Model Description

We model the Earth as a flat disc of uniform thickness $h$ and density $\rho$. The disc rotates with angular velocity $\omega$ about a vertical axis passing through its center (North Pole).

• North Pole: Center of the disc ($r=0$).
• South Pole: Circular edge ($r=R$).
• Longitudes: Radial lines extending from the center.
• Latitudes: Concentric circles centered at the pole.
• East: Tangential direction along the rotation.

Part (a): Thickness of the Disc

We are given that the radius of the disc is very large ($R \to \infty$) compared to the distance from the surface. At the center (North Pole), the gravitational field is equivalent to that near an infinite sheet of uniform mass density $\rho$ and thickness $h$.

The surface mass density is $\sigma = \rho h$. The gravitational field due to an infinite sheet is given by: $$ g_0 = 2\pi G \sigma = 2\pi G (\rho h) $$

Solving for thickness $h$:

$$ h = \frac{g_0}{2\pi \rho G} $$

Part (b): Firing along a Longitude

A shell is fired with velocity $u$ at an angle $\theta$ with the longitude. This means the projection of the velocity on the horizontal plane is along the radial line.

1. Analysis in Rotating Frame (Coriolis Effect)

Let the firing point be the origin of a local coordinate frame rotating with the disc.

• $\hat{i}$ (x-axis): Radial direction (along Longitude).
• $\hat{j}$ (y-axis): Tangential direction (East).
• $\hat{k}$ (z-axis): Vertical direction.

The angular velocity vector is $\vec{\omega} = \omega \hat{k}$.
The initial velocity vector is $\vec{v} = u \cos\theta \hat{i} + u \sin\theta \hat{k}$.

The Coriolis acceleration is given by $\vec{a}_{cor} = -2\vec{\omega} \times \vec{v}$. $$ \vec{a}_{cor} = -2(\omega \hat{k}) \times (u \cos\theta \hat{i} + u \sin\theta \hat{k}) $$ Since $\hat{k} \times \hat{k} = 0$ and $\hat{k} \times \hat{i} = \hat{j}$: $$ \vec{a}_{cor} = -2\omega u \cos\theta \hat{j} $$

The acceleration is in the $-\hat{j}$ direction (West). The displacement (deflection) $d$ is: $$ d = \left| \frac{1}{2} a_{cor} t^2 \right| = \omega u \cos\theta t^2 $$

2. Time of Flight

For the vertical motion, assuming gravity $g$ is constant: $$ t = \frac{2 u_z}{g} = \frac{2 u \sin\theta}{g} $$

3. Final Deflection

Substituting $t$ into the deflection equation:

$$ d = \omega u \cos\theta \left( \frac{2 u \sin\theta}{g} \right)^2 = \frac{4 \omega u^3 \sin^2\theta \cos\theta}{g^2} $$

The direction is West (to the right of the direction of motion relative to the ground).

Part (c): Firing along a Latitude

The shell is fired with velocity $u$ at an angle $\theta$ with the latitude. This means the horizontal component of velocity is along the East-West direction.

Velocity Components:
Radial velocity $v_r = 0$.
Tangential velocity $v_{tan} = u \cos\theta$ (Eastward).
Vertical velocity $v_z = u \sin\theta$.

1. Effective Radial Force

The particle experiences a change in the required centripetal force and the Coriolis force. Both act in the radial direction.
Coriolis Acceleration: $\vec{a}_{cor} = -2\vec{\omega} \times \vec{v}$. $$ \vec{a}_{cor} = -2(\omega \hat{k}) \times (u \cos\theta \hat{j}) $$ Since $\hat{k} \times \hat{j} = -\hat{i}$ (where $\hat{i}$ is radial outward): $$ \vec{a}_{cor} = 2\omega u \cos\theta \hat{i} $$

This acceleration is directed radially outward (South).

2. Deflection Calculation

Using the same time of flight $t = \frac{2 u \sin\theta}{g}$, the radial deflection is: $$ d = \frac{1}{2} a_{cor} t^2 = \frac{1}{2} (2\omega u \cos\theta) \left( \frac{2 u \sin\theta}{g} \right)^2 $$ $$ d = \omega u \cos\theta \frac{4 u^2 \sin^2\theta}{g^2} $$

The “$\pm$” in the original answer key corresponds to firing East (+) or West (-). If fired West, the Coriolis force directs inward.

Result: $\displaystyle \frac{4 \omega u^3 \sin^2\theta \cos\theta}{g^2}$ away from the latitude.