College

We appreciate your visit to For a 99 kg person standing at the equator what is the magnitude of the angular momentum about Earth s center due to Earth s. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

For a 99 kg person standing at the equator, what is the magnitude of the angular momentum about Earth's center due to Earth's rotation?

Answer :

Te magnitude of the angular momentum about Earth's center due to Earth's rotation for a 99 kg person standing at the equator is approximately[tex]\( 2.894 \times 10^{11} \text{ kg} \cdot \text{m}^2/\text{s} \).[/tex]

The moment of inertia[tex]\( I \)[/tex] for a point mass at a distance [tex]\( r \)[/tex] from the axis of rotation is [tex]\( I = mr^2 \)[/tex], where [tex]\( m \)[/tex] is the mass of the object. For a person standing at the equator, [tex]\( r \)[/tex] is the radius of Earth, which is approximately[tex]\( 6.371 \times 10^6 \)[/tex] meters.

The angular velocity [tex]\( \omega \)[/tex] of Earth's rotation is [tex]\( 2\pi \)[/tex] radians per sidereal day. A sidereal day is approximately [tex]\( 23 \)[/tex] hours, [tex]\( 56 \)[/tex] minutes, and [tex]\( 4.0916 \)[/tex] seconds, or [tex]\( 86164 \)[/tex] seconds. Therefore, [tex]\( \omega = \frac{2\pi}{86164 \text{ s}} \).[/tex]

Now, let's calculate the angular momentum:

1. Calculate the moment of inertia [tex]\( I \)[/tex]:

[tex]\[ I = mr^2 = (99 \text{ kg})(6.371 \times 10^6 \text{ m})^2 \] \[ I = 99 \times 6.371^2 \times 10^{12} \text{ kg} \cdot \text{m}^2 \] \[ I \approx 99 \times 40.586 \times 10^{12} \text{ kg} \cdot \text{m}^2 \] \[ I \approx 3.972 \times 10^{15} \text{ kg} \cdot \text{m}^2 \][/tex]

2. Calculate the angular velocity[tex]\( \omega \)[/tex] :

[tex]\[ \omega = \frac{2\pi}{86164 \text{ s}} \] \[ \omega \approx \frac{2 \times 3.14159}{86164} \text{ rad/s} \] \[ \omega \approx \frac{6.28318}{86164} \text{ rad/s} \] \[ \omega \approx 7.292 \times 10^{-5} \text{ rad/s} \][/tex]

3. Calculate the angular momentum[tex]\( L \)[/tex] :

[tex]\[ L = I\omega \] \[ L = (3.972 \times 10^{15} \text{ kg} \cdot \text{m}^2)(7.292 \times 10^{-5} \text{ rad/s}) \] \[ L \approx 2.894 \times 10^{11} \text{ kg} \cdot \text{m}^2/\text{s} \][/tex]

Thanks for taking the time to read For a 99 kg person standing at the equator what is the magnitude of the angular momentum about Earth s center due to Earth s. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada

Final answer:

The angular momentum of a person standing at the equator is found using the formula L = Iω with the approximation of Earth as a uniform sphere. By utilizing the Earth's rotation rate and the radius, and knowing the mass of the person, the angular momentum can be estimated as a rough approximation.

Explanation:

To calculate the angular momentum of a 99 kg person standing at the equator due to Earth's rotation, we will utilize the formula L = Iω where I is the moment of inertia and ω is the angular velocity. Earth is approximated as a uniform spherical body, so the moment of inertia for a person standing at the equator can be estimated using I = mR2, with mass m and radius R of the Earth. The Earth completes one rotation per day, hence ω = 2π radians/day. Converting this rotational period to seconds and using Earth's radius of 6.38 × 106 meters, the angular momentum for a person can be found.

Using the provided Earth's mass and radius, and the known angular velocity converted to radians per second (ω = 2π/86400 s-1), where the day has 86400 seconds, we substitute the values into the equation for angular momentum: L = 0.4×mass×radius2×ω.

However, as a tutor on Brainly, it's important to note that this method of calculation does not account for the distribution of mass in the person's body or the extended nature of the body's geometry, so the result is a very rough approximate.