Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 mass of the asteroid is \( \frac{M}{3} \).

Step by step solution

01

Understand the Problem

We are asked to determine the mass of an asteroid that, upon collision with Earth, will increase the length of a day by 25%. Currently, a day is 24 hours, so we want it to become 30 hours (since 25% of 24 is 6).
02

Apply Conservation of Angular Momentum

The total angular momentum before and after the collision must remain constant, since no external torques act on the system. Initially, the angular momentum, \( L_i \), of the Earth is \( I \omega \), where \( I = \frac{2}{5}MR^2 \) is the moment of inertia of Earth and \( \omega = \frac{2\pi}{86400 \text{ s}} \) is the initial angular velocity. After the collision, the final angular momentum, \( L_f \), is \( (I + mR^2) \omega_f \). Thus, \( I \omega = (I + mR^2) \omega_f \).
03

Set Up Equation for Angular Velocities

Relate the angular velocities before and after the collision. Since the length of the day becomes 30 hours, the new angular velocity \( \omega_f = \frac{2\pi}{108000 \text{ s}} \). Substitute this into \( I \omega = (I + mR^2) \omega_f \) to get:\[\frac{2MR^2\pi}{5 \times 86400} = \left( \frac{2}{5}MR^2 + mR^2 \right)\frac{2\pi}{108000}\]
04

Simplify and Solve for the Mass of the Asteroid

Cancel \(2\pi R^2\) from both sides and simplify to solve for \(m\):\[\frac{2M}{5 \times 86400} = \left(\frac{2}{5}M + m \right)\frac{1}{108000}\]Cross-multiply and solve for \(m\):\[m = \frac{M}{3}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Conservation of Angular Momentum
In physics, the conservation of angular momentum is a principle stating that if no external torque acts on a system, the total angular momentum of that system remains constant. This principle is especially vital when analyzing collision scenarios, such as an asteroid impacting Earth. When the asteroid collides with Earth, it changes how the Earth's mass and speed are distributed, effectively altering Earth's angular momentum. However, due to the conservation principle, the initial angular momentum (before the collision) should be equal to the final angular momentum (after the collision), if there are no external disturbances.To express it mathematically: the initial angular momentum is calculated as \( L_i = I \omega \), where \( I \) is Earth's moment of inertia and \( \omega \) is its angular velocity. After the collision, the combined system's angular momentum is given by \( L_f = (I + mR^2) \omega_f \). By setting these two equal, one can solve for unknown quantities, such as the mass of the asteroid in this exercise.
Moment of Inertia
The moment of inertia is an important concept to grasp when analyzing rotational motion and it represents an object's resistance to changes in its rotation. For the Earth, assuming it's a perfect sphere, the formula \( I = \frac{2}{5}MR^2 \) is used to calculate its moment of inertia.Understanding this property is pivotal during an asteroid collision scenario because it affects how energy is distributed.
  • An object with a larger moment of inertia requires more energy to change its state of rotation compared to one with a smaller moment of inertia.
  • This idea is analogous to trying to spin a computer chair versus a small paper fan; the chair, having more inertia, is harder to rotate quickly.
In our exercise, considering the Earth and the asteroid, changes in the moment of inertia post-collision explain why Earth's rotation period extends after the impact.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins around an axis. In this exercise, the initial and final angular velocities signify how the speed of Earth's rotation changes due to an asteroid impact.Before the collision, Earth's angular velocity \( \omega \) is given by \( \omega = \frac{2\pi}{86400 \, \text{s}} \), representing a full rotation every 24 hours. After the collision, the desired change for Earth's day to become 30 hours dictates a new angular velocity \( \omega_f = \frac{2\pi}{108000 \, \text{s}} \).The change in angular velocity from \( \omega \) to \( \omega_f \) is matched proportionally by changes in rotational inertia and accounts for the increase in day length from 24 to 30 hours. The relationship highlights the direct impact asteroids can have on a planet's rotational speed when they interact with it.
Physics Problem Solving
Solving physics problems involving concepts like angular momentum requires a strategic approach:1. **Understand the Problem:** Determine what changes after the event, such as an impact. In this exercise, it's the length of Earth's day.2. **Formulate the Conservation Equations:** Utilize conservation principles to equate initial and final states, such as angular momentum \( L_i = L_f \).3. **Identify Variables:** Clearly distinguish known quantities and solve for unknowns using given relations like moment of inertia and angular velocity.4. **Mathematical Simplification:** Break down the equations. In our example, cancelling terms like \( 2\pi R^2 \) simplifies the process of isolating the asteroid's mass \( m \).Each step involves critical thinking and methodical application of physics concepts to arrive at a solution. Persistence and practice go hand-in-hand with understanding the underlying physics to effectively solve such problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are designing a slide for a water park. In a sitting position, park guests slide a vertical distance \(h\) down the waterslide, which has negligible friction. When they reach the bottom of the slide, they grab a handle at the bottom end of a 6.00-m-long uniform pole. The pole hangs vertically, initially at rest. The upper end of the pole is pivoted about a stationary, frictionless axle. The pole with a person hanging on the end swings up through an angle of 72.0\(^\circ\), and then the person lets go of the pole and drops into a pool of water. Treat the person as a point mass. The pole's moment of inertia is given by \(I = \frac{1}{3} ML^2\), where \(L =\) 6.00 m is the length of the pole and \(M =\) 24.0 kg is its mass. For a person of mass 70.0 kg, what must be the height h in order for the pole to have a maximum angle of swing of 72.0\(^\circ\) after the collision?

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 angular speed 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, spherical shell with mass 2.00 kg rolls without slipping down a 38.0\(^\circ\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to 4.00 kg?

An electric motor consumes 9.00 kJ of electrical energy in 1.00 min. If one- third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2500 rpm?

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 N is applied to the rim of the wheel. The wheel has radius 0.120 m. Starting from rest, the wheel has an angular speed of 12.0 rev/s after 2.00 s. What is the moment of inertia of the wheel?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free