Question

If you travel by starship to the center of the galaxy such that you arrive within a subjective human lifespan, how much radiation damage will you suffer from the interstellar vacuum?

Assume special relativity holds and ignore the problems of accelerating and decelerating the spacecraft.

At what point in the journey through the interstellar vacuum would you will suffer too much radiation damage to go on?

How did you determine the amount of radiation damage? What equations did you need? What values did you use for the variables? How did you determine those values?

Answer 1

In order to estimate the radiation damage, we need to estimate the number of atoms in our path and the damage done by each atom. The solar system rotates around the center of the galaxy at a radius of about 10 kiloparsecs or about 3 x 10 ^-4 light years. Since 1 ly = (3 x 10 ⁸ m/s) X (? x 10 ⁷ s) = 10 ¹⁶ m, the distance traveled is d = 3x10 ²⁰ m. The typical density of interstellar space is 1 atom per cubic centimeter, primarily hydrogen. (By comparison, air at STP has a density of 6 x 10 ²³ molecules/2 x 10 ⁴ cm³ = 3 x 10¹⁹ cm^-3 .) If we consider a longitudinal slice of the spacecraft with surface area A = 1 cm² , it will encounter 3 x 10 ²² atoms on its journey.

The distance within a subjective human lifetime, the spacecraft must achieve a speed very close to the speed of light with a relativistic gamma factor of about 10³. The transformation to the center-of-mass system of the spacecraft, each interstellar atom hits at almost the speed of light with ? = 10³. Each atom will thus have a total energy of E = ?mc² = 10³× 1 GeV = 1 TeV. The bare proton will then transfer energy to the spacecraft material at a rate of 2 MeV per g/cm². It is used extensively in calculating the effects of the passage of high-energy charged particles through matter. Before stopping the proton can pass through about or about 10⁵ cm = 10³ m = 1 km of metal. It will thus not be possible to shield against these protons using mass shielding.

x = 10¹² eV / 2 x 10⁶ eV / (g/cm² ) = 5 x 10⁵ g/cm²

E = (3 x10²² protons/cm² ) x (2 x10⁶ eV-cm² /(g-proton))

= 6 x10²⁸ eV/g

= (6 x10²⁸ eV/g) x (1.6x10^-19 J/eV)

= 10¹⁰ J/g = 10¹³ J/kg.

The latent heat of vaporization of water is only 2×10⁶ J/kg, we would be vaporized more than 10⁶ times over. In a 30-year journey lasting 10⁹ s, we would be turned to steam within the first 10³ s or 20 min.

Radiation dose is measured in grays, where 1 Gy corresponds to an energy deposition of 1 J/kg. A lethal dose of radiation is about 10 Gy = 103 rads or about 10 J/kg. On accumulating the dose in 10^-12 of our journey, or in the first millisecond.