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Determine the ratio of the centrifugal acceleration to the gravitational acceleration at the Earth's equator.

Short Answer

Expert verified
The ratio of centrifugal acceleration to gravitational acceleration at the Earth's equator is approximately 0.0035.

Step by step solution

01

Understanding the Formula for Centrifugal Acceleration

At the Earth's equator, an object experiences centrifugal acceleration due to the Earth's rotation. The formula for centrifugal acceleration is given by \( a_c = \frac{v^2}{r} \), where \( v \) is the tangential velocity and \( r \) is the radius of Earth at the equator. Another expression for this is \( a_c = \omega^2 r \), where \( \omega \) is the angular velocity.
02

Calculating Angular Velocity

The angular velocity \( \omega \) of the Earth can be calculated from the fact that the Earth makes one full rotation (\( 2\pi \) radians) in about 24 hours. Thus, \( \omega = \frac{2\pi}{24 \times 3600} \) rad/s, as there are 3600 seconds in an hour.
03

Determining the Earth's Radius at the Equator

The radius of the Earth at the equator is approximately 6378 kilometers. For our calculations, we convert this to meters, giving us \( r = 6378000 \) meters.
04

Calculating Centrifugal Acceleration

Using \( a_c = \omega^2 r \), we substitute in the values: \( a_c = \left(\frac{2\pi}{24 \times 3600}\right)^2 \times 6378000 \). Calculating this gives \( a_c \approx 0.034 \text{ m/s}^2 \).
05

Understanding Gravitational Acceleration

The gravitational acceleration \( g \) at Earth's surface is approximately \( 9.81 \text{ m/s}^2 \). This is the acceleration due to Earth's gravitational pull.
06

Calculating the Ratio

The ratio of centrifugal acceleration to gravitational acceleration is then given by \( \frac{a_c}{g} = \frac{0.034}{9.81} \), which simplifies to approximately 0.0035.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Acceleration
Gravitational acceleration, denoted as \( g \), is the acceleration experienced by an object due to the gravitational pull of the Earth. On the surface of the Earth, this acceleration is approximately \( 9.81 \text{ m/s}^2 \). It acts downwards, towards the center of the Earth.

Gravitational acceleration is a constant value that affects everything near the Earth equally, regardless of its mass. This means that if you drop two objects from the same height, they will reach the ground at the same time, as long as air resistance is negligible.

Key points about gravitational acceleration:
  • It is responsible for keeping objects grounded.
  • Its value varies slightly depending on where you are on Earth due to factors like altitude.
  • It's crucial in calculations involving free-fall and projectile motion.
Earth's Rotation
Earth's rotation refers to the spinning of the Earth around its own axis. This rotation occurs from west to east and is responsible for the cycle of day and night.

The Earth completes one full rotation approximately every 24 hours. As it rotates, different parts of the globe experience sunlight, creating the day-night pattern.

The rotational motion of the Earth induces some effects worth noting:
  • Cyclical patterns: Understanding Earth's rotation is essential for grasping day length and time zones.
  • Rotational speed: At the equator, Earth rotates at about 1670 kilometers/hour. This speed contributes to the phenomena like centrifugal acceleration.
  • Coriolis effect: It's responsible for the deflection of objects due to Earth's rotation, which affects weather patterns and ocean currents.
Angular Velocity
Angular velocity \( \omega \) represents how fast an object rotates or revolves relative to another point, usually in terms of radians per second. For Earth, angular velocity is linked to its rotation.

To calculate Earth's angular velocity, we note that it completes a full turn or rotation (equal to \( 2\pi \) radians) in 24 hours. Therefore, the angular velocity of Earth can be mathematically expressed as:
\[ \omega = \frac{2\pi}{24 \times 3600} \text{ radians/second} \]
Feel free to perform this calculation to find its precise value.

Important points about angular velocity:
  • It differs from linear velocity because it describes rotation, not translation.
  • Angular velocity is crucial for understanding rotational mechanics and phenomena like centrifugal forces.
  • It connects seamlessly with other rotational formulas such as those for calculating centrifugal acceleration \( a_c \).

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Most popular questions from this chapter

Assume a large geoid anomaly with a horizontal scale of several thousand kilometers has a mantle origin and its location does not change. Because of continental drift the passive margin of a continent passes through the anomaly. Is there a significant change in sea level associated with the passage of the margin through the geoid anomaly? Explain your answer.

A volcanic plug of diameter \(10 \mathrm{~km}\) has a gravity anomaly of \(0.3 \mathrm{~mm} \mathrm{~s}^{-2}\). Estimate the depth of the plug assuming that it can be modeled by a vertical cylinder whose top is at the surface. Assume that the plug has density of \(3000 \mathrm{~kg} \mathrm{~m}^{-3}\) and the rock it intrudes has a density of \(2800 \mathrm{~kg} \mathrm{~m}^{-3}\).

Assuming that the difference in moments of inertia \(C-A\) is associated with a near surface density \(\rho_{m}\) and the mass \(M\) is associated with a mean planetary density \(\bar{\rho}\), show that $$J_{2}=\frac{2}{5} \frac{\rho_{m}}{\bar{\rho}} f$$ Determine the value of \(\rho_{m}\) for the Earth by using the measured values of \(J_{2}, \bar{\rho},\) and \(f .\) Discuss the value obtained.

For a point on the surface of the Moon determine the ratio of the acceleration of gravity due to the mass of the Earth to the acceleration of gravity due to the mass of the Moon.

Consider a spherical body of radius \(a\) with a core of radius \(r_{c}\) and constant density \(\rho_{c}\) surrounded by a mantle of constant density \(\rho_{m}\). Show that the moment of inertia \(C\) and mass \(M\) are given by $$\begin{aligned}C &=\frac{8 \pi}{15}\left[\rho_{c} r_{c}^{5}+\rho_{m}\left(a^{5}-r_{c}^{5}\right)\right] \\\M &=\frac{4 \pi}{3}\left[\rho_{c} r_{c}^{3}+\rho_{m}\left(a^{3}-r_{c}^{3}\right)\right] \end{aligned}$$ Determine mean values for the densities of the Earth's mantle and core given \(C=8.04 \times 10^{37}\) \(\mathrm{kg} \mathrm{m}^{2}, M=5.97 \times 10^{24} \mathrm{~kg}, a=6378 \mathrm{~km},\) and \(r_{c}=3486 \mathrm{~km}\)

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