91影视

It is given that $\mathrm{R}$is the radius of Earth, ${\mathrm{r}}_0$is the deepest point of tunnel and $\mathrm{A}$and $\mathrm{B}$are the two points on Earth鈥檚 surface.It鈥檚 figure is given by,

In the same way that geometry is the study of shape and algebra is the study of generalisations of arithmetic operations, calculus, sometimes known as infinitesimal calculus or "the calculus of infinitesimals," is the mathematical study of continuous change.. Differentiation and integration are the two main branches.

Write the gravitational potential energy of mass and radius r, $\frac{mg}{2R}\left(r^2-3R^2\right)$.

Write kinetic energy in terms of polar coordinates, $E_k=\frac{1}{2}m{\stackrel{藱}{r}}^2+\frac{1}{2}m{r}^2{\stackrel{藱}{胃}}^2$.

Write the energy conservation law assuming no initial velocity.

$E_{g_1}=E_{k_2}+E_{g_2}$

$-mgR=\frac{1}{2}m{\stackrel{藱}{r}}^2+\frac{1}{2}m{r}^2{\stackrel{藱}{胃}}^2+\frac{mg}{2R}\left(r^2-3R^2\right)$

Minimize the time integral and find the path through earth that require least time.

$鈭�dt=鈭�\frac{ds}{v}$ ......$\left(1\right)$

In above integral, write the velocity in form of energy conservation law.

$$\begin{gathered} \stackrel{鈫�}{v}=\stackrel{藱}{r}\hat{r}+r\stackrel{藱}{胃}\widehat{胃} \\ v=\left|\stackrel{鈫�}{v}\right| \\ v=\sqrt{{\stackrel{藱}{r}}^2+r{\stackrel{藱}{胃}}^2} \end{gathered}$$

Multiply the energy conservation law by a factor of $\frac{2}{\mathrm{m}}$to obtain the desired quantity.

$$\begin{gathered} -2gR={\stackrel{藱}{r}}^2+r^2{\stackrel{藱}{胃}}^2+\frac{g}{R}\left(r^2-3R^2\right) \\ {\stackrel{藱}{r}}^2+r^2\stackrel{藱}{胃}=gR-\frac{g}{R}{r}^2 \\ v=\sqrt{{\stackrel{藱}{r}}^2+r^2\stackrel{藱}{胃}} \\ v=\sqrt{gR-\frac{g}{R}{r}^2} \end{gathered}$$

Substitute value of velocity in equation (1).

$$\begin{gathered} 鈭�\frac{ds}{\sqrt{gR-{\displaystyle \frac{g}{R}}{r}^2}}=鈭�\sqrt{\frac{d{r}^2+r^{2}d{胃}^2}{gR-{\displaystyle \frac{g}{R}}{r}^2}} \\ =鈭�\sqrt{\frac{1+r^{2}d胃{\text{'}}^2}{gR{\displaystyle \frac{g}{R}}{r}^2}}dr \\ =鈭�\frac{\sqrt{R+R{r}^{2}d胃{\text{'}}^2}}{\sqrt{g\left(R^2-r^2\right)}}dr \end{gathered}$$

Let, $F=\frac{\sqrt{1+r^{2}d胃{\text{'}}^2}}{\sqrt{gR-{\displaystyle \frac{g}{R}}{r}^2}}.$

Write the Euler鈥檚 equation and find its derivatives.

role="math" localid="1664877126984" $$\begin{gathered} \frac{d}{dr}\frac{鈭�F}{鈭�胃\text{'}}-\frac{鈭�F}{鈭�胃}=0 \\ \frac{鈭�F}{鈭�胃\text{'}}=\frac{r^2胃\text{'}}{\sqrt{1+r^2胃{\text{'}}^2}\sqrt{gR-{\displaystyle \frac{g{r}^2}{R}}}} \\ \frac{鈭�F}{鈭�胃}=0 \end{gathered}$$

Find the first Euler鈥檚 integral.

$$\begin{gathered} \frac{d}{dr}\left[\frac{r^2胃\text{'}}{\sqrt{1+r^2胃{\text{'}}^2}\sqrt{gR-{\displaystyle \frac{g{r}^2}{R}}}}\right]=0 \\ \frac{r^2胃\text{'}}{\sqrt{1+r^2胃{\text{'}}^2}\sqrt{gR-{\displaystyle \frac{g{r}^2}{R}}}}=C \\ 胃{\text{'}}^2=\frac{C^2\left(gR-{\displaystyle \frac{g{r}^2}{R}}\right)}{r^2\left(r^2-C^2\left(gR-{\displaystyle \frac{g{r}^2}{R}}\right)\right)} \\ 胃\text{'}=\frac{C\sqrt{gR-{\displaystyle \frac{g{r}^2}{R}}}}{r\sqrt{r^2}\left(1+{\displaystyle \frac{g{C}^2}{R}}\right)-C^{2}gR} \end{gathered}$$

At the deepest point of tunnel $\mathrm{r}={\mathrm{r}}_0$and $胃\text{'}鈫�鈭�$. Substitute these values in the above differential equation to find the arbitrary constant.

