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Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth \(=6400 \mathrm{~km}\) and radius of the orbit of the earth about the sun \(=1 \cdot 5 \times 10^{8} \mathrm{~km}\).

Short Answer

Expert verified
The ratio of Earth's spinning angular momentum to its orbital angular momentum is extremely small, favoring orbital motion, approximately \(2.5 \times 10^{-5}\).

Step by step solution

01

Determine the Angular Momentum of Earth due to Spinning

The angular momentum due to spinning is calculated using the formula:\[ L_s = I_s \cdot \omega_s \]where \( L_s \) is the angular momentum due to spinning, \( I_s \) is the moment of inertia for a solid sphere, and \( \omega_s \) is the angular velocity.For a solid sphere, \( I_s = \frac{2}{5}mR^2 \), where \( m \) is Earth's mass and \( R \) is the Earth's radius.Assuming Earth's mass \( m = 6 \times 10^{24} \text{ kg} \) and converting Earth's radius to meters \( R = 6400 \times 10^3 \text{ m} \), we have:The Earth spins once a day, therefore,\[ \omega_s = \frac{2\pi}{86400} \text{ rad/s} \]Substitute the values to find:\[ L_s = \left(\frac{2}{5} \times 6 \times 10^{24} \times (6400 \times 10^3)^2\right) \times \frac{2\pi}{86400} \]
02

Calculate the Angular Momentum of Earth due to Orbital Motion

The angular momentum due to orbital motion is calculated using the formula:\[ L_o = mvr \]where \( L_o \) is the angular momentum due to orbital motion, and \( v \) is the orbital velocity.The orbital velocity can be calculated as:\[ v = \frac{2\pi r}{T} \]where \( r = 1.5 \times 10^8 \times 10^3 \text{ m} \) is the radius of the Earth's orbit around the sun, and \( T \) is one year, roughly \( 365.25 \times 86400 \text{ seconds} \).The angular momentum is:\[ L_o = m \cdot \left(\frac{2\pi \times 1.5 \times 10^8 \times 10^3}{365.25 \times 86400}\right) \cdot 1.5 \times 10^8 \times 10^3 \]
03

Calculate the Ratio of Angular Momenta

The ratio of the angular momentum for spinning to orbital motion is given by:\[ \text{Ratio} = \frac{L_s}{L_o} \]Substitute the values for \( L_s \) from Step 1 and \( L_o \) from Step 2 and simplify the expression to calculate the ratio.

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Key Concepts

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

Spinning Motion
Spinning motion refers to an object rotating around its own axis. Picture the Earth rotating around an imaginary line that stretches from the North Pole to the South Pole. This is Earth's spinning motion and it happens once every 24 hours, giving us day and night.
It's important because it contributes to Earth's angular momentum, a measure of Earth's rotational motion.

For any spinning object, the angular momentum (\(L_s\)) is calculated by multiplying the moment of inertia (\(I_s\)) by the angular velocity (\(\omega_s\)).
  • The moment of inertia indicates how mass is distributed around the rotation axis.
  • Angular velocity is the rate of change of rotational angle per unit of time.
For example, Earth is approximated as a solid sphere. The moment of inertia for a sphere is \(\frac{2}{5}mR^2\), where \(m\) is mass and \(R\) is radius. Earth’s mass and radius can be applied to this formula to calculate its spinning angular momentum.
Orbital Motion
Orbital motion occurs when an object circles around another in a path called an orbit. Think of Earth moving around the Sun – that’s orbital motion, completing one cycle approximately every 365 days.
This motion is key to understanding seasons and the passage of years.

The angular momentum of Earth due to its orbital motion (\(L_o\)) is derived from the product of its mass, velocity, and the radius of its orbit.
This formula is represented as:\[L_o = m \cdot v \cdot r\]where:
  • \(m\) is the Earth’s mass.
  • \(v\) is the orbital velocity, calculated as \(\frac{2\pi r}{T}\), where \(T\) is the Earth's orbital period around the Sun.
  • \(r\) is the distance from the Earth to the Sun.
By knowing these quantities, one can determine Earth's orbital angular momentum, showing how its mass and speed around the solar center impact this measure.
Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics, representing the resistance of an object to changes in its rotation.
In a way, it’s the rotational equivalent of mass in linear motion. Consider a compact disk spinning – different weight distributions change how it spins.

The moment of inertia depends on:
  • The object's shape and mass distribution relative to the axis of rotation.
  • How far the mass is spread from the axis, with greater distances resulting in larger moments of inertia.
For spheres such as Earth, \(I\) can be calculated using \(\frac{2}{5}mR^2\). Here \(m\) is the mass and \(R\) is the radius.
Understanding this concept clarifies why Earth's spinning and orbital motions influence different angular momenta differently. The distribution of mass in every rotating object determines how much energy is needed to change its velocity, i.e., how swiftly or slowly it spins.

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Most popular questions from this chapter

Three particles, each of mass \(200 \mathrm{~g}\), are kept at the corners of an equilateral triangle of side \(10 \mathrm{~cm}\). Find the moment of inertia of the system about an axis (a) joining two of the particles and (b) passing through one of the particles and perpendicular to the plane of the particles.

A body rotating at \(20 \mathrm{rad} / \mathrm{s}\) is acted upon by a constant torque providing it a deceleration of \(2 \mathrm{rad} / \mathrm{s}^{2} .\) At what time will the body have kinetic energy same as the initial value if the torque continues to act?

Figure (10-E4) shows two blocks of masses \(m\) and \(M\) connected by a string passing over a pulley. The horizontal table over which the mass \(m\) slides is smooth. The pulley has a radius \(r\) and moment of inertia \(I\) about its axis and it can freely rotate about this axis. Find the acceleration of the mass \(M\) assuming that the string does not slip on the pulley.

A boy is seated in a revolving chair revolving at an angular speed of 120 revolutions per minute. Two heavy balls form part of the revolving system and the boy can pull the balls closer to himself or may push them apart. If by pulling the balls closer, the boy decreases the moment of inertia of the system from \(6 \mathrm{~kg}-\mathrm{m}^{2}\) to \(2 \mathrm{~kg} \cdot \mathrm{m}^{2}\) what will be the new angular speed?

A wheel of mass \(10 \mathrm{~kg}\) and radius \(20 \mathrm{~cm}\) is rotating at an angular speed of 100 rev/min when the motor is turned off. Neglecting the friction at the axle, calculate the force that must be applied tangentially to the wheel to bring it to rest in 10 revolutions.

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