Answer
a)The average density of the Earth is about 5.51$\frac{g}{cm^{3}}$.
b)The density of our sun as a white drawf is about $1.1\times10^{6}\frac{g}{cm^{3}}$.
c)The density of a typical neutron star is about $4.7\times10^{14}\frac{g}{cm^{3}}$
Work Step by Step
We assume the Earth ,sun and the neutron star to be a sphere.
We also know that, Density(d)=$\frac{mass(m)}{volume(v)}$
where,v=$\frac{4}{3}.pi.r^{3}$
a)Given: mass of the Earth(m)=$5.97\times10^{24}kg$
Radius of the Earth(v)=$6.37\times10^{6}$m
Now,density,d=$\frac{m}{v}$
d=$\frac{5.97\times10^{24}}{\frac{4}{3}.pi.(6.37\times10^{6})^{3}}\frac{kg}{m^{3}}$
=$\frac{5.97\times10^{24}\times10^{3}}{\frac{4}{3}.pi.(6.37\times10^{6})^{3}\times(10^{2})^{3}}\frac{g}{cm^{3}}$
=5.51$\frac{g}{cm^{3}}$
The average density of the Earth is 5.51$\frac{g}{cm^{3}}$.
b)Given: Our sun in about 5 billion years will end up as a white dwarf with
mass,m=$1.99\times10^{30}kg$ and radius,r=$\frac{15000}{2}$km=7500km=$7.5\times10^6$
Its density,d=$\frac{m}{v}$
d=$\frac{1.99\times10^{30}}{\frac{4}{3}.pi.(7.5\times10^{6})^{3}}\frac{kg}{m^{3}}$
=$\frac{1.99\times10^{30}\times10^{3}}{\frac{4}{3}.pi.(7.5\times10^{6})^{3}\times(10^{2})^{3}}\frac{g}{cm^{3}}$
=$1.1\times10^{6}\frac{g}{cm^{3}}$
The density of the white dwarf is $1.1\times10^{6}\frac{g}{cm^{3}}$.
c)Given:For a typical neutron star,
mass,m=$1.99\times10^{30}$kg and radius,r=10km=$1.0\times10^4$
Now,density,d=$\frac{m}{v}$
d=$\frac{1.99\times10^{30}}{\frac{4}{3}.pi.(1.0\times10^{4})^{3}}\frac{kg}{m^{3}}$
=$\frac{1.99\times10^{30}\times10^{3}}{\frac{4}{3}.pi.(1.0\times10^{4})^{3}\times(10^2)^{3}}\frac{g}{cm^{3}}$
=$4.7\times10^{14}\frac{g}{cm^3}$
The density of a typical neutron star is $4.7\times10^{14}\frac{g}{cm^3}$
