Question

A bullet of mass \(m\) is fired with muzzle speed \(v_{\mathrm{o}}\) horizontally and due north from a position at colatitude \(\theta\). Find the direction and magnitude of the Coriolis force in terms of \(m, v_{\mathrm{o}}, \theta,\) and the earth's angular velocity \(\Omega .\) How does the Coriolis force compare with the bullet's weight if \(v_{\mathrm{o}}=1000 \mathrm{m} / \mathrm{s}\) and \(\theta=40\) deg?

Short answer

The Coriolis force acts eastward; its magnitude is relatively small compared to the bullet's weight.

Step-by-step solution

Understand the Problem

We need to find both the direction and magnitude of the Coriolis force acting on a bullet fired horizontally due north from a certain position on Earth, defined by colatitude \(\theta\). The relevant parameters include the mass of the bullet \(m\), its muzzle velocity \(v_{\mathrm{o}}\), the colatitude \(\theta\), and the Earth's angular velocity \(\Omega\). We also need to compare this force with the gravitational force (weight) acting on the bullet.

Determine the Direction of the Coriolis Force

The direction of the Coriolis force is given by the right-hand rule for the cross product of the velocity vector and the angular velocity vector. Since the bullet is fired horizontally due north, its velocity vector \(\mathbf{v}\) points north. The Earth's rotation vector \(\boldsymbol{\Omega}\) is directed along the axis of rotation (from south to north pole, align your fingers with it). Using the right-hand rule, the Coriolis force will act to the right of the motion, which is towards the east, at a colatitude \(\theta\).

Calculate the Magnitude of the Coriolis Force

The magnitude of the Coriolis force \(F_c\) is given by:\[ F_c = 2 m v_{\mathrm{o}} \Omega \sin(\theta) \]Substitute the known values and expression to find the magnitude explicitly.

Compute the Comparison with Gravitational Force

First, calculate the gravitational force (weight) \(W\):\[ W = m g \]where \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)).Next, compare the magnitude of the Coriolis force \(F_c\) and the weight \(W\) for given \(v_{\mathrm{o}} = 1000 \, \text{m/s}\) and \(\theta = 40^\circ\).

Perform the Calculations

Given \(v_{\mathrm{o}} = 1000 \, \text{m/s}\), \(\theta = 40^\circ\), and approximate \(\Omega = 7.29 \times 10^{-5} \, \text{rad/s}\), calculate:\[ F_c = 2 \cdot m \cdot 1000 \cdot 7.29 \times 10^{-5} \cdot \sin(40^\circ) \]For the comparison, simplify this expression and compare it to \(m \cdot 9.81\) after substituting \(\sin(40^\circ)\) with its approximate value (0.6428).

Key concepts

Angular Velocity

Angular velocity is a crucial concept to grasp when analyzing the effects, like the Coriolis force, due to rotation. It describes how fast something is rotating in terms of the angle through which it turns. Angular velocity, denoted as \( \Omega \), is a vector quantity:

• It has both magnitude and direction.
• The magnitude tells us how fast the rotation is happening.
• Its direction is given by the right-hand rule, which we will discuss more below.

In the context of Earth, the angular velocity is the constant rate at which the Earth rotates around its axis. The approximate value is \( 7.29 \times 10^{-5} \, \text{rad/s} \). This small value reflects a gradual rotation, but its effects become significant over large scales, such as atmospheric motion or bullets in flight!

Gravitational Force

Gravitational force, often referred to as weight, is the force exerted by Earth's gravity on an object. It is responsible for keeping objects grounded. The gravitational force \( W \) is determined using:

• The mass of the object \( m \).
• The acceleration due to gravity \( g \), approximately \( 9.81 \, \text{m/s}^2 \) at Earth's surface.

The equation for calculating weight is straightforward:\[ W = m g\]This force acts vertically downward towards the center of the Earth. In our exercise, this fundamental force is compared to the Coriolis force to understand their relative magnitudes.

Colatitude

Colatitude is a lesser-known geographical term, but it's quite important for analyzing forces on a rotating Earth. It is simply the complementary angle to the latitude.

• Latitude measures north-south position between the poles and the equator in degrees.
• Colatitude is calculated as: \( \text{Colatitude} = 90^\circ - \text{Latitude} \).

This concept comes into play when computing effects of the Coriolis force because it determines the component of a point on the Earth's surface that is perpendicular to the axis of rotation. For instance, at a latitude of \( 40^\circ \), the colatitude is \( 50^\circ \). This angle is critical in calculating the Coriolis effect, as the force is proportional to \( \sin(\text{Colatitude}) \).

Right-Hand Rule

The right-hand rule is a simple and reliable mnemonic to determine the direction of vectors resulting from cross product operations. In the setting of Coriolis force, it's used to find the direction of this force.

• Point your right hand's fingers in the direction of the velocity vector.
• Align your palm perpendicular to the angular velocity vector (Earth's north along rotation axis).
• Your thumb then points in the direction of the Coriolis force.

This method helps visualize complex three-dimensional motion interactions simply by using your hand. In the given exercise, it shows the resulting Coriolis force acts due east on a northward-flying bullet.