Question

Imagine a smooth tunnel along a chord of nonrotating earth at a distance $\frac{R}{2}$ from the centre. R is the radius of the earth. A projectile is fired along the tunnel from the centre of the tunnel at a speed $V_o = \sqrt{gR}$ [g is acceleration due to gravity at the surface of the earth]. (a) Is the angular momentum [about the centre of the earth] of the projectile onserved as it moves along the tunnel? (b) Calculate the maximum distance of the projectile from the centre of the earth during its course of motion.

Answer 1

(a) No (b) R

Answer 2

The problem involves a projectile moving in a smooth tunnel along a chord of the Earth. We need to analyze the angular momentum and the maximum distance from the Earth's center.

(a) Is the angular momentum [about the centre of the earth] of the projectile conserved as it moves along the tunnel?

1. Forces acting on the projectile:

   • Gravitational Force ($\vec{F}_g$): Assuming uniform density of the Earth, the gravitational force inside the Earth is directly proportional to the distance from the center, i.e., $F_g = \frac{GMm}{R^3}r = mg\frac{r}{R}$, where $r$ is the distance from the center. The force is always directed towards the center of the Earth. Since $\vec{F}_g$ is a central force, the torque due to gravity about the center of the Earth ($\vec{\tau}_g = \vec{r} \times \vec{F}_g$) is zero.
   • Normal Force ($\vec{N}$): The tunnel is a physical constraint that forces the projectile to move along the chord. This constraint is maintained by a normal force from the walls of the tunnel.

2. Analysis of Normal Force and its Torque:

   Let the center of the Earth be the origin (0,0). The tunnel is along a chord at a distance $d = R/2$ from the center. Let's align the coordinate system such that the chord lies along the x-axis, and its center is at $(0, d)$. So, the position of the projectile is $\vec{r} = x\hat{i} + d\hat{j}$.

   The normal force $\vec{N}$ must be perpendicular to the tunnel, so it acts along the y-axis: $\vec{N} = N_y \hat{j}$.

   The gravitational force is $\vec{F}_g = -mg\frac{\vec{r}}{R} = -mg\frac{x\hat{i} + d\hat{j}}{R}$.

   Since the projectile is constrained to move along the chord, its y-coordinate is constant ($y=d$), meaning $\ddot{y}=0$. The net force in the y-direction must be zero:

   $F_{gy} + N_y = 0$

   $-mg\frac{d}{R} + N_y = 0$

   $N_y = mg\frac{d}{R}$.

   So, the normal force is $\vec{N} = \left(mg\frac{d}{R}\right)\hat{j}$. Since $d=R/2$, $\vec{N} = (mg/2)\hat{j}$.

   Now, let's calculate the torque due to this normal force about the center of the Earth:

   $\vec{\tau}_N = \vec{r} \times \vec{N} = (x\hat{i} + d\hat{j}) \times \left(mg\frac{d}{R}\hat{j}\right)$

   $\vec{\tau}_N = x\left(mg\frac{d}{R}\right)(\hat{i} \times \hat{j}) + d\left(mg\frac{d}{R}\right)(\hat{j} \times \hat{j})$

   $\vec{\tau}_N = x\left(mg\frac{d}{R}\right)\hat{k}$

   Since $d=R/2$, $\vec{\tau}_N = x\left(mg\frac{R/2}{R}\right)\hat{k} = x\left(\frac{mg}{2}\right)\hat{k}$.

   As the projectile moves along the tunnel, its x-coordinate changes. Therefore, $\vec{\tau}_N$ is generally non-zero (unless $x=0$).

   Since there is a non-zero net torque about the center of the Earth, the angular momentum of the projectile about the center of the Earth is not conserved.

(b) Calculate the maximum distance of the projectile from the centre of the earth during its course of motion.

1. Equation of motion along the tunnel:

   The projectile is constrained to move along the x-axis (the chord). The net force along the x-axis is only the x-component of the gravitational force:

   $m\ddot{x} = F_{gx}$

   From $\vec{F}_g = -mg\frac{x\hat{i} + d\hat{j}}{R}$, we have $F_{gx} = -mg\frac{x}{R}$.

   So, $m\ddot{x} = -mg\frac{x}{R}$

   $\ddot{x} = -\frac{g}{R}x$

   This is the equation for Simple Harmonic Motion (SHM) with angular frequency $\omega = \sqrt{g/R}$.

