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 3×104 light years. Since 1 ly = (3×10⁸ m/s)×(?×10⁷ s) =10¹⁶ m, the distance traveled is d = 3×10²⁰ m. The typical density of interstellar space is 1 atom per cubic centimeter, primarily hydrogen. If we consider a longitudinal slice of the spacecraft with surface area A = 1 cm², it will encounter 3×10²² atoms on its journey. In order to travel this distance spacecraft must achieve a speed very close to the speed of light with a relativistic gamma factor of about 10³. This means that, transforming to the center of mass system of the spacecraft, each interstellar atom hits at almost the speed of light with ? = 10³. Each hydrogen atom will thus have a total energy of E = ?mc² = 10³×1 GeV = 1 TeV.

At these energies, when the hydrogen atom strikes the spacecraft it will lose its electron very quickly. The bare proton will then transfer energy to the spacecraft material (or to us) 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. This means that, before stopping, the proton can pass through about

x=10¹²(ev)/2×10⁶(ev/(g/cm²)=5×10⁵g/cm² or about 10⁵cm = 10³ m = 1 km of metal. It will thus not be possible to shield against these protons using mass shielding. Now we can consider the damage done by these protons. Consider a volume of 1 cm³ within our bodies. Our bodies have a density of about 1 g/cm² so that this volume has a mass of 1g. This means that each proton passing through this volume transfers 2MeV (through interactions with the atomic electrons). Thus, the total energy per gram transferred by all the protons is

E=(3×10²²protons/cm²)×(2×10⁶evcm²/gproton)

=6×10²⁸ev/g

=10¹⁰J/g=10¹³J/kg

Since 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.