Question

Recent findings in astrophysics suggest that the observable universe can be modeled as a sphere of radius R = 13.7 × 109 light-years = 13.0 × 1025 m with an average total mass density of about 1 × 10-26 kg/m3. Only about 4% of total mass is due to "ordinary" matter (such as protons, neutrons, and electrons). Part A Estimate how much ordinary matter (in kg) there is in the observable universe.

Answer 1

Answer:

$3.7\times 10^{51})$ kg

Explanation:

$R$ = radius of the sphere modeled as universe = $13\times 10^{25}$ m

Volume of sphere is given as

$V = \frac{4\pi R^{3}}{3}$

$V = \frac{4(3.14) (13\times 10^{25})^{3}}{3}$

$V = 9.2\times 10^{78}$ m³

$\rho$ = average total mass density of universe = $1\times 10^{-26}$ kg/m³

$m$ = Total mass of the universe = ?

We know that mass is the product of volume and density, hence

$m = \rho V$

$m = (1\times 10^{-26}) (9.2\times 10^{78})$

$m = 9.2\times 10^{52}$ kg

$M$ = mass of "ordinary" matter  = ?

mass of "ordinary" matter is only about 4% of total mass, hence

$M = (0.04) m$

$M = (0.04)(9.2\times 10^{52})$

$M = 3.7\times 10^{51}$ kg