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NASA SP-413

Nov 17, 1975 **...** The concept of spacecraft Earth, a **sphere** of finite resources and ominous ... **for**
**ten** weeks to construct a convincing picture of how ... Appendix K — The Lunar
**Gas Gun** Mass Driver . ... that space colonization offers large **potential** benefits
and ... kPa or ~**100 mm Hg**) **for** good respiration yet **low** enough to.

For more information, see NASA SP-413

I found an answer from en.wikipedia.org

Glossary of **physics** - Wikipedia

This glossary of **physics** is a list of definitions of terms and concepts relevant to
**physics**, its sub-disciplines, and related **fields**, including mechanics, materials ...

For more information, see Glossary of **physics** - Wikipedia

Given that

Potential of gun collector = 100 V

Magnetic fiels B = 2.83 * 10^{-4} T

Radius of the curved path r = 12.0 cm = 12 *10^2 m

Step 1: Determine the velocity of an ejected electron from gun due to voltage

Kinetic energy of electron K.E = \frac{1}{2} mv^2

v^2 = \frac{2 K.E}{m}

According to energy conversion K.E = energy of each electron = eV

v^2 = \frac{2 eV}{m} ....................(1)

Step 2: Velocity of electron in a spherical bulb due to magnetic field.

Electron moving in a circular path so it acquires centripetal force

F = \frac{mv^2}{r}

Force on the electron in a magnetic field

F =evB \because \theta = 90\degree

So, from the two equations we can write, evB = \frac{mv^2}{r}

eB = \frac{mv}{r}

v = \frac{eBr}{m}

v^2 = \frac{e^2B^2r^2}{m^2} ....................(2)

Step 3: Calculating the specific charge ratio

From equation (1) and (2)

\frac{e^2B^2r^2}{m} = \frac{2 eV}{m}

\frac{e}{m} = \frac{2V}{B^2 r^2}

Plug in the values in the above equation.

[math] \frac{e}{m} = \frac{2 * 100}{[2.83 * 10^{-4}]^2 (12*10^2)^2} [/math]

\frac{e}{m} = 1.734 * 10^{11} C/kg

Hence, the specific charge ratio \frac{e}{m} = 1.734 * 10^{11} C/kg