/15/2007 07:20 pm »
But Kevlar is stretchy.

Offline khallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #53 on: 08/16/2007 02:29 am »
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A_M_Swallow - 15/8/2007  10:00 AM

On the Moon spans of several hundred miles are possible using Kevlar and its rivals because Kevlar is over 10 times as strong as steel, much lighter and the Moon's gravity is 1/6 of the Earth's.  This allows enormous structures.

In theory, your calculations are correct. In practice such a length would probably require towers considerably taller than your 56 km high examples or higher tension since the Moon curves considerably over that length.

Getting back to your comment about my rails proposal.

Quote
A_M_Swallow - 14/8/2007  7:37 AM

Quote
khallow - 14/8/2007  1:00 PM

Good of you to spot the strength in my proposal. Presumably it'll be a while before the tether can be manufactured on the Moon. So it'll have to be brought from Earth. On the other hand, low tech aluminum rails can be made on site. Just bring a starter factory and you can eventually produce these rails with no further input needed from Earth (aside from customers for the rail).

Incidentally, a lot of these high tech designs as used on Earth depend on having a lot of volume to fund the trains. I don't see that existing on the Moon for some time. A low tech rail solution which can be made on the Moon provides good results for low investment.

The big problem with rail on the moon is that trains operate on the flat, anything over 2 degrees causes problems.  On Earth this is solved by using bridges, embankments and tunnels.  We do not know how to automate the building of these and there is insufficient man power on the Moon.  Now if we can design a train that runs along "rope" bridges with 100 mile spans...

Keep in mind, rails can carry a lot more load than cable. On Earth, the carrying capacity of a rail is on the order of 40 tons per 40 feet, ie, a ton a foot (or about 3*10^5 N per meter of track). Installation (and perhaps construction) of your towers are going to require non-automated labor as well. So I don't see the labor advantage.

Second, there are practical solutions to both our labor problems. Namely, labor is very scarce on the Moon, but far less so on Earth. There are probably hundreds of thousands of people who have experience operating heavy earth moving or mining equipment. Send up or construct on site, the necessary lunar analogues to bulldozers, earthmovers, tunnellers, etc. And have them driven Earth-side by relatively cheap Earth teleoperators.
Karl Hallowell

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #54 on: 08/16/2007 06:11 am »
The best towers are the mountains.  We can fairly cheaply joint the tops of 2 mountains together.  The practical limit is the sag.  The stretch will have to be allowed for in the design, possibly by using winches.

Rail tracks can be made out of Kevlar or one of its rivals.  A metal coating may permit Maglev vehicles to use them.

Remote control of mining and construction machines from Earth should reduce costs.  Such machines can be tested on the Earth.


Offline clongton

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Re: High speed Lunar Magnetic Levitation Train
« Reply #55 on: 08/16/2007 11:16 am »
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A_M_Swallow - 16/8/2007  2:11 AM

Rail tracks can be made out of Kevlar or one of its rivals.  A metal coating may permit Maglev vehicles to use them.
Perhaps the metal coating could be applied in a sort-of "extrusion" process.
Chuck - DIRECT co-founder
I started my career on the Saturn-V F-1A engine

Offline meiza

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Re: High speed Lunar Magnetic Levitation Train
« Reply #56 on: 08/16/2007 11:47 am »
Carbon fiber is much less stretchy than kevlar

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #57 on: 08/16/2007 02:10 pm »
For those who wanted to combine the sling and the Maglev train.

Central pole around which an arm can rotate at 3.94 RPM.  The arm is shaped like the second hand of an analogue watch.

The long side of the arm carries the cargo.  It is 5.774 km in length and supplies 100g acceleration.

The short side of the arm is 0.2695 km long and takes 4.67 g.  This end is attached via a cable to a train driving around a 1.694 km circular track.

To fire the cargo at 2.38 km/s release when the train is doing exactly 400 km/h and pointing in the correct direction.

The cable between the train and the arm needs to be sufficiently long that the arm can bounce up and down without damaging the train.
Supply the train with solar electric power via the tracks "third rail".

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #58 on: 08/19/2007 06:20 pm »
Quote
A_M_Swallow - 15/8/2007  6:00 PM
The maximum span for a material is a function of the maximum cable length and the tower height.

Lmax = M/(Dg) cos (A)

Where
Lmax = maximum length of cable in m
M = maximum stress in N/m/m.
D = density in kg/m/m/m
g = gravity in m/s/s
A = angle between the cable and the vertical at the tower in radians

Assume that the sag to length ratio is tower Height/Length is 1/10, so that A = 1.2 radians,

Repeating for S-glass type fibreglass, which can be made on the Moon.

Tensile strength of structural fibreglass (M) = 3400 MPa = 3.4 E 9 N/m^2
Density of fibreglass = 2.6 g/cm^3 = 2,600 kg/m^2
Moon gravity = 1.622 m/s/s

Lmax =  M/(Dg) cos (A) = 3.4 E 9 / (2600 * 1.622) * cos(1.2) = 292140 m = 292 km (or 180 miles)


Since the listed lunar mountains are between 0.5 km and 4 km high limiting the drop to 1 in 100 gives an angle
of 88.85 degrees or 1.5507 radians.
[Comment things are getting too flat when 5 significant figures are needed.]

Lmax = 3.4E9 / (2600*1.622) * cos(1.5507) = 16765 m = 16 km (or 9.9 miles)

Offline A_M_Swallow

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Re: High speed Lunar Magnetic Levitation Train
« Reply #59 on: 08/20/2007 05:42 pm »
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renclod - 14/8/2007  4:20 PM

A_M_  what's killing your concept is the 3g limitation and that is because you think manned vehicles.
But people might not need so frequent launches from Luna as lunar-made LOX (LLOX), for example.
I like the circular maglev rail idea, as opposed to linear maglev. I like the gradual buid-up in velocity.
What about the 7km radius rail, 2.4km/s velocity to low energy escape, 82g - a case we are discussing in the neighbour lunar sling thread.
What if we can find a good candidate crater for this ?
And build on the crater's rim ?


A 7km radius crater is a little small for human purposes but would allow the throwing of cargo into space,
82.5 g to Lunar escape velocity and 37.3g to Low Lunar Orbit (LLO)

Escape velocity = 2.38km/s
LLO velocity = 1.6km/s

A 17 km radius crater can give people at 4g a Delta_V of 0.8167 km/s.  That is half of the Delta_V to LLO which is a major saving in propellent.  Stage 2 of the launch can be provided by a rocket.
Cargo staying on the train suffers 34g to escape velocity or 15.4g to LLO.
See the attached spread sheet for details.

A 22 km radius crater supplies half LLO at 3g.


A 66 km radius crater can send people to LLO at 4g.
This would permit the lander to rendezvous with a ship going to the Earth or Mars.
There are several craters about that size on the Moon.

A 87 km radius crater can send people to LLO at 3g.  Cargo can goto escape velocity at 6.64g.

Launch times using t = E/P = m(v^2 - u^2)/2P
Mass = 50mT and Power = 8MW

To Delta_V of 0.8km/s (half LLO)
t = 50,000 (800*800 - 0)/(2 * 8,000,000) = 2,000 sec =  0.55 hours


To Low Lunar Orbit, where m = 50mT, v = 1.6km/s, u=0, P=8MW

t = 50,000 (1,600*1,600 - 0)/(2 * 8,000,000) = 8,000 sec = 2.2 hours

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