2. Initial Conditions:

   The projectile is fired from the center of the tunnel. In our coordinate system, the center of the tunnel is at $(0, d) = (0, R/2)$. So, the initial x-coordinate is $x(0) = 0$.

   The initial speed is $V_o = \sqrt{gR}$ along the tunnel. So, the initial x-velocity is $\dot{x}(0) = V_o = \sqrt{gR}$.

3. Amplitude of SHM:

   For an SHM starting from equilibrium ($x(0)=0$) with initial velocity $\dot{x}(0)=V_o$, the solution is $x(t) = A\sin(\omega t)$, where $A$ is the amplitude.

   The velocity is $\dot{x}(t) = A\omega\cos(\omega t)$.

   At $t=0$, $\dot{x}(0) = A\omega$.

   So, $A = \frac{\dot{x}(0)}{\omega} = \frac{\sqrt{gR}}{\sqrt{g/R}} = \sqrt{gR} \cdot \sqrt{R/g} = R$.

   If the tunnel were infinitely long, the projectile would oscillate between $x=-R$ and $x=R$.

4. Limits of the Tunnel:

   The tunnel is a chord at a distance $d=R/2$ from the center. The ends of the chord lie on the surface of the Earth (distance $R$ from the center).

   Let $x_{max\_tunnel}$ be the half-length of the chord. Using Pythagoras theorem:

   $x_{max\_tunnel}^2 + d^2 = R^2$

   $x_{max\_tunnel}^2 + (R/2)^2 = R^2$

   $x_{max\_tunnel}^2 = R^2 - R^2/4 = 3R^2/4$

   $x_{max\_tunnel} = \frac{\sqrt{3}}{2}R$.

   The tunnel extends from $x = -\frac{\sqrt{3}}{2}R$ to $x = \frac{\sqrt{3}}{2}R$.

5. Maximum distance from the center:

   The amplitude of the projectile's SHM is $A=R$. Since $A=R > x_{max\_tunnel}=\frac{\sqrt{3}}{2}R$, the projectile will reach the ends of the tunnel before completing its full SHM oscillation.

   Therefore, the maximum x-coordinate reached by the projectile is the end of the tunnel, $x_{max} = \frac{\sqrt{3}}{2}R$.

   The y-coordinate of the projectile is always $y=R/2$.

   The distance of the projectile from the center of the Earth is $r = \sqrt{x^2 + y^2}$.

   To find the maximum distance, we use the maximum x-coordinate:

   $r_{max} = \sqrt{x_{max}^2 + (R/2)^2}$

   $r_{max} = \sqrt{\left(\frac{\sqrt{3}}{2}R\right)^2 + \left(\frac{R}{2}\right)^2}$

   $r_{max} = \sqrt{\frac{3}{4}R^2 + \frac{1}{4}R^2}$

   $r_{max} = \sqrt{\frac{4}{4}R^2} = \sqrt{R^2} = R$.

   The maximum distance of the projectile from the center of the Earth is $R$. This occurs when the projectile reaches either end of the tunnel, which are points on the Earth's surface.

Explanation of the solution:

(a) Angular momentum about the Earth's center is not conserved because the normal force exerted by the tunnel walls creates a non-zero torque about the Earth's center. While gravity is a central force, the constraint force (normal force) is not.

(b) The motion along the tunnel (x-direction) is Simple Harmonic Motion (SHM) due to the x-component of gravity. The equation of motion is $\ddot{x} = -\frac{g}{R}x$, with angular frequency $\omega = \sqrt{g/R}$. Given initial conditions $x(0)=0$ and $\dot{x}(0)=V_o=\sqrt{gR}$, the amplitude of this SHM is $A = V_o/\omega = R$. The tunnel is a chord at distance $R/2$ from the center, so its ends are at $x = \pm\sqrt{R^2 - (R/2)^2} = \pm \frac{\sqrt{3}}{2}R$. Since the SHM amplitude ($R$) is greater than the tunnel's half-length ($\frac{\sqrt{3}}{2}R$), the projectile will reach the ends of the tunnel. The maximum distance from the center of the Earth occurs at these ends, calculated as $r_{max} = \sqrt{(\frac{\sqrt{3}}{2}R)^2 + (R/2)^2} = \sqrt{\frac{3}{4}R^2 + \frac{1}{4}R^2} = R$.