Question

Due to rotation of the earth the direction of vertical at a place is not along the radius of the earth and actually makes a small angle $\phi$ with the true vertical (i.e. with radius). At hat latitude $\Theta$ is this angle $\phi$ maximum?

• A. 0
• B. 30
• C. 45
• D. 60

Answer 1

Answer: (C)

45^\circ

Answer 2

The effective acceleration due to gravity ($\vec{g}_{eff}$) at a latitude $\Theta$ is the vector sum of the true gravitational acceleration ($\vec{g}_0$, directed towards the Earth's center) and the centrifugal acceleration ($\vec{a}_c$, directed outwards from the axis of rotation).

The true vertical is along the direction of $\vec{g}_0$. The apparent vertical is along the direction of $\vec{g}_{eff}$. The angle $\phi$ is between these two directions.

Let's resolve $\vec{g}_{eff}$ into components along the true vertical and true horizontal.

1. True Gravitational Acceleration ($\vec{g}_0$): Acts radially inwards, magnitude $g_0$.

2. Centrifugal Acceleration ($\vec{a}_c$): Magnitude $a_c = \omega^2 R \cos\Theta$, where $R$ is Earth's radius and $\omega$ is its angular velocity. This force acts outwards from the axis of rotation.

   • The component of $\vec{a}_c$ along the true horizontal (tangent to the surface, towards the equator) is $g_{eff,H} = a_c \sin\Theta = \omega^2 R \cos\Theta \sin\Theta$.
   • The component of $\vec{a}_c$ along the true vertical (radially outwards) is $a_c \cos\Theta = \omega^2 R \cos^2\Theta$. This component acts opposite to $\vec{g}_0$.

The effective components of acceleration are:

• Horizontal component: $g_{eff,H} = \omega^2 R \sin\Theta \cos\Theta$.
• Vertical component: $g_{eff,V} = g_0 - \omega^2 R \cos^2\Theta$.

The angle $\phi$ between the apparent vertical and the true vertical is given by:

$$\tan\phi = \frac{g_{eff,H}}{g_{eff,V}} = \frac{\omega^2 R \sin\Theta \cos\Theta}{g_0 - \omega^2 R \cos^2\Theta}$$

To find the latitude $\Theta$ where $\phi$ (and thus $\tan\phi$) is maximum, we differentiate this expression with respect to $\Theta$ and set it to zero.

Let $K = \frac{\omega^2 R}{g_0}$. Since $g_0 \gg \omega^2 R$, $K$ is a small quantity.

$$\tan\phi = \frac{K \sin\Theta \cos\Theta}{1 - K \cos^2\Theta} = \frac{\frac{K}{2} \sin(2\Theta)}{1 - K \cos^2\Theta}$$

Differentiating and setting the derivative to zero (ignoring the denominator which is never zero):

$$\frac{d}{d\Theta}\left(\frac{K}{2} \sin(2\Theta)\right)(1 - K \cos^2\Theta) - \frac{K}{2} \sin(2\Theta) \frac{d}{d\Theta}(1 - K \cos^2\Theta) = 0$$ $$K\cos(2\Theta)(1 - K \cos^2\Theta) - \frac{K}{2} \sin(2\Theta) (-K \cdot 2\cos\Theta (-\sin\Theta)) = 0$$ $$K\cos(2\Theta)(1 - K \cos^2\Theta) - K^2 \sin(2\Theta) \sin\Theta \cos\Theta = 0$$

Divide by $K$ (since $K \ne 0$):

$$\cos(2\Theta)(1 - K \cos^2\Theta) - K \sin(2\Theta) \sin\Theta \cos\Theta = 0$$

Substitute $\sin(2\Theta) = 2\sin\Theta \cos\Theta$:

$$\cos(2\Theta)(1 - K \cos^2\Theta) - K (2\sin\Theta \cos\Theta) \sin\Theta \cos\Theta = 0$$ $$\cos(2\Theta)(1 - K \cos^2\Theta) - 2K \sin^2\Theta \cos^2\Theta = 0$$

Substitute $\cos(2\Theta) = 2\cos^2\Theta - 1$ and $\sin^2\Theta = 1 - \cos^2\Theta$:

$$(2\cos^2\Theta - 1)(1 - K \cos^2\Theta) - 2K (1 - \cos^2\Theta)\cos^2\Theta = 0$$

Let $c^2 = \cos^2\Theta$.

$$(2c^2 - 1)(1 - Kc^2) - 2K (1 - c^2)c^2 = 0$$ $$2c^2 - 2Kc^4 - 1 + Kc^2 - 2Kc^2 + 2Kc^4 = 0$$ $$(2+K-2K)c^2 - 1 = 0$$ $$(2-K)c^2 - 1 = 0$$ $$c^2 = \frac{1}{2-K}$$

Since $K = \frac{\omega^2 R}{g_0}$ is very small ($K \approx 0.00345$), we can approximate $2-K \approx 2$.

$$\cos^2\Theta \approx \frac{1}{2}$$ $$\cos\Theta = \frac{1}{\sqrt{2}}$$

This implies $\Theta = 45^\circ$.

The angle $\phi$ is maximum at a latitude of $45^\circ$.