91Ó°ÊÓ

All the physical quantities are defined for matter, in general, have mass, volume, and density. These three are an object's most fundamental properties. Density is calculated by dividing mass by volume.

The given data and data from Appendix B can be listed below,

The earth is taken as a sphere, so its volume is given by,

Here, $r_E$ is the radius of the earth.

Substitute values in the above expression, we have

The earth's density can be expressed as,

Here, $M_E$ is the mass of the earth.

Substitute the values in the above equation.

Thus, the earth's average density is $5.52\mathrm{g}/{\mathrm{cm}}^3$.

The white dwarf is taken as a sphere, so its volume is given by,

$V_w=\frac{4\mathrm{Ï€}}{3}{r}_w^3$

Here,$r_w$ is the white dwarf's radius.

Substitute value in the above expression.

The white dwarf"s density is expressed as,

Here, $M_W$ is the white dwarf's mass.

Substitute the values in the above equation.

Thus, the white dwarf's density is $1.12×{10}^6\mathrm{g}/{\mathrm{cm}}^3$.

The volume of the neutron star, which is taken as a sphere is given by,

$V_N=\frac{4\mathrm{Ï€}}{3}{r}_N^3$

Here, $r_N$ is the neutron star's radius.

Substitute value in the above,

The white dwarf's density is expressed as,

Here, $M_N$ is the mass of the neutron star.

Substitute the values in the above equation.

Thus, the neutron star's density is $4.74×{10}^{14}\mathrm{g}/{\mathrm{cm}}^3$.

The average density of the earth, the white dwarf and the neutron star is found to be $5.52\mathrm{g}/{\mathrm{cm}}^3,1.12×{10}^6\mathrm{g}/{\mathrm{cm}}^3$and $4.74×{10}^{14}\mathrm{g}/{\mathrm{cm}}^3$respectively.