Find the approximate mass of the dark and luminous matter in the Milky Way galaxy. Assume the luminous matter is due to approximately \({\rm{1}}{{\rm{0}}^{{\rm{11}}}}\) stars of average mass \({\rm{1}}{\rm{.5}}\) times that of our Sun, and take the dark matter to be \({\rm{10}}\) times as massive as the luminous matter.

Short Answer

Expert verified

The mass is obtained as: \({M_{tot}}{\rm{ = 3}}{\rm{.28 x 1}}{{\rm{0}}^{{\rm{42}}}}{\rm{ kg}}\).

Step by step solution

01

Define Cosmology

Cosmology is the study of the character and evolution of the universe.The two most important features of the universe are the cosmological red shifts of its galaxies being proportional to distance and its cosmic microwave background (CMBR).

02

Evaluating the mass

The mass of luminous matter of the value \({\rm{m}}\) in our galaxy is:

\(\begin{array}{c}{M_L} &= {\rm{1}}{{\rm{0}}^{{\rm{11}}}}{\rm{ \times 1}}{\rm{.5 \times }}{{\rm{m}}_{\rm{s}}}\\ & = {\rm{1}}{{\rm{0}}^{{\rm{11}}}}{\rm{ \times 1}}{\rm{.5 \times (1}}{\rm{.99 \times 1}}{{\rm{0}}^{{\rm{30}}}}{\rm{kg)}}\\ &= {\rm{2}}{\rm{.98 \times 1}}{{\rm{0}}^{{\rm{41}}}}{\rm{kg}}\end{array}\)

The value of\({{\rm{m}}_{\rm{s}}}\)denotes the mass of the Sun.

We then know that the mass of dark matter is:

\({M_D} = 10{M_L}\)....... (i)

The total mass is then obtained as:

\(\begin{array}{c}{M_{tot}} &= {M_L} + {M_D}\\ &= {\rm{11}} {M_L}\\ &= {\rm{ 3}}{\rm{.28 \times 1}}{{\rm{0}}^{{\rm{42}}}}{\rm{ kg}}\end{array}\)\(\)

Therefore, the total mass is: \({\rm{3}}{\rm{.28 \times 1}}{{\rm{0}}^{{\rm{42}}}}{\rm{ kg}}\).

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(a) Estimate the mass of the luminous matter in the known universe, given there are\({\rm{1}}{{\rm{0}}^{{\rm{11}}}}\)galaxies, each containing\({\rm{1}}{{\rm{0}}^{{\rm{11}}}}\)stars of average mass\({\rm{1}}{\rm{.5}}\)times that of our Sun. (b) How many protons (the most abundant nuclide) are there in this mass? (c) Estimate the total number of particles in the observable universe by multiplying the answer to (b) by two, since there is an electron for each proton, and then by\({\rm{1}}{{\rm{0}}^{\rm{9}}}\), since there are far more particles (such as photons and neutrinos) in space than in luminous matter.

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