Question

The earth may be considered as a spherically shaped uniform core of density ${{\rho }_{1}}$ and radius $\dfrac{R}{2}$ surrounded by a uniform shell of thickness $\dfrac{R}{2}$and density ${{\rho }_{2}}$. Find the ratio of$\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}$ if the acceleration due to gravity is the same as the surface as at depth $\dfrac{R}{2}$ from the surface.

Answer 1

We know that the density of the earth varies with the depth. We need to understand the variation of the gravitational forces due to this density changes with depth caused by the change in mass to solve the ratio between the densities.

Complete answer:
We are given the two parts of the earth with different densities for which we have to find the ratio between the densities between them.
Now, we are given the densities of two parts of the earth. We can find their mass by using the relation between the mass, volume and density as –

& \text{Mass, m = Volume }\\!\\!\times\\!\\!\text{ Density} \\\ & \Rightarrow m=\dfrac{4}{3}\pi {{r}^{3}}\times \rho \\\ \end{aligned}$$ Now, we can find the mass of each of the two parts of the earth. (i)The mass of the core with radius $$\dfrac{R}{2}$$ can be given as – $$\Rightarrow {{m}_{core}}=\dfrac{4}{3}\pi {{(\dfrac{R}{2})}^{3}}\times {{\rho }_{1}}$$ (ii)The mass of the outer shell of radius $$\dfrac{R}{2}$$ can be given as – $$\begin{aligned} & {{m}_{shell}}=\dfrac{4}{3}\pi [{{R}^{3}}-{{(\dfrac{R}{2})}^{3}}]\times {{\rho }_{2}} \\\ & \Rightarrow {{m}_{shell}}=\dfrac{4}{3}\pi \dfrac{7{{R}^{3}}}{8}\times {{\rho }_{2}} \\\ \end{aligned}$$ We can find the total mass of the Earth from the above to as – $$\begin{aligned} & M={{m}_{core}}+{{m}_{shell}} \\\ & \Rightarrow M=\dfrac{4}{3}\pi {{\dfrac{R}{8}}^{3}}\times {{\rho }_{1}}+\dfrac{4}{3}\pi \dfrac{7{{R}^{3}}}{8}\times {{\rho }_{2}} \\\ & \therefore M=\dfrac{4\pi {{R}^{3}}}{24}({{\rho }_{1}}+7{{\rho }_{2}}) \\\ \end{aligned}$$ Now, we know that the acceleration due to gravity of the earth up to the half of the radius is ‘g’ itself. It can be given as – $$\begin{aligned} & g=\dfrac{GM}{{{R}^{2}}} \\\ & \Rightarrow g=\dfrac{G4\pi {{R}^{3}}}{{{R}^{2}}24}({{\rho }_{1}}+7{{\rho }_{2}}) \\\ & \therefore g=G\dfrac{4\pi R}{24}({{\rho }_{1}}+7{{\rho }_{2}}) \\\ \end{aligned}$$ Now, the acceleration due to gravity at the half of the radius can be given as – $$\begin{aligned} & g'=\dfrac{Gm}{{{r}^{2}}} \\\ & \Rightarrow g'=\dfrac{G\dfrac{4}{3}\pi {{(\dfrac{R}{2})}^{3}}\times {{\rho }_{1}}}{{{(\dfrac{R}{2})}^{2}}} \\\ & \therefore g'=G\dfrac{4\pi R}{6}{{\rho }_{1}} \\\ \end{aligned}$$ Now, we can find the ratio between the two acceleration due to gravities to find the ratio between the densities of the two parts. According to the question, the acceleration due to gravity is the same for both. So, we can equate them as – $$\begin{aligned} & g'=g \\\ & \Rightarrow G\dfrac{4\pi R}{6}{{\rho }_{1}}=G\dfrac{4\pi R}{24}({{\rho }_{1}}+7{{\rho }_{2}}) \\\ & \Rightarrow {{\rho }_{1}}=\dfrac{{{\rho }_{1}}+7{{\rho }_{2}}}{4} \\\ & \Rightarrow 3{{\rho }_{1}}=7{{\rho }_{2}} \\\ & \therefore \dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\dfrac{7}{3} \\\ \end{aligned}$$ The ratio of the densities of the two parts of the earth is, therefore, given as 7:3. This is the required solution. **Note:** The acceleration due to gravity of the earth is highly dependent on the depth of the earth from where it is considered. Here, we assume as per the question that the acceleration due to gravity remains constant up to half the radius, which is not practical.