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Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth \(=12800 \mathrm{~km}\) and it takes 24 hours for the earth to complete one revolution about its axis.

Short Answer

Expert verified
The acceleration due to Earth's rotation at the equator is approximately 0.0337 m/s².

Step by step solution

01

Calculate the Radius of the Earth

The diameter of the Earth is given as 12800 km. To find the radius, we divide the diameter by 2. \[\text{Radius of Earth} = \frac{12800}{2} = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}\] Converting kilometers to meters for standard units.
02

Determine the Circumference of the Earth

Using the formula for the circumference of a circle \(C = 2\pi r\), where \(r\) is the radius:\[C = 2 \pi \times 6.4 \times 10^6 \text{ m}\] This gives us the total distance a point on the equator travels in one full rotation.
03

Calculate the Velocity of a Point on the Equator

The Earth makes one full rotation in 24 hours. To find the linear velocity \(v\), we use the formula \(v = \frac{C}{T}\), where \(T\) is the period of rotation in seconds (24 hours = 86400 seconds):\[v = \frac{2 \pi \times 6.4 \times 10^6}{86400}\] Simplifying this gives the velocity of a point on the equator.
04

Calculate the Centripetal Acceleration

The formula for centripetal acceleration \(a\) is given by \(a = \frac{v^2}{r}\), where \(v\) is the linear velocity and \(r\) is the radius of the Earth's rotation:\[a = \frac{\left(\frac{2 \pi \times 6.4 \times 10^6}{86400}\right)^2}{6.4 \times 10^6}\] Simplifying this gives the centripetal acceleration at the equator due to Earth's rotation.

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Key Concepts

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

Earth's Rotation
The rotation of the Earth is a fascinating phenomenon with profound effects on our daily lives. Imagine the Earth as a giant spinning top, rotating around its axis from west to east. One full rotation takes exactly 24 hours, known as a solar day. This spin is responsible for the cycle of day and night, as different parts of the Earth face the Sun and then move into shadow.

Since the Earth is rotating, every point on its surface is actually moving in a circular path. This is especially noticeable at the equator, where the Earth's circumference is greatest. The rotation introduces the concept of centripetal acceleration—since the Earth is rotating, objects are constantly changing direction, even though they are not moving in a straight line.
  • The Earth's rotation is uniform, meaning it happens at a constant pace.
  • This uniform rotation generates a consistent linear velocity at different points along the equator.
  • A significant consequence of rotation is the creation of the Coriolis effect, which affects weather patterns and ocean currents.
Understanding Earth's rotation is crucial for calculating various phenomena, such as the apparent change of stars in the sky and the mechanics of satellites orbiting the planet.
Equatorial Velocity
Equatorial velocity is the speed at which a point on the Earth's equator travels due to the planet's rotation. Since the Earth rotates fully in 24 hours, each point along the equator covers a vast distance, completing a circular journey around the Earth's axis.

To calculate this velocity, we first find the circumference of the Earth by using the formula: circumference = 2\(\pi\) \(r\), where \(r\) is the Earth's radius. If we consider the Earth's radius to be approximately 6,400,000 meters, the circumference becomes the path length for any point on the equator:

\[ C = 2 \pi \times 6.4 \times 10^6 \text{ m} \approx 4.014 \times 10^7 \text{ m} \]

The period \(T\) for one rotation is 86400 seconds (24 hours). Therefore, the equatorial velocity \(v\) is given by:
  • \(v = \frac{C}{T} = \frac{4.014 \times 10^7}{86400} \approx 463 \text{ m/s} \)
This velocity means that, on average, a point on the equator moves at nearly 463 meters per second. It's fascinating to consider how fast we are traveling due to Earth's motion, even when standing still! This rotation introduces the necessity of centripetal force to maintain this motion.
Radius of the Earth
The radius of the Earth is a crucial measurement used in many scientific calculations, particularly those involving its rotation and gravitational effects. Defined as the distance from the Earth's center to its surface, this radius is approximately 6,400 kilometers (or 6.4 million meters) at the equator, due to the planet's oblate shape—wider at the equator than between the poles.

The simple relation \(\text{Radius} = \frac{\text{Diameter}}{2}\) is used to calculate this value when the Earth's diameter is known. Given a diameter of approximately 12,800 kilometers, dividing this by two provides the familiar quoted radius. The Earth’s rotation causes this equatorial bulge, making the radius greater there than at the poles.

  • This radius is used in calculating the Earth's circumference: \(C = 2 \pi r\).
  • It provides a foundational constant for calculating equatorial velocity and centripetal acceleration.
  • Knowledge of the Earth's radius is essential for space sciences and geodesy.
Accurate knowledge of the Earth's radius allows for precise applications in fields such as geosciences, astronomy, and aviation.

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Most popular questions from this chapter

Two blocks each having a mass of \(3 \cdot 2 \mathrm{~kg}\) are connected by a wire \(C D\) and the system is suspended from the ceiling by another wire \(A B\) (figure 15 -E5). The linear mass density of the wire \(A B\) is \(10 \mathrm{~g} \mathrm{~m}^{-1}\) and that of \(\mathrm{CD}\) is \(8 \mathrm{~g} \mathrm{~m}^{-1}\). Find the speed of a transverse wave pulse produced in \(A B\) and in \(C D\).

A stone is fastened to one end of a string and is whirled in a vertical circle of radius \(R\). Find the minimum speed the stone can have at the highest point of the circle.

A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is \(1 \cdot 0 \mathrm{~cm}\) and the displacement becomes zero 200 times per second. The linear mass density of the string is \(0 \cdot 10 \mathrm{~kg} \mathrm{~m}^{-1}\) and it is kept under a tension of \(90 \mathrm{~N}\). (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive \(x\) -direction and at \(t=0\), the end \(x=0\) is at its positive extreme position. Write the wave equation. (c) Find the velocity and acceleration of the particle at \(x=50 \mathrm{~cm}\) at time \(t=10 \mathrm{~ms}\).

A wave propagates on a string in the positive \(x\) -direction at a velocity \(v\). The shape of the string at \(t=t_{0}\) is given by \(g\left(x, t_{0}\right)=A \sin (x / a)\). Write the wave equation for a general time \(t .\)

A wire, fixed at both ends is seen to vibrate at a resonant frequency of \(240 \mathrm{~Hz}\) and also at \(320 \mathrm{~Hz} .\) (a) What could be the maximum value of the fundamental frequency? (b) If transverse waves can travel on this string at a speed of \(40 \mathrm{~m} \mathrm{~s}^{-1}\), what is its length ?

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