Question

Mountain pulls.A large mountain can slightly affectthe direction of “down” as determined by a plumb line. Assumethat we can model a mountain as a sphere of radius$\mathbf{R}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{km}$and density (mass per unit volume).$\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$Assume alsothat we hang a 0.50mplumb line at a distance offrom the sphere’s centre and such that the sphere pulls horizontally on thelower end. How far would the lower end move toward the sphere?

Short answer

The lower end of the mountain moves toward the sphere at a distance of $8.2\times {10}^{-6}\mathrm{m}$.

Step-by-step solution

The given data

• Radius of the sphere is $R=2000\mathrm{m}$
• Density of sphere is$p=2.6\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$
• Hanging distance of plumb is 3R
• Length of the plumb line, I =0.5m

Understanding the concept of Gravitational force and density

We use the concept of density to find the mass of the sphere and Newton’s second law to write the equation of net force for equilibrium. We can write the net force equation for x and y directions and by solving them, we got in terms of mass and radius. We can write the length component to find the value of x.

Formulae:

Density of a material,$p=\frac{M}{V}$P (i)

Force according to Newton’s law of motion, F=ma (ii)

Gravitational Force, $F=\frac{GMm}{r^2}$ (iii)

Calculation of the distance at which the lower end will move towards sphere

We have density and radius of the sphere. Using this information, we can calculate mass,

$$\begin{gathered} M=pV\left(\mathrm{using}\mathrm{eduation}\left(\mathrm{i}\right)\right) \\ =p\left(\frac{4\pi }{3}\right)R^3 \\ =2.6\mathrm{kg}/{\mathrm{m}}^3{10}^3\left(4\times \frac{3.14}{3}\right){\left(2\times {10}^3\mathrm{m}\right)}^3 \\ =8.7\times {10}^{13}\mathrm{kg} \end{gathered}$$

Using equation (iii), the force between spherical mountain and plumb is,

$F=\frac{GMm}{r^2}$

For equilibrium net force along x should be zero, hence, we can write the equation as:

$$\begin{gathered} \underset{}{\sum F_{xnet}=0} \\ T\sin \theta -F=0\left(\mathrm{from}\mathrm{body}\mathrm{diagram}\right) \end{gathered}$$

Using the value of equation (iii) we get,

$T\sin \theta -F=0....................\left(1\right)$

Similarly, we can write the net force along y direction,

$$\begin{gathered} \sum _{}{F}_{ynet}=0 \\ T\cos -mg=0\left(\mathrm{from}\mathrm{body}\mathrm{diagram}\right) \\ T=\frac{mg}{\cos \theta }......................\left(2\right) \end{gathered}$$

Putting equation (2) in equation (1) we get,

$$\begin{gathered} \frac{mg}{\cos \theta }\sin \theta -\frac{GMm}{r^2}=0 \\ mg\tan \theta =\frac{GMm}{r^2} \\ \tan \theta =\frac{GM}{g{r}^2}.....................\left(3\right) \end{gathered}$$

Lower end of the plumb moves towards the sphere,

$$\begin{gathered} x=I\tan \theta \\ =\frac{IGM}{g{r}^2}\left(\mathrm{from}\mathrm{equation}\left(3\right)\right) \\ =\frac{\left(0.50\right)\left(6.67\times \times {10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(8.7\times {10}^{13}\mathrm{kg}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2{\left(3\times 2.00\times {10}^3\mathrm{m}\right)}^2} \\ =8.2\times {10}^{-6}\mathrm{m} \end{gathered}$$

Hence,

the lower end moves at a distance of $8.2\times {10}^{-6}\mathrm{m}$.