Documents / Report

Defense Intelligence Reference Document Negative Mass Propulsion

Defense Intelligence Agency · 43 pages · text from the file's own layer

This Defense Intelligence Reference Document from the Defense Intelligence Agency is dated 3 January 2011. It was produced under the Advanced Aerospace Weapon System Applications (AAWSA) Program as part of a series of advanced technology reports. The report asks whether negative mass exists and whether it could be used for propulsion. It concludes that one route might be an ultra-light form of matter that could have gathered at the Moon's center. Reaching it would take a tunnel dug with thermonuclear shape charges.

  • p. 8 …the Planck length ~ 10-33 cm), then according to Heisenberg's uncertainty principle special relativity would…
  • p. 11 …is not fictitious but represents physical reality. According to Einstein's E = mc2 one obtains for…
  • p. 12 …2c (21) According to (21) the positive mass as the source of the Newtonian potential is…
  • p. 14 …account the gravitational field of this configuration. According to (27) the gravitational potential of a point…
  • p. 15 …it therefore must have negative mass components. According to SchrOdinger 4, it is these negative mass…
  • p. 16 …The velocity-of-light result was first obtained by Breit 6, according to which Dirac's…
  • p. 18 …very large negative m- =-lm+I mass. According to this hypothesis one would have for the…
  • p. 20 …8FFll!l*L 1!1!11! t!IIU:!Y According to Newtonian mechanics and Planck's…
  • p. 21 …Variation with regard to n according to ( aL)- a ( aL )=oen ar an/er (53) leads…
  • p. 24 …proper convention about the synchronization of clocks. According to Einstein, two clocks, A and B, are…
  • p. 26 …out to be equal to c, because according to (72) 2d' (74) Rather than using a…
  • p. 28 …established if one sets m • =--v· 2e According to (78) and (82), <ti and A shift…
  • p. 31 …the angular velocity of a rotating platform. According to (93), one has for the gravitational vector…
  • p. 32 …larger than in any rotating platform experiment. According to (102), the velocity reaches a maximum at…
  • p. 41 …be very different than for solid rocks. According to eqn (112) the 37 UNCLASSIFIED/ ;CFQA: QFFI…
UNCLASSIFIED/ ,'l"e" Cl"l"!e1,it tl!!L e11t I
If rs is the horizontal radius up to which the rocks at a certain depth have to be
shattered (cylindrical case), one finds by equating p(r) in (115) and (117)
(116)
The total energy required to shatter the rocks to make a tunnel from the center at the
moon where r = 0, to its surface where r = R, is then given by summing up over the
slices with radius rs and thickness dr
(117)
where as in eqn (110) s,. ~1010 erg/cm3 is the cohesive binding energy of the rocks. For
(119) one can write
'
E=llc 2 Rt:, (p,,,,)p 0 )"" f(1-x 2 )"
0
dx. (118)
0
With the help of Euler's betafunction one has
'( ')'N (! !!J .,J;;f 1-(xr dx=(ll2)B -,- ec-.
0 2 9 2 (119)
Hence,
(120)
finds that E;:;:2xl0 24 erg:::;,5xl04 kiloton=50Megaton.
It must be emphasized that this energy must be quite nonuniformly released along the
tunnel shaft. Nuclear fusion explosions below a yield of 10 kiloton become
uneconomical with only a fraction of the energy in the fissionable material (needed to
36
UNCLASSIFIED/ ,'P811. 8PPUiiliR.k Wilii &•lkl/'

Not linked to a story yet.

About this file

Report, from the dia collection. The PDF is mirrored here; the original link is under it. 43 pages are in the text index: search them above, or from the library's search.