Chapter 5, Problem 75GP

(a)

To determine

The approximation formula for $\mathrm{\theta }$ in terms of the mass of the mountain, distance to its center, the radius and mass of Earth.

Answer to Problem 75GP

The approximation formula for $\mathrm{\theta }$ in term of the mass of the mountain is $\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{m}}_{\mathrm{M}}{\mathrm{R}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}{}^2}{{\mathrm{D}}_{\mathrm{M}}{}^2{\mathrm{M}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}\right)$ .

Explanation of Solution

Given:

The given figure is shown below.

  

Formula used:

The gravitational force of attraction is

  $\mathrm{F}=\mathrm{G}\frac{\mathrm{m}\mathrm{M}}{{\mathrm{r}}^2}$

  $\mathrm{F}$ is a gravitational force, $\mathrm{G}$ is the gravitation constant and m , M are the two masses and r is the distance between the two.

Calculation:

Consider the free body diagram of plumb bob, whose mass is m , as shown below.

  

Here, $\mathrm{\theta }$ is the angle by which bob is deflected from the vertical due to nearby massive mountain. ${\mathrm{F}}_{\mathrm{M}}$ is the gravitational force of attraction between mountain and bob, F_T is the tension in the string and mg is the weight of bob.

The gravitational force of attraction between plumb bob and mountain is given as,

  ${\mathrm{F}}_{\mathrm{M}}=\mathrm{G}\frac{\mathrm{m}{\mathrm{m}}_{\mathrm{M}}}{{\mathrm{D}}_{\mathrm{M}}{}^2}$  (1)

Here, m is the mass of bob, m_m is mountain mass and ${\mathrm{D}}_{\mathrm{M}}$ is the distance between plumb bob and mountain.

It is given that bob is not accelerating, so the net force in horizontal as well as vertical direction will be zero.

In vertical direction net force is,

  $\begin{array}{l}{\mathrm{F}}_{\mathrm{T}}\cos \mathrm{\theta }-\mathrm{m}\mathrm{g}=0\\ {\mathrm{F}}_{\mathrm{T}}=\frac{\mathrm{m}\mathrm{g}}{\cos \mathrm{\theta }}\end{array}$

In horizontal direction, net force is

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{M}}-{\mathrm{F}}_{\mathrm{T}}\sin \mathrm{\theta }=0\\ {\mathrm{F}}_{\mathrm{M}}={\mathrm{F}}_{\mathrm{T}}\sin \mathrm{\theta }\\ =\frac{\mathrm{m}\mathrm{g}}{\cos \mathrm{\theta }}\sin \mathrm{\theta }\\ =\mathrm{m}\mathrm{g}\tan \mathrm{\theta }\end{array}$  (2)

The expression for g at Earth’s surface is,

  $\mathrm{g}=\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}{{\mathrm{R}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}{}^2}$

Substituting the value in equation (2)

  $\begin{array}{c}{\mathrm{F}}_{\mathrm{M}}=\mathrm{m}\mathrm{g}\tan \mathrm{\theta }\\ =\mathrm{m}\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}{{\mathrm{R}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}{}^2}\tan \mathrm{\theta }\end{array}$  (3)

From (1) and (3),

  $\begin{array}{l}\mathrm{m}\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}{{\mathrm{R}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}{}^2}\tan \mathrm{\theta }=\mathrm{G}\frac{\mathrm{m}{\mathrm{m}}_{\mathrm{M}}}{{\mathrm{D}}_{\mathrm{M}}{}^2}\\ \tan \mathrm{\theta }=\frac{{\mathrm{m}}_{\mathrm{M}}{\mathrm{R}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}{}^2}{{\mathrm{D}}_{\mathrm{M}}{}^2{\mathrm{M}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}\\ \mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{m}}_{\mathrm{M}}{\mathrm{R}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}{}^2}{{\mathrm{D}}_{\mathrm{M}}{}^2{\mathrm{M}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}\right)\end{array}$

Conclusion:

Therefore, the approximation formula for $\mathrm{\theta }$ in term of the mass of the mountain is $\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{m}}_{\mathrm{M}}{\mathrm{R}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}{}^2}{{\mathrm{D}}_{\mathrm{M}}{}^2{\mathrm{M}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}\right)$ .

(b)

To determine

The mass of Mt. Everest.

Answer to Problem 75GP

The mass of Mt. Everest is $5\times {10}^{13}\text{kg}$ .

Explanation of Solution

Given:

The given figure is shown below.

  

The shape of Mt Everest is cone whose base diameter is 4000 m and height is 4000 m. Density or mass per unit volume is 3000 kg per m³.

Formula used:

The mass is given by,

  $\text{Mass}=\text{volume}\times \text{density}$

Calculation:

If Mt Everest shape is cone, then is volume is given as,

  $\mathrm{V}=\frac{1}{3}\mathrm{\pi }{\mathrm{r}}^2\mathrm{H}$

So, mass of Mt Everest will be,

  $\begin{array}{c}{\mathrm{M}}_{\mathrm{E}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{t}}=\mathrm{V}\times \mathrm{\rho }\\ =\frac{1}{3}\times \mathrm{\pi }\times {\mathrm{r}}^2\times \mathrm{H}\times \mathrm{\rho }\end{array}$

Substituting the values,

  $\begin{array}{c}{\mathrm{M}}_{\mathrm{E}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{t}}=\frac{1}{3}\times 3.14\times {\left(\frac{4000}{2}\right)}^2\times 4000\times 3000\\ =5\times {10}^{13}\text{kg}\end{array}$

Conclusion:

Therefore, the mass of Mt. Everest is $5\times {10}^{13}\text{kg}$ .

(c)

To determine

The angle $\mathrm{\theta }$ of the plumb bob.

Answer to Problem 75GP

The angle $\mathrm{\theta }$ of the plumb bob if it is $5\text{kg}$ from the center is $\text{8}\times {\text{10}}^{-4}\text{degrees}$

Explanation of Solution

Given:

The given figure is shown below.

  

Distance of bob from center of Mt Everest is 5 km.

The shape of Mt Everest is cone whose base diameter is 4000 m and height is 4000 m. Density or mass per unit volume is 3000 kg per m³.

Formula used:

From part (a), the angle $\mathrm{\theta }$ is given

  $\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{m}}_{\mathrm{M}}{\mathrm{R}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}{}^2}{{\mathrm{D}}_{\mathrm{M}}{}^2{\mathrm{M}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}\right)$

Calculation:

From part (b), mass of Mt. Everest is $5\times {10}^{13}\text{kg}$ .

The value of $\mathrm{\theta }$ is,

  $\begin{array}{c}\mathrm{\theta }={\tan }^{-1}\left(\frac{{\mathrm{m}}_{\mathrm{M}}{\mathrm{R}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}{}^2}{{\mathrm{D}}_{\mathrm{M}}{}^2{\mathrm{M}}_{\mathrm{E}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}\right)\\ ={\tan }^{-1}\left(\frac{5\times {10}^{13}\times {\left(6.38\times {10}^6\right)}^2}{5.97\times {10}^{24}{\left(5000\right)}^2}\right)\\ =7.8\times {10}^{-4}\text{degrees}\\ \approx \text{8}\times {\text{10}}^{-4}\text{degrees}\end{array}$

Conclusion:

Therefore, the angle $\mathrm{\theta }$ of the plumb bob if it is $5\text{kg}$ from the center is $\text{8}\times {\text{10}}^{-4}\text{degrees}$ .