[extropy-chat] MARS: Because it is hard

[Dan Clemmensen]

Thanks, Alan. OK, we'll make it 150 Km. in a vacuum and
ignoring the cost of land, the extra length is no big deal. We
can use the same linear induction technology for launch and
landing orbits and for local transportation.

Alan Eliasen wrote:

>Dan Clemmensen wrote:
>  
>
>>Now all you need is a way to land on the moon without using fuel you
>>must carry.
>>Since it's a vacuum, It is theoretically possible to reverse the linear
>>motor. Of course,
>>I don't want to be too close to the thing when they test it... 
>>Basically, you use an
>>ion drive to lower the orbit to nearly ground level, and then make sure
>>you don't miss
>>the mouth of the linear motor. The motor decelerates you from ELLO
>>(Extremely low
>>Lunar Orbit) to zero over a 100Km track. Using regenerative braking, you
>>could launch
>>an outbound capsule on a separate track as you landed an inbound capsule.
>>
>>Spike: how fast is ELLO? Surely 100KM is long enough to stay below 1G?
>>    
>>
>
>   I'm not Spike, but here's how to figure it.
>
>   Angular acceleration when you're going in a circle is given by:
>
>   a = v^2 / r
>
>   Where v is your velocity and r is the radius.
>
>   Obviously, if you're in a circular orbit, the acceleration should be equal
>to the acceleration of gravity, so g = a.  This means that velocity is:
>
>    v = sqrt[g r]
>
>   The moon's (or any spherical body's) gravitational acceleration is given by
>g = G m / r^2
>
>   G is the gravitational constant
>
>   Thus, the equation becomes:
>
>   v = sqrt[G m / r]
>
>   Plugging these numbers into Frink, a calculating tool I've written, (
>http://futureboy.homeip.net/frinkdocs/ , you can plug these numbers in online
>at http://futureboy.homeip.net/frink/ ), this all becomes simply (in Frink
>notation):
>
>   v = sqrt[moonmass G / moonradius]
>
>   (Frink knows a lot about different planetary bodies.  "G", "moongravity"
>and "moonradius" are defined units in its data file.
>
>   This gives a velocity of about 1680 m/s.
>
>   To find out how long it would take to decelerate to a stop from this
>velocity, (again, in Frink)
>
>   t = v / gravity
>
>   Where "gravity" or "gee" is one earth gravity.  This is about 171 seconds.
>
>   If you accelerate smoothly over this time, your average velocity will be
>1/2 v.  To find the distance you travel in a time t, in Frink, this is:
>
>   1/2 t v -> km
>
>   Gives a result of about 144 km that you'll need to stop.  So, your
>accelerator will either have to be more than 144 km, or you'll need to brake
>harder.
>
>   You can use Frink to plug in your own numbers.
>
>  
>