Chapter 30, Problem 7P

To determine

(a) To Determine:

The radius of Earth if it has density of nucleus.

Answer to Problem 7P

Solution:

$R_e=183.50\text{m}$

Explanation of Solution

Given:

Density of Earth is same as density of nuclei.

Formula used:

${\rho }_e={\rho }_n=\frac{M_e}{V_e}$

Calculation:

$\begin{array}{l}{M}_e=5.97\times {10}^{24}\text{kg}\\ {\rho }_n=2.3\times {10}^{17}{\text{kg/m}}^{\text{3}}\end{array}$

Density of Earth is same as density of nuclei. Thus, it is given by:

$\begin{array}{l}{\rho }_e={\rho }_n=\frac{M_e}{V_e}=\frac{M_e}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\pi R_e^3}\\ R_e^3=\frac{M_e}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\pi {\rho }_n}\\ R_e=183.50\text{m}\end{array}$

To determine

(b) To Determine:

The radius of ${}_{92}{}^{238}U$ if it had Earth’s density.

Answer to Problem 7P

Solution:

$3.46\times {10}^4$

Explanation of Solution

Given:

Density of Uranium = Density of Earth

Formula used:

$Density=\rho =\frac{M}{V}$

Calculation:

Density of Earth is equal to density of uranium. Thus,

$\begin{array}{l}{\rho }_e=\frac{M_u}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\pi R_u^3}\\ R_u=\frac{3.98\times {10}^{-25}\text{kg}}{5.515\times {10}^3{\text{kg/m}}^{\text{3}}}\\ R_u=2.58\times {10}^{-10}{\text{m}}^{\text{3}}\\ \text{Increase}\text{in}\text{Radius}\text{of}\text{Uranium}=\frac{2.58\times {10}^{-10}{\text{m}}^{\text{3}}}{7.44\times {10}^{-15}}=3.46\times {10}^4\end{array}$