# Assuming that the Hubble constant H0=74 km/s/mock the critical-density of the universe is 1.0 *10^336-334-31256kg/m^3. What formula do I use to find: How much mass is contained in a volume of...

Assuming that the Hubble constant H0=74 km/s/mock the critical-density of the universe is 1.0 *10^336-334-31256kg/m^3.

What formula do I use to find:

How much mass is contained in a volume of 1cubic light year?

What is the length of a cube that would enclose 1 earth mass of material?

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The formula that is used to evaluate the density is the following,such that:

rho = m/V

m represents the mass of the substance

V reprsents the volume

Hence, if given the density and volume, the mass is evaluated such that:

m = V*rho

On the other hand, one light year measures 8.46*10^47 m^3 and the critical density is 1*10^(-26) (Kg)/(m^3).

m = 8.46*10^47 m^3 * 1*10^(-26) (Kg)/(m^3)

Reducing the similar units of measure yields:

m = 8.46*10^(47-26) Kg

m = 8.46*10^21 Kg

You need to evaluate the length of the cube that encloses one Earth mass, hence, you need to use the volume formula, such that:

V = l^3 => l = root(3)(V)

You need to evaluate the volume of Earth, such that:

V = m/(rho)

The Earth's mass is considered as being 6.4*10^24 Kg.

V = (6.4*10^246.4*10^24 Kg)/(5520*10^3(Kg)/m^3)

V = 1200*10^15 m^3

l = root(3)(1200*10^15 m^3) => l = 1.6*10^6 m