Chapter 7, Problem 41A

a.

To determine

The mass of the Earth.

Answer to Problem 41A

  $6.018\times {10}^{24}\text{kg}$

Explanation of Solution

Given:

The radius of Earth is $6.4\times {10}^6\text{m}$ .

Mass, $m=1.0\text{kg}$

Weight, $w=9.8\text{N}$

Formula used:

Gravitational force:

  $F=\frac{G{m}_1{m}_2}{r^2}$

Where, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between the masses.

Calculation:

The value of gravitational constant is:

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Now, the gravitational attractive force exerted on the given mass by the Earth will be equal to the weight of the mass on Earth’s surface. Then using the given values of the mass and the radius of the Earth, the mass of the Earth will be,

  $\begin{array}{c}F=\frac{G{m}_{E}m}{r^2}\\ 9.8=\frac{6.67\times {10}^{-11}\times m_E\text{}\times 1.0}{{\left(6.4\times {10}^6\right)}^2}\\ m_E=6.018\times {10}^{24}\text{kg}\end{array}$

Conclusion:

Thus, the mass of the Earth is $6.018\times {10}^{24}\text{kg}$ .

b.

To determine

The average density of the Earth.

Answer to Problem 41A

  $5.4\times {10}^3{\text{kg/m}}^{\text{3}}$

Explanation of Solution

Given:

The radius of Earth is $6.4\times {10}^6\text{m}$ .

From part (a), the mass of the Earth is $6.018\times {10}^{24}\text{kg}$ .

Formula used:

  $\text{Density = }\frac{\text{Mass}}{\text{Volume}}$

Calculation:

The volume of the Earth will be,

  $V=\frac{4}{3}\pi r_{Earth}^3$

Then, the density of the Earth will be,

  $\begin{array}{c}\text{Density = }\frac{{\left(\text{Mass}\right)}_{Earth}}{\text{Volume}}\\ =\frac{6.018\times {10}^{24}}{\frac{4}{3}\pi {\left(6.4\times {10}^6\right)}^3}\\ =5.4\times {10}^3{\text{kg/m}}^{\text{3}}\end{array}$

Conclusion:

Thus, the average density of the Earth is $5.4\times {10}^3{\text{kg/m}}^{\text{3}}$ .