Question

The Earth has an angular speed of \(7.272 \cdot 10^{-5} \mathrm{rad} / \mathrm{s}\) in its rotation. Find the new angular speed if an asteroid \(\left(m=1.00 \cdot 10^{22} \mathrm{~kg}\right)\) hits the Earth while traveling at a speed of \(1.40 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\) (assume the asteroid is a point mass compared to the radius of the Earth) in each of the following cases: a) The asteroid hits the Earth dead center. b) The asteroid hits the Earth nearly tangentially in the direction of Earth's rotation. c) The asteroid hits the Earth nearly tangentially in the direction opposite to Earth's rotation.

Short answer

Question: Calculate the new angular speed of Earth in the three cases (dead center, tangentially in the direction of Earth's rotation, and tangentially opposite Earth's rotation) after the collision with the asteroid of mass \(1.00 \cdot 10^{22} kg\) and speed \(1.40 \cdot 10^3 m/s\), considering Earth has an initial angular speed of \(7.272 \cdot 10^{-5} rad/s\). Answer: a) For the case of dead center: \(\omega_{final\_a}\) b) For the case of tangentially in the direction of Earth's rotation: \(\omega_{final\_b}\) c) For the case of tangentially opposite Earth's rotation: \(\omega_{final\_c}\) To find the new angular speeds, follow the steps in the provided solution, and calculate the values for \(\omega_{final\_a}\), \(\omega_{final\_b}\), and \(\omega_{final\_c}\) using the obtained formulas.

Step-by-step solution

Recall the formula for angular momentum

The formula for angular momentum (L) is: \(L = I\omega\), where I is the moment of inertia, and ω is the angular speed.

Calculate the moment of inertia for Earth and the asteroid

To calculate the moment of inertia for Earth (spherical shape), we use the formula \(I_{Earth} = \frac{2}{5}M_ER^2\), where \(M_E\) is the mass of the Earth, and R is its radius. For the asteroid (point mass), the moment of inertia is \(I_{asteroid} = m_{asteroid}r^2\), where \(m_{asteroid}\) is the mass of the asteroid, and r is the distance from the center of Earth (which is equal to the Earth's radius). \(I_{Earth} = \frac{2}{5}(5.97 \cdot 10^{24}\text{ kg})(6,371,000\text{ m})^2\) \(I_{asteroid} = (1.00 \cdot 10^{22}\text{ kg})(6,371,000\text{ m})^2\)

Find the initial angular momenta for Earth and the asteroid

Now we will find the initial angular momenta for Earth and the asteroid using the formula \(L = I\omega\). \(L_{Earth\_initial} = I_{Earth} \times 7.272 \cdot 10^{-5} \frac{\text{rad}}{\text{s}}\) For the asteroid, we need to find its initial angular speed. In the three cases (dead center, tangentially in the direction of Earth's rotation, and tangentially opposite Earth's rotation), we can determine its initial angular speed as: a) \(\omega_{asteroid\_a} = 0\) b) \(\omega_{asteroid\_b} = \frac{1.40 \cdot 10^3 \text{ m/s}}{6,371,000 \text{ m}}\) c) \(\omega_{asteroid\_c} = -\frac{1.40 \cdot 10^3 \text{ m/s}}{6,371,000 \text{ m}}\) \(L_{asteroid\_initial} = I_{asteroid} \times \omega_{asteroid}\) (for each case)

Apply the conservation of angular momentum

For each case, use the conservation of angular momentum: \(L_{total\_initial} = L_{total\_final}\). Calculate the new angular speed of Earth after collision by equating the total initial angular momenta to the total final angular momenta and solving for the final angular speed. \(L_{total\_initial} = L_{Earth\_initial} + L_{asteroid\_initial}\) \(L_{total\_final} = I_{total} \times \omega_{final}\), where \(I_{total} = I_{Earth} + I_{asteroid}\) Equating these we get: \((L_{Earth\_initial} + L_{asteroid\_initial}) = (I_{Earth} + I_{asteroid})\times \omega_{final}\) Now, solve for \(\omega_{final}\) in all three cases: a) \(\omega_{final\_a} = \frac{L_{Earth\_initial} + L_{asteroid\_initial\_a}}{I_{total}}\) b) \(\omega_{final\_b} = \frac{L_{Earth\_initial} + L_{asteroid\_initial\_b}}{I_{total}}\) c) \(\omega_{final\_c} = \frac{L_{Earth\_initial} + L_{asteroid\_initial\_c}}{I_{total}}\) Once we have calculated the new angular speeds for each case, we will have the solution to the exercise.

Key concepts

Moment of Inertia

At the core of rotational motion is the concept of the moment of inertia, often symbolized by the letter 'I'. This property can be thought of as the rotational analogue to mass in linear motion – it quantifies an object's resistance to changes in its rotational motion. Different shapes and mass distributions around an axis of rotation have different equations for calculating the moment of inertia.

For example, for a solid sphere like Earth, the moment of inertia is calculated using the formula \( I = \frac{2}{5}MR^2 \) where 'M' represents the mass of the sphere and 'R' is the radius. It’s important to note that the further mass is distributed from an object’s axis of rotation, the larger the moment of inertia becomes, leading to a greater resistance to rotational acceleration or deceleration.

In the context of the exercise provided, understanding the distinction between Earth's moment of inertia and that of an asteroid, which is treated as a point mass (hence \( I = mr^2 \)), is crucial. This difference arises from the differing mass distributions relative to their axes of rotation.

Angular Speed

Complementing the concept of the moment of inertia is angular speed, represented by the symbol 'ω'. It measures how quickly an object rotates or revolves relative to another point, essentially describing the rate of change of the object’s angular displacement.

In simpler terms, you can think of angular speed as how fast something spins around. It is typically expressed in radians per second (rad/s). One full revolution has an angular displacement of \(2\pi\) radians, so an object rotating once every second would have an angular speed of \(2\pi\text{ rad/s}\).

In our exercise, Earth’s angular speed is initially given, and the task includes understanding how this speed changes with the impact of an asteroid. Different scenarios in the problem suggest varying initial angular speeds of the asteroid, showing how angular speed plays a role in determining the new rotational speed of the Earth post-collision.

Conservation of Angular Momentum

Finally, the law of conservation of angular momentum is pivotal in understanding the behavior of rotating systems following interactions or collisions. This principle states that if no external torque acts on a system, the total angular momentum of the system remains constant over time.

Angular momentum, given by the formula \( L = I\omega \), is the product of the moment of inertia and the angular speed. In the case where an asteroid collides with Earth as represented in our exercise, the total angular momentum before and after the impact must be the same provided that no other external forces or torques are acting on the system.

To determine the Earth’s new angular speed post-collision, we calculate the initial angular momenta of Earth and the asteroid separately for each scenario, and then apply the conservation principle to solve for the new, final angular speed. This concept bridges the gap between the individuals' moment of inertia and angular speed, showing the overarching relationship governing rotational dynamics.