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Asteroid Collision! Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth's mass \(M\) , for the day to become 25.0\(\%\) longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Short Answer

Expert verified
The asteroid must have a mass equal to one-fourth of Earth's mass.

Step by step solution

01

Understand the Problem

The problem asks us to determine the mass of an asteroid (in terms of Earth's mass \( M \)) that would cause Earth's day (rotation period) to increase by 25% after a collision. Currently, Earth's day is approximately 24 hours, so we want it to be 30 hours.
02

Apply Angular Momentum Conservation

Because there are no external torques acting on the Earth-asteroid system, angular momentum is conserved. The initial angular momentum of Earth (\( L_i = I_i \omega_i \)), where \( I_i \) is Earth's initial moment of inertia and \( \omega_i \) is its initial angular velocity, is equal to the final angular momentum of the system \( L_f = I_f \omega_f \).
03

Expression for Moment of Inertia

Assuming Earth is a uniform sphere, its moment of inertia is \( I_i = \frac{2}{5} M R^2 \) where \( M \) is Earth's mass and \( R \) its radius. After the collision, the moment of inertia increases to \( I_f = \frac{2}{5} M R^2 + mR^2 \), where \( m \) is the asteroid's mass.
04

Calculate Angular Velocities

The initial angular velocity \( \omega_i \) is \( \frac{2\pi}{T_i} = \frac{2\pi}{24 \text{ hours}} \). The final angular velocity \( \omega_f \) must result in a period \( T_f = 1.25 \, T_i = 30 \text{ hours} \), so \( \omega_f = \frac{2\pi}{30}\text{ hours}^{-1} \).
05

Set Up and Solve for Asteroid Mass

Using the conservation of angular momentum \( L_i = L_f \), i.e., \( I_i \omega_i = I_f \omega_f \):\[ \frac{2}{5} M R^2 \left( \frac{2\pi}{24} \right) = \left( \frac{2}{5} M R^2 + m R^2 \right) \left( \frac{2\pi}{30} \right) \]Solving for \( m \), we get:\[ m = M \left( \frac{1}{5} \right) \left( \frac{5}{4} \right) \]\[ m = \frac{M}{4} \]
06

Interpret the Result

The mass of the asteroid must be \( \frac{M}{4} \), meaning the asteroid's mass must equal one-fourth of Earth's mass to lengthen the day to 30 hours due to a direct collision.

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Key Concepts

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

Moment of Inertia
The moment of inertia is a crucial concept in understanding rotational motion. It is a measure of an object's resistance to changes in its rotation. You can think of it as the rotational equivalent of mass in linear motion. For solid spheres like Earth, the moment of inertia, denoted as \( I \), is calculated using the formula \( I = \frac{2}{5} MR^2 \). Here, \( M \) is the mass of the object (in this case, Earth) and \( R \) is the radius.When an asteroid impacts Earth, this distribution of mass changes, altering the moment of inertia. After the collision, the new moment of inertia becomes larger due to the added mass of the asteroid. This is expressed as \( I_f = \frac{2}{5} MR^2 + mR^2 \), where \( m \) is the mass of the asteroid. This increment in moment of inertia affects how Earth spins after the collision.
Angular Velocity
Angular velocity, represented by the symbol \( \omega \), refers to how fast an object rotates around a specific axis. For Earth, it involves how quickly it spins around its axis, dictating the length of a day. Mathematically, angular velocity is calculated using the formula \( \omega = \frac{2\pi}{T} \), where \( T \) represents the time taken for one complete rotation. In this exercise, Earth's initial angular velocity is \( \omega_i = \frac{2\pi}{24} \, \text{hours}^{-1} \), corresponding to a 24-hour day. After the asteroid collision, we aim to find a new \( \omega_f \) such that the day extends by 25%, making it \( T_f = 30 \, \text{hours} \). This changes the angular velocity to \( \omega_f = \frac{2\pi}{30} \, \text{hours}^{-1} \). This reduced angular velocity is key to explaining the slower rotation of Earth post-collision.
Impact on Earth's Rotation
The collision of an asteroid with Earth can drastically alter its rotation. This impacts not just the length of a day but can affect climate and geography over time. Due to the conservation of angular momentum, Earth's spin changes when external forces change its mass distribution. When the asteroid adds its mass, Earth's moment of inertia increases, slowing its rotation to conserve angular momentum. This phenomenon is likened to a figure skater extending their arms to spin more slowly. For our problem, the increase in Earth's day length from 24 to 30 hours is a reflection of this physical principle, which works to balance the planet’s angular momentum with its new mass and size.
Collision Dynamics
Collision dynamics encompasses the study of how objects behave when they crash into one another. In the context of our exercise, the asteroid impacts Earth—causing physical and kinetic changes. These changes can be explained through physics laws like the conservation of angular momentum. When the asteroid collides at the equator, it effectively adds its mass to Earth's, altering the planet's rotation characteristics. The dynamics involve assessing how the redistribution of mass affects Earth's motion. In our scenario, we calculated that the asteroid required to change the day length by 25% must be one-fourth of Earth's mass. This example illustrates how significant an impact can be and highlights the importance of understanding collision dynamics in planetary science.

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Most popular questions from this chapter

A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 \(\mathrm{m}\) and a total mass of 120 \(\mathrm{kg}\) . The turntable is initially rotating at 3.00 \(\mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a 70.0 -kg parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

A uniform disk with mass 40.0 \(\mathrm{kg}\) and radius 0.200 \(\mathrm{m}\) is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force \(F=30.0 \mathrm{N}\) is applied tangent to the rim of the disk. (a) What is the magnitude \(v\) of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution? (b) What is the magnitude \(a\) of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolution?

A uniform rod of length \(L\) rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed \(v\) strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angeed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

A hollow, thin-walled sphere of mass 12.0 \(\mathrm{kg}\) and diameter 48.0 \(\mathrm{cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value 1.10 . (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time 3.00 s, find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

A child rolls a 0.600 -kg basketball up a long ramp. The basketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.0 \(\mathrm{m} / \mathrm{s} .\) When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.0 \(\mathrm{m} / \mathrm{s}\) . Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase of the ball as it rolls up the ramp.

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