In the standard model of the expansion of the universe,the total energy of a standard mass m is assumed to be zero.When we solve the resulting energy equation for the critical mass density of the universe,we find that the critical mass density is given by
A) $\rho _ { c } = \frac { H ^ { 2 } } { 8 \pi G }$ .
B) $\rho _ { c } = \frac { 3 H ^ { 2 } } { 8 \pi G }$ .
C) $\rho _ { c } = \frac { 3 H ^ { 2 } } { 2 \pi G }$ .
D) $\rho _ { c } = \frac { 3 H ^ { 2 } } { \pi G }$ .
E) $\rho _ { c } = \frac { 3 H ^ { 2 } } { G }$ .

Question

Quizplus · Accepted Answer

The critical mass density is given by the equation:
ρ_c = 3H^2/8πG
where ρ_c is the critical mass density, H is the Hubble constant and G is the gravitational constant.
The energy equation for a standard mass m is given by:
E = mc^2 - GMm/r
where E is the total energy, c is the speed of light, G is the gravitational constant, M is the mass of the universe and r is the radius of the universe.
In the standard model, the total energy E is assumed to be zero. Therefore, we can write:
mc^2 - GMm/r = 0
Solving for r, we get:
r = GM/c^2
Substituting this value of r in the equation for the critical mass density, we get:
ρ_c = 3c^2/8πGGr^2
Substituting the value of r, we get:
ρ_c = 3c^2/8πG(GM/c^2)^2
Simplifying, we get:
ρ_c = 3/(8πGt^2)
where t is the age of the universe.
Substituting the values of G and t, we get:
ρ_c = 1.88 x 10^-29 g/cm^3
which corresponds to choice B.

In the Standard Model of the Expansion of the Universe,the ρc=H28πG\rho _ { c } = \frac { H ^ { 2 } } { 8 \pi G }ρc​=8πGH2​

In the Standard Model of the Expansion of the Universe,the $\rho _ { c } = \frac { H ^ { 2 } } { 8 \pi G }$