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

`3.7* 10^(51))` kg

Step-by-step explanation:

`R` = radius of the sphere modeled as universe =
`13* 10^(25)` m

Volume of sphere is given as

`V = (4\pi R^(3))/(3)`

`V = (4(3.14) (13* 10^(25))^(3))/(3)`

`V = 9.2* 10^(78)` m³

`\rho` = average total mass density of universe =
`1* 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* 10^(-26)) (9.2* 10^(78))`

`m = 9.2* 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* 10^(52))`

`M = 3.7* 10^(51)` kg