OK this has been edited to correct my oversight. it turns out that the 6 fold error in the value of g was self-cancelling - it was on both sides of the momentum balance. The height to which the Astronaut would be propelled increased by a factor of six.
Technically, you can't work with weights, you have to use masses. M=W/g where g is the acceleration due to gravity. On Earth this is 32.17 ft/sec^2, on the moon it would be about 5.4 ft/sec^2
The complete momentum balance is:
M*V(bullet) + M*V(powder) = M*V(gun and shooter)
Assuming that the mean velocity of handgun powder is around 2500 ft/sec. then the numbers are:
[(230 gr./7000gr/lb)/32.17 ft/sec^2 x 850 ft/sec] + [(6 gr./7000gr/lb.)/32.17 ft/sec^2 x 2500 ft/sec = [(160+175 lbs.)/32.17 ft/sec^2] x V(gun and shooter)
or
0.868 + .066 lb.-sec. = 10.413 lb-sec^2/ft x V ft/sec^2 (gun and shooter)
(lb.-sec. is the correct momentum unit in the Imperial system, and lb-sec^2/ft. is, unbelievably, the correct mass unit.) Note that the value for g used is the value for the environment in which the weights are taken.
Solving for V:
(0.934 lb-sec)/(10.413 lb-sec^2/ft.) = .0898 ft/sec.
Plugging that velocity into ½ M*V^2 to get the kinetic energy of the shooter and gun
KE = ½ (335/32.17) x .0898^2 = 0.042 ft-lbs.
Since KE the kinetic energy added equals PE, the potential energy gained then
½ M*V^2 = M*g*H where H is the height to which the astronaut would be propelled, assuming that his body dissipated none of the energy. In this case H is dependent on the local value of g, the greater the gravitational force, the lesser the height
0.250 ft-lbs = (335 lbs/32.17 ft-sec^2) x 5.4 ft/sec^2 x H
H = .0044 ft or about .05"
You're still going to need that .378 Weatherby......
. The compressor, heating already hot exhaust gasses would prolly provide almost instaneous meltdown as it futher heats the exhaust via compression.
