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Chapter 3: Problem 29

The earth has a radius of 6380 \(\mathrm{km}\) 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 \(a_{n d}\) at the equator is greater than \(g\) , objects would 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) Radial acceleration is approximately 0.034 m/s虏 or 0.0035g. (b) The Earth would need to have a rotation period of around 1.4 hours for objects to fly off.

Step by step solution

01

Calculate the Circumference of the Earth

The circumference of the Earth at the equator is given by the formula \( C = 2 \pi r \). Given the radius \( r = 6380 \text{ km} \), we can convert it to meters: \( r = 6380 \times 1000 \text{ m} \). Hence, the circumference \( C = 2 \pi (6380 \times 1000) \text{ m} \).
02

Determine the Tangential Velocity

The tangential velocity \( v \) at the equator can be calculated by dividing the circumference by the time it takes for one full rotation, which is 24 hours. First, convert the time to seconds: \( 24 \text{ hr} = 86400 \text{ s} \). Then, \( v = \frac{C}{T} = \frac{2 \pi (6380 \times 1000) \text{ m}}{86400 \text{ s}} \).
03

Calculate the Radial Acceleration

The radial (centripetal) acceleration \( a \) is given by \( a = \frac{v^2}{r} \). Use the velocity from Step 2 and the radius in meters to find \( a \).
04

Express Radial Acceleration as a Fraction of g

The acceleration due to gravity \( g \) is approximately \( 9.8 \text{ m/s}^2 \). Express the radial acceleration \( a \) from Step 3 as a fraction of \( g \) by \( \frac{a}{g} \).
05

Determine Rotation Period for 'Fly Off' Condition

For objects to fly off, the radial acceleration should equal \( g \). Using \( a = \frac{v^2}{r} = g \), find the new rotation period \( T \) when \( v = \sqrt{g \cdot r} \). Solve for \( T = \frac{C}{v} \).

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Key Concepts

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

Tangential Velocity
Tangential velocity is the speed of an object moving along a circular path. If you're at the equator, imagine the Earth spinning like a big merry-go-round. Tangential velocity is how fast you're moving sideways as the Earth spins beneath your feet.

To find this speed, you must know two things:
  • The circumference of the circle you're moving along - in this case, the equator, calculated by the formula: \( C = 2 \pi r \).
  • The time it takes to complete one full circle - Earth's rotation takes 24 hours.
Converting this time into seconds gives us \( T = 86400 \text{ s} \). The formula for tangential velocity \( v \) is then: \[ v = \frac{C}{T} \]Plug in the values to get the actual speed at which you are moving at the equator on Earth's surface.
Centripetal Acceleration
Centripetal acceleration is what keeps objects moving in a curve instead of shooting off in a straight line. It acts towards the center of the circle, helping maintain circular motion.

On Earth, this type of acceleration helps keep everything, including us, grounded as the planet rotates.

The formula for centripetal acceleration \( a \) is: \[ a = \frac{v^2}{r} \]where \( v \) is the tangential velocity and \( r \) is the radius of the Earth.

Using the velocity from the tangential velocity calculation and Earth's radius, you can determine this acceleration. The result is quite small on Earth compared to gravitational acceleration, hence we don鈥檛 feel it overtly.
Rotation Period
The rotation period is simply how long it takes for a planet to make one complete turn on its axis.

For Earth, this period is 24 hours. However, if the rotation period were shorter (meaning Earth spins faster), the centripetal acceleration at the equator would increase.

If it increased enough to equal gravitational force, objects would no longer "stick" to the surface. This hypothetical rotation period can be found by setting the centripetal acceleration equal to gravity \( g \) and solving for \( T \) again using \( v = \sqrt{g \cdot r} \). This would provide the exact time it would take for objects to start flying off due to increased angular velocity.
Gravitational Force
Gravitational force is the attraction between two bodies with mass. On Earth, this force gives us weight and keeps us grounded. This force is essentially the "glue" that holds us and everything else on the surface of the Earth.

