Problem 4 - Classical Mechanics

Question

Find the velocity relative to an inertial frame (in which the centre of the Earth is at rest) of a point on the Earth's equator. An aircraft is flying above the equator at \(1000 \mathrm{~km} \mathrm{~h}^{-1}\). Assuming that it flies straight and level (i.e., at a constant altitude above the surface) what is its velocity relative to the inertial frame (a) if it flies north, (b) if it flies west, and (c) if it flies east?

Step-by-Step Solution

Answer

Answer: (a) The aircraft's velocity when flying north is approximately 1950 km/h in a direction orthogonal to Earth's rotational velocity. (b) The aircraft's velocity when flying west is approximately 670 km/h opposite to Earth's rotational velocity. (c) The aircraft's velocity when flying east is approximately 2670 km/h in the same direction as Earth's rotational velocity.

Step 1: Calculate Earth's equatorial speed

First, we will calculate the speed of a point on Earth's equator due to its rotation. Earth's circumference is given by \(C = 2\pi R\), where \(R\) is Earth's radius, approximately \(6371 \mathrm{~km}\). Earth takes about 24 hours to complete one rotation, so the speed of a point on the equator is: v = \frac{C}{T} = \frac{2\pi R}{24\mathrm{~h}} = \frac{2\pi(6371)~\mathrm{km}}{24\mathrm{~h}} v \approx 1670 \mathrm{~km}\mathrm{~h}^{-1} Where v is the speed of a point of the Earth's equator.

Step 2: Find the aircraft's velocity in each case

Now we will use vector addition to find the aircraft's velocity in each case. Let's denote the aircraft's speed of \(1000 \mathrm{~km} \mathrm{~h}^{-1}\) as \(v_a\). (a) If the aircraft flies north: In this case, the aircraft's velocity is orthogonal to Earth's rotational velocity, so the total velocity is the vector sum of the aircraft's northward speed and Earth's eastward speed. Using the Pythagorean theorem, we get: v_t = \sqrt{v^2 + v_a^2} = \sqrt{(1670\mathrm{~km}\mathrm{~h}^{-1})^2 + (1000\mathrm{~km}\mathrm{~h}^{-1})^2} v_t \approx 1950 \mathrm{~km}\mathrm{~h}^{-1} (b) If the aircraft flies west: In this case, the aircraft's velocity is opposite to Earth's rotational velocity. So, we will subtract Earth's equatorial speed from the aircraft's speed to get the total velocity: v_t = v - v_a = 1670\mathrm{~km}\mathrm{~h}^{-1} - 1000\mathrm{~km}\mathrm{~h}^{-1} v_t = 670 \mathrm{~km}\mathrm{~h}^{-1} (c) If the aircraft flies east: In this case, the aircraft's velocity is in the same direction as Earth's rotational velocity. We will add Earth's equatorial speed to the aircraft's speed to get the total velocity: v_t = v + v_a = 1670\mathrm{~km}\mathrm{~h}^{-1} + 1000\mathrm{~km}\mathrm{~h}^{-1} v_t = 2670 \mathrm{~km}\mathrm{~h}^{-1} In conclusion, the aircraft's velocity relative to the inertial frame (in which Earth's center is at rest) is approximately (a) \(1950 \mathrm{~km}\mathrm{~h}^{-1}\) if it flies north, (b) \(670 \mathrm{~km}\mathrm{~h}^{-1}\) if it flies west, and (c) \(2670 \mathrm{~km}\mathrm{~h}^{-1}\) if it flies east.

Key Concepts

Inertial Frame

An inertial frame is a reference frame where an object either remains at rest or continues to move at a constant velocity unless acted upon by a force. In simpler terms, it’s like being in a car moving smoothly on a straight road. You feel no forces acting on you unless the car accelerates or brakes.
In the realm of physics and relative velocity, an inertial frame simplifies calculations by allowing us to assume that no external forces are acting on the system in question. This is especially handy when considering velocities relative to Earth or, as in this exercise, a point on Earth’s equator.
In this exercise, you’re asked to find the velocity of an aircraft relative to an inertial frame where the center of Earth is stationary. This means we remove the complexity added by Earth's rotation, focusing only on the straightforward path around the sun.

Earth's Rotation

Earth's rotation affects relative velocity calculations because points on Earth's surface, such as the equator, move with the planet as it spins. This rotational movement is around the Earth’s axis, taking about 24 hours to complete a full turn.
For points on the equator, this rotation can be visualized as moving along Earth's circumference. Since the radius of Earth is approximately 6371 km, and using the formula for the circumference of a circle, we calculate this rotational speed like so:

• Circumference, \( C = 2\pi R \)
• Equatorial speed, \( v = \frac{C}{T} = \frac{2\pi(6371)\ \text{km}}{24\ \text{h}} \approx 1670\ \text{km/h} \)

This speed needs to be considered when calculating the aircraft's velocity relative to the inertial frame at Earth’s center.

Vector Addition

Vector addition is a mathematical method used to combine different velocities or forces that an object might experience. In the context of this exercise, consider the aircraft's velocity relative to both the Earth's rotation and the point it's moving towards.
Depending on the direction the aircraft travels:

• **Northward Flight:** The aircraft's velocity is perpendicular to the equatorial speed, meaning you use the Pythagorean theorem to add them: \( v_t = \sqrt{(1670\ \text{km/h})^2 + (1000\ \text{km/h})^2} \).
• **Westward Flight:** The aircraft's velocity is in the opposite direction to Earth's rotation, resulting in subtraction: \( v_t = 1670\ \text{km/h} - 1000\ \text{km/h} \).
• **Eastward Flight:** The aircraft's velocity supplements Earth's rotation, necessitating addition: \( v_t = 1670\ \text{km/h} + 1000\ \text{km/h} \).

Using vector addition, we determine how these different directions and speeds work together (or against each other) to produce the total velocity of the aircraft in relation to the inertial frame.