Answer
The density of Mars to that of the Earth is $~~0.735$
Work Step by Step
The radius of the Earth is $\frac{1.3\times 10^4~km}{2} = 6.5\times 10^3~km$
We can find the volume of the Earth:
$V_E = \frac{4}{3}\pi~R^3$
$V_E = \frac{4}{3}\pi~(6.5\times 10^3~km)^3$
$V_E = 1.15\times 10^{12}~km^3$
Let $M_E$ be the mass of the Earth.
We can find an expression for the density of the Earth:
$\rho_E = \frac{M_E}{V_E} = \frac{M_E}{1.15\times 10^{12}~km^3}$
The radius of Mars is $\frac{6.9\times 10^3~km}{2} = 3.45\times 10^3~km$
We can find the volume of Mars:
$V_M = \frac{4}{3}\pi~R^3$
$V_M = \frac{4}{3}\pi~(3.45\times 10^3~km)^3$
$V_M = 1.72\times 10^{11}~km^3$
Let $M_E$ be the mass of the Earth.
Then the mass of Mars is $~~0.11~M_E$
We can find an expression for the density of Mars:
$\rho_M = \frac{M_M}{V_M} = \frac{0.11~M_E}{1.72\times 10^{11}~km^3}$
We can find the ratio of the density of Mars to that of the Earth:
$\frac{\rho_M}{\rho_E} = \frac{\frac{0.11~M_E}{1.72\times 10^{11}~km^3}}{\frac{M_E}{1.15\times 10^{12}~km^3}}$
$\frac{\rho_M}{\rho_E} = \frac{(0.11)(1.15\times 10^{12}~km^3)}{1.72\times 10^{11}~km^3}$
$\frac{\rho_M}{\rho_E} = 0.735$
The density of Mars to that of the Earth is $~~0.735$
