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Chapter 3: Problem 25

The earth has a radius of 6380 \(\mathrm{km}\) and turns around once on its axis in 24 \(\mathrm{h}\) . (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of \(g .\) (b) If \(a_{\text { rad }}\) at the equator is greater than \(g,\) objects will fly off the earth's surface and into space. (We will see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

Short Answer

Expert verified
(a) \( a_{\text{rad}} \approx 0.034 \text{ m/s}^2 \), which is approximately 0.0035 times \( g \). (b) The rotation period must be about 1.41 hours for objects to fly off.

Step by step solution

01

Convert Radius to Meters

The given radius of the Earth is \( 6380 \text{ km} \). First, convert this to meters by multiplying by 1000: \( 6380 \text{ km} \times 1000 = 6.38 \times 10^6 \text{ m} \).
02

Calculate the Radial Acceleration Formula

The radial acceleration for an object moving in a circle is given by \( a_{ ext{rad}} = \frac{v^2}{r} \), where \( v \) is the tangential speed and \( r \) is the radius.
03

Find the Tangential Speed at the Equator

The tangential speed \( v \) at the equator can be found using the formula \( v = \frac{2\pi r}{T} \), where \( T \) is the period of rotation (24 hours or 86400 seconds). Thus, \( v = \frac{2\pi \times 6.38 \times 10^6}{86400} \approx 465.1 \text{ m/s} \).
04

Compute the Radial Acceleration

Substitute the values into \( a_{ ext{rad}} = \frac{v^2}{r} \): \( a_{ ext{rad}} = \frac{(465.1)^2}{6.38 \times 10^6} \approx 0.034 \text{ m/s}^2 \).
05

Express Radial Acceleration as a Fraction of Gravity

The acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \). So, the fraction is \( \frac{0.034}{9.8} \approx 0.0035 \).
06

Determine Rotation Period for Critical Acceleration

For objects to begin flying off, \( a_{ ext{rad}} \) must equal \( g \). Set \( a_{ ext{rad}} = g \), so \( \frac{v^2}{r} = g \). Using \( v = \frac{2\pi r}{T} \), solve \( \frac{(\frac{2\pi r}{T})^2}{r} = g \). This simplifies to \( T = \frac{2\pi \sqrt{r}}{\sqrt{g}} \).
07

Calculate the New Rotation Period

Using \( r = 6.38 \times 10^6 \) and \( g = 9.8 \), calculate \( T = \frac{2\pi \sqrt{6.38 \times 10^6}}{\sqrt{9.8}} \approx 5067 \text{ s} \) or about 1.41 hours.

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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 and integral part of what makes life possible on our planet. The Earth completes a full rotation on its axis approximately every 24 hours. This rotation is what gives us the changing cycle of day and night, as different parts of the Earth face the Sun and then turn away from it.
The concept of Earth's rotation is crucial when discussing radial acceleration, particularly at the equator, where the effects of this rotation are most pronounced due to the large circumference of the Earth at this latitude.
It's important to remember that this rotation happens at a consistent angular velocity, meaning that every location on Earth turns with the same rotational speed, although the linear speed could vary depending on how far a point is from the axis. Understanding Earth's rotation helps to frame the bigger picture of how it affects movements and accelerations on and near the planet's surface.
Tangential Speed
Tangential speed is the linear speed of something moving along a circular path. For an object on Earth's surface, this speed is greatest at the equator due to the maximum distance from the axis of rotation.
At the equator, the tangential speed can be found using the formula: \[ v = \frac{2\pi r}{T} \]
where \( r \) is the Earth's radius and \( T \) is the rotational period (24 hours or 86400 seconds). In our exercise, we've computed this to be approximately 465.1 m/s.
This speed is crucial for determining the radial acceleration, as the faster an object is moving along its path, the greater the force required to keep it on that path, leading us to the next concept.
  • Tangential speed varies with the distance from the rotation axis.
  • It is an essential component of calculating radial acceleration.
  • At the poles, tangential speed is minimal compared to at the equator.
Acceleration Due to Gravity
Acceleration due to gravity, often denoted by \( g \), is approximately 9.8 m/s虏 on Earth's surface. This acceleration affects all objects that are near Earth's surface and plays a significant role in everyday physics.
In the context of our exercise, when calculating the radial acceleration at the Earth's equator, it is important to compare it with gravity to determine its significance. This comparison helps us understand under what circumstances an object might break free from the Earth's grasp.
For an object at the equator to be launched into space, its radial acceleration would need to match or exceed the acceleration due to gravity. This concept gives insights into how fast the Earth would need to spin for such a phenomenon to occur, which, according to the computations, would reduce the day's length significantly.
Understanding gravity is pivotal as it's the force that keeps us grounded and affects everything from walking and driving to the orbits of satellites. In this scenario, it acts as a reference point for evaluating other forces and movements on Earth's surface.

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Most popular questions from this chapter

The radius of the earth's orbit around the sun (assumed to be circular) is \(1.50 \times 10^{8} \mathrm{km},\) and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in \(\mathrm{m} / \mathrm{s} ?\) (b) What is the radial acceleration of the earth toward the sun, in \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius = \(5.79 \times 10^{7} \mathrm{km}\) , orbital period \(=88.0\) days \() .\)

Tossing Your Lunch. Henrietta is going off to her physics class, jogging down the sidewalk at 3.05 \(\mathrm{m} / \mathrm{s} .\) Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is 38.0 \(\mathrm{m}\) above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally 9.00 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. (a) With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? (b) Where is Henrietta when she catches the bagels?

A student sits atop a platform a distance \(h\) above the ground. He throws a large firecracker horizontally with a speed \(v .\) However, a wind blowing parallel to the ground gives the firecracker a constant horizontal acceleration with magnitude a. This results in the firecracker reaching the ground directly under the student. Determine the height \(h\) in terms of \(v, a,\) and \(g .\) You can ignore the effect of air resistance on the vertical motion.

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going \(90.0 \mathrm{km} / \mathrm{h},\) to his enemy's car, which is going 110 \(\mathrm{km} / \mathrm{h}\) . The enemy's car is 15.8 \(\mathrm{m}\) in front of the hero's when he lets go of the grenade. If the hero throws the grenade so its initial velocity relative to him is at an angle of \(45^{\circ}\) above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance. Find the magnitude of the velocity both relative to the hero and relative to the earth.

A car traveling on a level horizontal road comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side of the road the car is on is 21.3 \(\mathrm{m}\) above the river, while the opposite side is a mere 1.8 \(\mathrm{m}\) above the river. The river itself is a raging torrent 61.0 \(\mathrm{m}\) wide. (a) How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands on the other side?
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