Man-made Stutterwarp Discharge Points

Introduction

The idea of creating a man-made stutterwarp discharge point seems to appear on the mailing list at least once a year.  Gamemasters and/or players come up with the idea of hauling loads of "asteroidal rock" ( or other materials) to a spot in deep space 3.85 light-years from a solar system to create an artificial ( man-made) discharge point.  This allows them to break the 7.7 lightyear range limit without using tugships.

I hate to disappoint you, but I am afraid the job is so big, that achieving it within a practical time span is impossible.

A Brief Review of Density

When discussing the following, several people have confused density and atomic mass ( atomic weight).

Atomic weight does not convert into density.  Density is mass/standard volume.  Atomic mass is mass/mole, but a mole of each element has a different volume from a mole of another element ( in other words the volumes vary).

Quick and Dirty Proof using Director's Guide

First to discharge you need a 0.1 g field ( see 2300 AD Director's Guide ( DG ) p 62)) at the height you are orbiting.

Second turn to page 91 of the DG ( World Gravity Table).  To have an orbit with a gravity level of 0.1g, the surface gravity must be higher than 0.1g.  But to use this table we first need the density in terms of Earth density.

There is not much lead in space, iron is more common so let us try to use that ( assuming you could even find enough iron asteroids).  Let us see how big a man-made chunk of iron needs to be to have just a surface gravity of 0.1g.

Turning to page 88 of the DG ( down in the lower left corner the Equivalent Densities table).  Iron has a density of 7.9 g/cc, which is the equivalent of 1.43 Earth densities.

Going back to the World Gravity table ( DG p 91) and looking down the rightmost column ( using the 1.4 density column -- the closest match) we see that a surface gravity of 0.108 ( the first entry) is a ball 1000 km in diameter.

Now going to the prior page ( DG p 90) we can get its mass off the World Mass table.  A density of 1.4 Earths cross-referenced with 1000 km gives us 0.0006 Earth masses.

Earth masses approximately 6 x 10^24 kg ( a 6 followed by 24 zeros), which is 6 x 10^21 tons.  So our above man-assembled iron stutterwarp body/point masses:

0.0006 times 6 x 10^21 = 3.6 x 10^18 tons = 3,600,000,000,000,000,000 tons of iron

A thousand ships, each hauling a million tons each trip, each making ten trips a year, would still take 360 million years to create a man-made discharge point with a surface gravity of 0.1g!!

For a ship in a safe orbit ( tens of kilometers above the surface), this hypothetical lump of iron still would not work.  The orbiting ship is still not exposed to a high enough gravity field.  You would need to make it even larger than the above calculation.

Only if the ship landed onto the surface could it discharge, but then you have the problem of carrying enough fuel to land and take-off safely ( this extra fuel mass would either slow the ship, or reduce its cargo carrying ability).

So you can see, it is much quicker to search for Brown Dwarfs.

Formula-based Proof

One or two people have wanted to make a large space station with a "heavy center" which creates a 0.1g field at the end of a 100 meter long docking boom.  So I have done the math for a ship just 100 meters ( far far too close for a safe orbit) from the surface of spheres made of various elements.

Material  Density  Diameter   Mass
          (t/m^3)    (km)     (metric tons)
Water       1.0     7015.6    1.8x10^20 = 180,000,000,000,000,000,000   
Iron        7.8      899.8    3.0x10^18 =   3,000,000,000,000,000,000
Lead       11.35     618.5    1.4x10^18 =   1,400,000,000,000,000,000
Uranium    18.95*    371.6    5.1x10^17 =     510,000,000,000,000,000
Gold       19.3      363.9    4.9x10^17 =     490,000,000,000,000,000
Platinum   21.4      328.2    4.0x10^17 =     400,000,000,000,000,000

* a dangerous mix of all isotopes at far more than critical mass

As you can plainly see, it really takes a lot of matter ( mass) to generate that large a gravity field.

It also shatters the space station with a "heavy center" idea, since it would have to be the size of a planetoid.

So you see that moving that much mass in less than multimillions of years is still quite impossible for humanity in 2300AD.

Formulas for this proof

Given:
  Volume of a sphere = 4/3* pi * r^3
    and
  Density Ratio ( in kilograms/Cubic Meter) = grams/cubic centimeter * 1000

Then:
  Mass = volume * density in kg/cubic meter

Combining the above into one formula, becomes:
  M = 4/3* pi * r^3 * ( grams/cc * 1000)

Given:
  a = GM/(9.8*(r^2))

But often you want the value of gravity at a certain orbital height 
above the surface of a planet, so instead you can use.

  a = GM/(9.8*((p+h)^2))

Where:
a = acceleration in terms of Earth gravities ( where 1 = 9.8 m/sec^2)
G = universal gravity constant 6.67x10^-11
r = distance in meters from center of mass to object being attracted
p = planet's radius in meters
h = orbital height in meters above the planet's surface
M = mass in kilograms of the object doing the pulling

So to figure out 0.1g discharge set "a" = 0.1 and then solve for "r" or "p".

After finding "r" or "p", you need to divide M by 1000 to get mass in tons.

Conclusion

A gravity well is really easy to create, all it takes is *a lot* of matter to make a significant gravitational field.  The problem is that the job will take hundreds of millions of years to complete ( unless you move things at near-relativistic speeds -- but that creates more new problems than it solves).

In 2300 AD, you really need a decent sized moon ( like our Moon) or a planet or a star for discharge.