Question

The density of a certain planet varies with radial distance as: rho(r)= rho0(1-αr/R0) where R0= 25.12 x 10^6 m is the radius of the planet, rho0= 3800.0 kg/m^3 is its central density, and α = 0.24. What is the total mass of this planet ?

Answer 1

Answer:

The total mass of this planet is $2.0689\times10^{26}\ kg$

Explanation:

Given that,

Radius $R_{0}=25.12\times10^{6}\ m$

Density $\rho_{0}=3800.0\ kg/m^3$

Central density$\alpha=0.24$

The density of a certain planet varies with radial distance as

$\rho(r)=\rho_{0}(1-\dfrac{\alpha r}{R_{0}})$

We need to calculate the total mass of this planet

Using formula of density

$\rho=\dfrac{M}{V}$

$M=\rho\times V$

On integrating

$M=\int_{0}^{R_{0}}{\rho(r)\times4\pi r^2 dr}$

Put the value of $\rho{r}$ into the formula

$M=\int_{0}^{R_{0}}{\rho_{0}(1-\dfrac{\alpha r}{R_{0}})\times4\pi r^2 dr}$

$M=\rho_{0}\times 4\pi\int_{0}^{R_{0}}{(r^2-\dfrac{\alpha r^3}{R_{0}})dr}$

$M=\rho_{0}\times 4\pi\times(\dfrac{r^3}{3}-\dfrac{\alpha\times r^4}{4\times R_{0}})_{0}^{R_{0}}$

$M=4\pi\times\rho_{0}\times R_{0}^3(\dfrac{4-3\alpha}{12})$

Put the value into the formula

$M=4\pi\times3800.0\times(25.12\times10^{6})^3(\dfrac{4-3\times0.24}{12})$

$M=2.0689\times10^{26}\ kg$

Hence, The total mass of this planet is $2.0689\times10^{26}\ kg$