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(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model it as a particle? (b) Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles, modeling it as a uniform sphere. Consult Appendix \(\mathrm{E}\) and the astronomical data in Appendix F.

Short Answer

Expert verified
The Earth's angular momentum due to its orbit around the sun is approximately \(2.66 \times 10^{40} kg.m^2/s\), while due to its rotation around its own axis, it's about \(7.07 \times 10^{33} kg.m^2/s\). Thus, the Earth's angular momentum is dominated by its orbital motion around the sun.

Step by step solution

01

Calculate the Earth's Angular Momentum in its Orbit Around the Sun

The angular momentum \(L\) of an object moving in a circle of radius \(r\) with linear speed \(v\) is given by the equation \(L = mvr\), where \(m\) is the mass of the object. Given that the Earth's mass \(m_e\) is \(5.972 \times 10^{24} kg\), its orbital radius \(r_e\) around the sun is about \(1.496 \times 10^{11} m\), and its orbital speed \(v_e\) is about \(2.978 \times 10^{4} m/s\), we can substitute these values into the equation to find \(L\).
02

Calculate the Earth's Angular Momentum About its Own Axis

Angular momentum of a rotating sphere is given by \(L = 2/5 mR^2 \omega\), where \(m\) is the mass, \(R\) is the radius, and \(\omega\) is the angular speed. The Earth's radius \(R_e\) is \(6.371 \times 10^{6} m\), and given that it completes one rotation in 24 hours or \(8.64 \times 10^{4} s\), the angular speed \(\omega\) is \(2\pi / T \) where \(T\) is the period. Substituting these values into the equation, we can find \(L\) of Earth due to its own rotation.

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Key Concepts

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

Circular Orbit Mechanics
When studying the movement of celestial bodies, such as planets, we often encounter circular orbit mechanics. One fundamental quantity in this field of mechanics is the angular momentum, which is a measure of the rotational motion of an object in orbit. To grasp the concept, imagine a planet like Earth moving along a circular path around the Sun. Angular momentum (L) in this context is the product of the Earth's mass (me), its velocity (ve), and the radius of its orbit (re).

Using the formula for angular momentum (L = mvr), where m is the mass, v the linear speed, and r the radius, we can calculate the angular momentum of Earth in its solar orbit. This calculation forms the basis for understanding how the Earth interacts in the solar system—considering gravitational influences, orbital stability, and conservation of angular momentum.

When forming a simplified model of the Earth's orbit, treating it as a particle can be reasonable if we are only interested in its motion around the Sun and not its rotation. As a particle, we consider only its center of mass and negate other factors such as its own rotation and shape. This simplification is often made in physics to streamline calculations under specific conditions where the detail is not required.
Rotational Dynamics
Rotational dynamics is concerned with the effects of forces acting on a rotating object and how these forces change the object's motion. It forms an essential part of understanding how the Earth rotates on its axis and revolves around the Sun. In rotational dynamics, much of our interest is on quantities like torque, angular velocity, and angular momentum.

Angular Velocity

Angular velocity (\(omega\)) is the rate at which an object rotates or revolves around an axis or point. It's measured in radians per second (rad/s), where one radian is the angle made at the center of a circle by an arc whose length is equal to the radius of the circle. Earth completes one rotation every 24 hours, giving us a way to calculate its angular velocity as \(omega = \frac{2\tri}{T}\), with T being the period of rotation.

Conservation of Angular Momentum

An essential principle in this field is the conservation of angular momentum, which states that if no external torque acts on a system, the total angular momentum of the system remains constant. This principle is why Earth continues to spin at a nearly constant rate and maintain its orbit around the Sun.
Uniform Spherical Body Rotation
When modeling the Earth as a uniform spherical body, we assume it has a symmetrical mass distribution about its axis. This simplified model aids in comprehending the Earth's rotation dynamics. The moment of inertia (I) is a key term in this context, representing the measure of an object's resistance to changes in its rotational motion.

For a solid sphere, the moment of inertia is given by \(I = \frac{2}{5} mR^2\), where m is the mass and R is the radius. Knowing the Earth's moment of inertia allows us to calculate its angular momentum for its spin around the axis through the north and south poles using the formula \(L = Iomega\). Here, \(omega\) is the angular velocity.

By integrating the formula for a uniform spherical body's rotation with that of a body in circular motion, a comprehensive understanding of the Earth's movement can be achieved. Whether it is the diurnal rotation around its axis or its annual orbit around the Sun, these principles not only explain the dynamics of Earth but also other astronomical bodies exhibiting similar motion.

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Most popular questions from this chapter

The Atwood's Machine. Figure P10.63 illustrates an Atwood's machine. Find the linear accelerations of blocks \(A\) and \(B,\) the angular acceleration of the wheel \(C,\) and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks \(A\) and \(B\) be 4.00 \(\mathrm{kg}\) and \(2.00 \mathrm{~kg},\) respectively, the moment of inertia of the wheel about its axis be \(0.220 \mathrm{~kg} \cdot \mathrm{m}^{2}\), and the radius of the wheel be \(0.120 \mathrm{~m}\).

An engine delivers 175 hp to an aircraft propeller at 2400 rev \(/\) min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

A wheel with radius \(0.0600 \mathrm{~m}\) rotates about a horizontal frictionless axle at its center. The moment of inertia of the wheel about the axle is \(2.50 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The wheel is initially at rest. Then at \(t=0 \mathrm{a}\) force \(F=(5.00 \mathrm{~N} / \mathrm{s}) t\) is applied tangentially to the wheel and the wheel starts to rotate. What is the magnitude of the force at the instant when the wheel has turned through 8.00 revolutions?

A \(55 \mathrm{~kg}\) runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude \(2.8 \mathrm{~m} / \mathrm{s}\). The turntable is rotating in the opposite direction with an angular velocity of magnitude \(0.20 \mathrm{rad} / \mathrm{s}\) relative to the earth. The radius of the turntable is \(3.0 \mathrm{~m},\) and its moment of inertia about the axis of rotation is \(80 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.

One force acting on a machine part is \(\overrightarrow{\boldsymbol{F}}=(-5.00 \mathrm{~N}) \hat{\imath}+\) \((4.00 \mathrm{~N}) \hat{\jmath} .\) The vector from the origin to the point where the force is applied is \(\vec{r}=(-0.450 \mathrm{~m}) \hat{\imath}+(0.150 \mathrm{~m}) \hat{\jmath}\). (a) In a sketch, show \(\vec{r}, \vec{F}\), and the origin. (b) Use the right-hand rule to determine the direction of the torque. (c) Calculate the vector torque for an axis at the origin produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

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