Expressed by the formula: \[ F_g = m imes g \]where \( m \) is mass and \( g \) is the gravitational acceleration (approximately \( 9.8 \text{ m/s}^2 \)).

In this context, the gravitational force ensures that the centripetal force required for circular motion is balanced. A quicker rotation would necessitate a greater centripetal force, potentially reaching a point where centripetal force equals gravitational force, leading to the discussed 'fly off' scenario.

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Most popular questions from this chapter

Two tanks are engaged in a training exercise on level ground. The first tank fires a paint-filled training round with a muzzle speed of 250 \(\mathrm{m} / \mathrm{s}\) at \(10.0^{\circ}\) above the horizontal while advancing toward the second tank with a speed of 15.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground. The second tank is retreating at 35.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground, but is hit by the shell. You can ignore air resistance and assume the shell hits at the same height above ground from which it was fired. Find the distance between the tanks (a) when the round was first fired and (b) at the time of impact.

For no apparent reason, a poodle is running at a constant speed of \(v=5.00 \mathrm{m} / \mathrm{s}\) in a circle with radius \(R=2.50 \mathrm{m} .\) Let \(\vec{v}_{1}\) be the velocity vector at time \(t_{1},\) and let \(\vec{v}_{2}\) be the velocity vector at time \(t_{2}\) . Consider \(\Delta \vec{v}=\vec{v}_{2}-\vec{v}_{1}\) and \(\Delta t=t_{2}-t_{1}\) . Recall that \(\vec{a}_{\mathrm{av}}=\Delta \vec{v} / \Delta t .\) For \(\Delta t=0.5 \mathrm{s}, 0.1 \mathrm{s},\) and 0.05 \(\mathrm{s}\) , calculate the magnitude (to four significant figures) and direction (relative to \(\vec{v}_{1} )\) of the average acceleration \(\vec{d}_{\mathrm{ev}}\) . Compare your results to the general expression for the instantaneous acceleration \(\vec{a}\) for uniform circular motion that is derived in the text.

On level ground a shell is fired with an initial velocity of 80.0 \(\mathrm{m} / \mathrm{s}\) at \(60.0^{\circ}\) above the horizontal and feels no appreciable air resistance. (a) Find the horizontal and vertical components of the shell's initial velocity. (b) How long does it take the shell to reach its highest point? (c) Find its maximum height above the ground. (d) How far from its firing point does the shell land? (e) At its highest point, find the horizontal and vertical components of its acceleration and velocity.

A rocket is launched vertically from rest with a constant upward acceleration of 1.75 \(\mathrm{m} / \mathrm{s}^{2} .\) Suddenly 22.0 \(\mathrm{s}\) after launch, an unneeded fuel tank is jettisoned by shooting it away from the rocket. Acrew member riding in the rocket measures that the initial speed of the tank is 25.0 \(\mathrm{m} / \mathrm{s}\) and that it moves perpendicular to the rocket's path. The fuel tank feels no appreciable air resistance and reels only the force of gravity once it leaves the rocket. (a) How feels only the force of gravity once it leaves the rocket. (a) How fast is the rocket moving at the instant the fuel tank is jettisoned? (b) What are the horizontal and vertical componed by (i) a crew unk's velocity just as it is jetisoned as measured by (i) a crew member in the rocket and (ii) a technician standing on the ground? (c) At what angle with respect to the horizontal does the jettisoned fuel tank initially move, as measured by a crew member in the rocket and (ii) a technician standing on the ground? (d) What maximum height above the launch pad does the jettisoned tank reach?

Two students are canoeing on a river. While heading effort? upstream, they accidentally drop an emply bottle oveabound. They then continue paddling for 60 minutes, reaching a point 20 \(\mathrm{km}\) farther upstream. At this point they realize that the bottle is missing and, driven by ecological awareness, they tum around and head downstram. They catch up with and retrieve the bottle (which has been moving along with the current) 5.0 \(\mathrm{km}\) down- stream from the tum-around point. (a) Assuming a constant paddling effort throughout, how fast is the river flowing? (b) What would the canoe speed in a still lake be for the same paddling effort?
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