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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 \(m\):\[m = \frac{M}{3}\]

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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.

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Most popular questions from this chapter

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00-m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The 0.800-kg rod has a light basket at the other end to catch the 0.163-kg puck. The puck slides across the ice with velocity \(\overrightarrow{v}\) (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then?

A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 110 kg and radius 4.0 m. Calculate the magnitude of the total angular momentum of the woman\(-\)disk system. (Assume that you can treat the woman as a point.)

A block with mass m is revolving with linear speed \(v_1\) in a circle of radius \(r_1\) on a frictionless horizontal surface (see Fig. E10.40). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_2\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_1\) and the radius \(r_1\). (b) Use \(W = \int_{r1}^{r2} \overrightarrow{T} (r) \cdot d\overrightarrow{r}\) to calculate the work done by \(\overrightarrow{T}\) when \(r\) changes from \(r_1\) to \(r_2\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s. The ball rebounds in the opposite direction with a speed of 6.00 m/s. (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

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