Problem 94 - Physics for Scientists and Engineers with Modern Physics

Question

Earth is not quite an inertial frame. We often make measurements in a reference frame fixed on the Earth, assuming Earth is an inertial reference frame. But the Earth rotates, so this assumption is not quite valid. Show that this assumption is off by 3 parts in 1000 by calculating the acceleration of an object at Earth's equator due to Earth's daily rotation, and compare to \(g=9.80 \mathrm{~m} / \mathrm{s}^{2},\) the acceleration due to gravity.

Step-by-Step Solution

Answer

The assumption of Earth as an inertial frame is off by about 3.46 parts in 1000.

Step 1: Understanding the Problem

To determine how the Earth's rotation affects the assumption of it being an inertial frame, we need to calculate the centripetal acceleration due to Earth's rotation at the equator and compare it with the gravitational acceleration, \(g = 9.80 \, \mathrm{m/s^2}\).

Step 2: Calculate the Centripetal Acceleration

The centripetal acceleration \(a_c\) for an object at the Earth's equator is given by the formula \(a_c = \omega^2 r\), where \(\omega\) is the angular velocity and \(r\) is the radius of the Earth. The Earth's angular velocity \(\omega\) is \(\frac{2\pi}{86400} \, \mathrm{rad/s}\) because the Earth completes one rotation in 86400 seconds (24 hours). The average radius \(r\) of the Earth is approximately \(6371 \, \mathrm{km}\) or \(6.371 \times 10^6 \, \mathrm{m}\).

Step 3: Perform Calculations

First, calculate \(\omega\):\[ \omega = \frac{2\pi}{86400} \, \text{rad/s} \approx 7.27 \times 10^{-5} \, \text{rad/s} \]Now, calculate \(a_c\):\[ a_c = \omega^2 r = (7.27 \times 10^{-5})^2 \times 6.371 \times 10^6 \, \text{m} \]\[ a_c \approx 0.0339 \, \mathrm{m/s^2} \]

Step 4: Compare Accelerations and Assess Error

To find the discrepancy due to Earth’s rotation, compare \(a_c\) to \(g\):\[ \text{Discrepancy ratio} = \frac{a_c}{g} = \frac{0.0339}{9.80} \approx 0.00346 \]Convert the ratio to parts per 1000:\[ 0.00346 \times 1000 \approx 3.46 \text{ parts per 1000} \]

Step 5: Conclusion

The effect of Earth's rotation introduces an error of approximately 3.46 parts per 1000 in the assumption of Earth as an inertial reference frame, which aligns with the question's statement that the assumption is off by about 3 parts in 1000.

Key Concepts

Centripetal Acceleration

When an object moves in a circular path, there is always a force pulling it towards the center of the circle. This force is known as centripetal force, and the acceleration caused by it is called centripetal acceleration. Imagine spinning a ball on a string; to keep it moving in a circle, you must constantly pull inward on the string. This pulling action, directed towards the center of the circle, generates centripetal acceleration. In mathematical terms, centripetal acceleration \(a_c\) is calculated using the formula \(a_c = \omega^2 r\), where \(\omega\) is the angular velocity of the object and \(r\) is the radius of the circular path. This concept is essential when studying the rotation of the Earth and how it affects objects on its surface.

Angular Velocity

Angular velocity describes how fast an object rotates around a specific axis. Consider how many turns a spinning record makes in a given time. Instead of measuring in usual speed units like meters per second, angular velocity is measured in radians per second (rad/s). This unit represents the angle covered in time. For Earth, which spins once every 24 hours, the angular velocity is calculated as \(\omega = \frac{2\pi}{86400} \, \text{rad/s}\), resulting in approximately \(7.27 \times 10^{-5} \, \text{rad/s}\). This number might seem small, but even tiny angular velocities are significant over large radii like that of Earth. The angular velocity is crucial in understanding Earth's rotation and its impact on equatorial objects.

Gravitational Acceleration

Gravitational acceleration refers to the acceleration that the force of gravity imparts on an object. On Earth, this is most commonly known as \(g\) and is approximately \(9.80 \, \text{m/s}^2\). This value means that an object falling freely under Earth's gravity will accelerate at \(9.80 \, \text{m/s}^2\) towards the surface. Gravitational acceleration is constant near Earth's surface and acts as a reference point when comparing other forces and accelerations, like centripetal acceleration due to Earth's rotation. Understanding gravitational acceleration helps greatly when assessing how significant Earth's rotation is to the assumption that Earth is an inertial reference frame.

Earth's Rotation

The Earth completes one rotation around its axis in roughly 24 hours, which significantly affects the dynamics of objects and systems on its surface. Earth's rotation causes objects at the equator to experience a centrifugal force that effectively reduces the gravitational pull experienced. This rotation-induced force results in a centripetal acceleration, affecting the assumption that Earth is a perfectly inertial frame of reference. The rotational motion means forces are not uniformly distributed, especially compared to non-rotating central reference points. This is why careful consideration of Earth's rotation is important in scientific calculations and assumptions, as it introduces accelerations that slightly modify expected outcomes.

Reference Frame Assumption

When analyzing motion and forces, scientists often use reference frames, which are fixed perspectives in space used for measurements and observations. Ideally, these frames are inertial, meaning they are either at rest or move with constant velocity. However, since Earth rotates, it's technically a non-inertial frame. Despite this, for most everyday purposes, Earth's surface is assumed to be an inertial frame because the errors introduced by its rotation are small. The exercise reveals that Earth's rotation causes an error of about 3 parts in 1000, highlighting why scientists account for such differences in precise calculations. Understanding the nuances of reference frame assumption is critical in physics as it dictates how forces and motions are calculated.