$鈭�=\frac{C\sqrt{gR-{\displaystyle \frac{g{r}_0^2}{R}}}}{r_0\sqrt{r_0^2}\left(1+{\displaystyle \frac{g{C}^2}{R}}\right)-C^{2}gR}$

$$\begin{gathered} r_0^2\left(1+\frac{g{C}^2}{R}\right)-C^{2}gR=0 \\ {r}_0^{2}R+r_0^{2}g{C}^2-C^{2}g{R}^2=0 \\ C=\sqrt{\frac{r_0^{2}R}{g(R^2-r_0^2)}} \end{gathered}$$

Substitute the value of arbitrary constant $\mathrm{C}$in the differential equation $\mathrm{胃}\text{'}$.

$胃\text{'}=\frac{\sqrt[C]{gR-{\displaystyle \frac{g{r}^2}{R}}}}{\sqrt[r]{r^2\left(1+{\displaystyle \frac{g{C}^2}{R}}\right)}-C^{2}gR}$

$$\begin{gathered} 胃\text{'}=\frac{\sqrt{{\displaystyle \frac{r_0^{2}R}{g\left(R^2-r_0^2\right)}}}\sqrt{gR-{\displaystyle \frac{g{r}^2}{R}}}}{\sqrt[r]{r^2\left(1+{\displaystyle \frac{g\left({\displaystyle \frac{r_0^{2}R}{g\left(R^2-r_0^2\right)}}\right)}{R}}\right)}-\left({\displaystyle \frac{r_0^{2}R}{g(R^2-r_0^2}}\right)gR} \\ 胃\text{'}=\frac{\sqrt[r_0]{{\displaystyle \frac{R^2-r^2}{R^2-r_0^2}}}}{\sqrt[rR]{{\displaystyle \frac{r^2-r_0^2}{R^2-r_0^2}}}} \\ 胃\text{'}=\frac{r_0}{rR}\sqrt{\frac{R^2-r^2}{r^2-r_0^2}} \end{gathered}$$

Substitute the value of $\mathrm{胃}\text{'}$in the time integral between the limits ${\mathrm{r}}_0$and $\mathrm{R}$.

$$\begin{gathered} T=鈭�dt \\ T=2{鈭�}_{r_0}^R\sqrt{\frac{r^2\left(R^2-r_0^2\right)}{gR\left(R^2-r^2\right)\left(r^2-r_0^2\right)}}dr \\ T=\mathrm{蟺}\sqrt{\frac{{\mathrm{R}}^2-{\mathrm{r}}_0^2}{\mathrm{gR}}} \end{gathered}$$

Find the time at ${\mathrm{r}}_0=0$.

$$\begin{gathered} T_{r_0-0}=蟺\sqrt{\frac{R^2-{\left(0\right)}^2}{gR}} \\ {T}_{r_0-0}=\mathrm{蟺}\sqrt{\frac{\mathrm{R}}{\mathrm{g}}} \\ {T}_{r_0-0}=2531s \\ {T}_{r_0-0}鈮�42\min \end{gathered}$$

Find the time at $r_0=0.99R$.

$$\begin{gathered} T_{r_0-0.99R}=蟺\sqrt{\frac{R^2-{\left(0.99R\right)}^2}{gR}} \\ {T}_{r_0-0.99R}=\mathrm{蟺}\sqrt{\frac{0.0199R}{\mathrm{g}}} \\ {T}_{r_0-0.99R}=357s \\ {T}_{r_0-0.99R}鈮�6\min \end{gathered}$$

Therefore, from the figure,

where, $\mathrm{R}$ is the radius of Earth, r₀ is the deepest point of tunnel and A and B are the two points on Earth鈥檚 surface, it is proved that the path through the earth requiring the least time is $T=\mathrm{蟺}\sqrt{\frac{{\mathrm{R}}^2-{\mathrm{r}}_0^2}{\mathrm{gR}}}$, and time at $r_0=0$and $r_0=0.99\mathrm{R}$are $42\min$and $6\min$, respectively.