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” (i.e., from the atoms in interstellar space)? Assume that special relativity holds and ignore the problems of accelerating and decelerating the spacecraft.

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 10kilo/sec or about 3*10⁴ light years. Since 1 light year=(3*10⁸m/s)*(pi*10⁷s)=10¹⁶m,the distance travelled is d=3*10²⁰m. The typical density of interstellar space is 1 atom per cubic centimeter,primarily hydrogen.(By comparison, air at STP has a density of 6*10²³ molecules/2*10⁴cm³=3*10¹⁹.)If we consider a longitudinal slice of the spacecraft with surface areaA=1cm²,it will encounter 3*10²² atoms on its journey.

In order to travel this distance within a subjective human lifetime,our 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 gamma rays r=10³.Each hydrogen atom will thus have a total energy of E=rmc²=10³*1GeV=1TeV.

At these energies,when the hydrogen atom strikes the spacecraft it will loose its energy very quickly. The bare proton will then transfer energy to the spacecraft material(or to us) at a rate iof 2MeV per g/cm².This means that,before stopping,the proton can pass through about

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

or about 10⁵cm=10³=1km 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 1cm³ within our bodies. Our bodies have a density of about 1g/cm² so that this volume has a mass of 1g/cm² so that this volume has a mass of 1g. This means that each proton passing through this volume transfers 2MeV. Thus, the total energy per gram transferred by all the protons is

E=(3*10²²protons/cm²)*(2*10⁶eV-cm²/(g-proton))

=6*10²⁸eV/g

=(6*10²⁸eV/g)*(1.6*10^-19J/eV)

=10¹⁰J/g=10¹³J/kg.

since the latent heat of vapourisation of water is only 2*10⁶J/kg,we should be vapourized more than 10⁶ times over.In a 30 year journey lasting 10⁹s,we would be turned to steam within the first 10³s or 20min.

unfortunately it is even worse than that.Radiation dose is measured in grays,where 1Gy corresponds to an energy deposition of 1J/kg.A lethal dose of radiation is about 10Gy=10³rads or about 10 J/kg. We will accumulate that dose in 10^-12 of our journey,or in the first millisecond