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Chapter 8: Problem 40

The earth has a radius of \(6.38 \times 10^{6} \mathrm{m}\) and turns on its axis once every \(23.9 \mathrm{h}\). (a) What is the tangential speed (in \(\mathrm{m} / \mathrm{s}\) ) of a person living in Ecuador, a country that lies on the equator? (b) At what latitude (i.e., the angle \(\theta\) in the drawing) is the tangential speed one-third that of a person living in Ecuador?

Short Answer

Expert verified
(a) 465 m/s, (b) 70.5掳 latitude.

Step by step solution

01

Calculate the Circumference of Earth

The circumference of Earth can be found using the formula \(C = 2 \pi r\), where \(r\) is the radius of Earth. Given \(r = 6.38 \times 10^6\) m, calculate:\[C = 2 \pi (6.38 \times 10^6) = 4.01 \times 10^7 \text{ m}\]
02

Find the Rotational Period in Seconds

Convert the rotational period from hours to seconds. Since there are 3600 seconds in an hour:\[23.9 \times 3600 = 86184 \text{ s}\]
03

Calculate the Tangential Speed at the Equator

The tangential speed at the equator is the circumference divided by the rotational period:\[v = \frac{C}{T} = \frac{4.01 \times 10^7}{86184} \approx 465 \text{ m/s}\]
04

Relate Tangential Speed to Latitude

Tangential speed at a latitude \(\theta\) is given by \(v_\theta = v_0 \cos(\theta)\), where \(v_0 = 465\) m/s.
05

Set Up Equation for One-Third Speed

We want the speed to be one-third of the speed at the equator, so:\[v_\theta = \frac{1}{3}v_0 = 155 \text{ m/s}\]
06

Solve for Latitude Angle

Set the equation from Step 4 to equal 155 m/s and solve for \(\theta\):\[155 = 465 \cos(\theta)\]\[\cos(\theta) = \frac{155}{465} = \frac{1}{3}\]\[\theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.5^\circ\]

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Key Concepts

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

Earth Radius
The Earth's radius is a fundamental measurement in understanding our planet's size and how it moves in space. The radius of Earth, which is the distance from the center of the Earth to the surface, measures approximately \(6.38 \times 10^6\) meters. This standard measure is crucial when performing calculations related to Earth's rotation and surface phenomena.Understanding the radius helps in:
  • Determining the circumference of Earth, which is the distance around the equator, calculated using the formula \(C = 2\pi r\).
  • Calculating distances and speeds related to Earth's rotation, such as tangential and rotational speeds.
The radius is often used in calculations involving geographical locations and satellite orbits, giving us insight into many practical and scientific fields.
Rotational Dynamics
Rotational dynamics is the study of how objects rotate and the forces that affect their rotation. For Earth, its rotation is a key aspect of its dynamics. To understand Earth's rotational dynamics:
  • The Earth rotates on its axis once approximately every 23.9 hours, which is known as the rotational period.
  • This rotation influences various phenomena such as day and night cycle and the apparent movement of the sun and stars across the sky.
  • Rotational dynamics involves calculating aspects like the angular velocity and tangential speed, which depend on Earth's size and rotation speed.
Grasping these aspects helps us understand everything from local weather patterns to satellite trajectories.
Latitude Calculation
Latitude is a geographic coordinate that specifies the north-south position of a point on Earth's surface. It affects the tangential speed at different locations due to Earth's spherical shape.To calculate latitude in the context of changing tangential speed:
  • The tangential speed at any latitude \(\theta\) is given by the formula \(v_\theta = v_0 \cos(\theta)\), where \(v_0\) is the tangential speed at the equator.
  • This formula shows that the speed decreases as one moves towards the poles, which is due to the cosine of the latitude angle.
  • For example, when the tangential speed is one-third of that at the equator, we can solve for \(\theta\) using \(\cos(\theta) = \frac{1}{3}\).
Understanding latitude calculations can help in various applications, from navigation to optimizing planet-based observations.
Equatorial Speed
The equatorial speed is the speed at which a point on the Earth's surface moves due to its rotation, specifically at the equator. This is the highest speed a point on Earth's surface can have due to its rotation. Key points about equatorial speed:
  • At the equator, the Earth's surface moves at approximately 465 meters per second (m/s).
  • This speed is calculated by dividing the Earth's circumference by its rotational period, demonstrating how rapidly Earth rotates even though it's not easily perceived.
  • Equatorial speed affects the strength of phenomena like the Coriolis effect, which influences weather patterns and ocean currents.
This concept significantly contributes to our understanding of global dynamics and the forces that shape our environment.

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Most popular questions from this chapter

A pitcher throws a curveball that reaches the catcher in 0.60 s. The ball curves because it is spinning at an average angular velocity of 330 rev/min (assumed constant) on its way to the catcher鈥檚 mitt. What is the angular displacement of the baseball (in radians) as it travels from the pitcher to the catcher?

A stroboscope is a light that flashes on and off at a constant rate. It can be used to illuminate a rotating object, and if the flashing rate is adjusted properly, the object can be made to appear stationary. (a) What is the shortest time between flashes of light that will make a three-bladed propeller appear stationary when it is rotating with an angular speed of \(16.7 \mathrm{rev} / \mathrm{s} ?\) (b) What is the next shortest time?

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is \(8.3 \mathrm{m}\) above the water. One diver runs off the edge of the cliff, tucks into a "ball," and rotates on the way down with an average angular speed of 1.6 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.

The table that follows lists four pairs of initial and fi nal angles of a wheel on a moving car. The elapsed time for each pair of angles is 2.0 s. For each of the four pairs, determine the average angular velocity (magnitude and direction as given by the algebraic sign of your answer). $$ \begin{array}{lcc} & \text { Initial angle } \theta_{0} & \text { Final angle } \theta \\ \hline \text { (a) } & 0.45 \mathrm{rad} & 0.75 \mathrm{rad} \\ \hline \text { (b) } & 0.94 \mathrm{rad} & 0.54 \mathrm{rad} \\ \hline \text { (c) } & 5.4 \mathrm{rad} & 4.2 \mathrm{rad} \\ \hline \text { (d) } & 3.0 \mathrm{rad} & 3.8 \mathrm{rad} \\ \hline \end{array} $$

Conceptual Example 2 provides some relevant background for this problem. A jet is circling an airport control tower at a distance of \(18.0 \mathrm{km}\) An observer in the tower watches the jet cross in front of the moon. As seen from the tower, the moon subtends an angle of \(9.04 \times 10^{-3}\) radians. Find the distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.
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