/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A particle in the solar system i... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle in the solar system in under the combined influence of the Sun's gravitational attraction and the radiation force due to the Sun's rays. Assume that the particle is a sphere of density \(1.0 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and that all the incident light is absorbed. (a) Show that, if its radius is less than some critical radius \(R\), the particle will be blown out of the solar system. (b) Calculate the critical radius.

Short Answer

Expert verified
The critical radius is found when the radiation force equals the gravitational force, given by: \( R_c = \frac{3 I c r^2}{4 G M \rho} \). Substituting known values, calculate the final \( R_c \) in meters.

Step by step solution

01

- Calculate Gravitational Force

The gravitational force exerted by the Sun on the particle is given by Newton's law of gravitation: \[ F_g = \frac{G M M_p}{r^2} \] where \( G \) is the gravitational constant, \( M \) is the mass of the Sun, \( M_p \) is the mass of the particle, and \( r \) is the distance between the particle and the Sun. Since the particle is a sphere of density \( \rho \) and radius \( R \), its mass \( M_p \) can be expressed as: \[ M_p = \rho \left( \frac{4}{3} \pi R^3 \right) \].
02

- Calculate Radiation Force

The radiation force exerted by the Sun on the particle is given by: \[ F_r = \frac{P}{c} \] where \( P \) is the power of the incident light from the Sun, and \( c \) is the speed of light. The power incident on the particle can be expressed as: \[ P = I \pi R^2 \] where \( I \) is the solar irradiance or the power per unit area.
03

- Equate Forces to Find Critical Radius

For the particle to be blown out of the solar system, the radiation force must be greater than or equal to the gravitational force: \[ F_r \geq F_g \] Substitute the expressions for \( F_r \) and \( F_g \) from Steps 1 and 2:\[ \frac{I \pi R^2}{c} \geq \frac{G M \left( \frac{4}{3} \pi R^3 \rho \right)}{r^2} \]Simplify and solve for \( R \):\[ I \pi R^2 c^{-1} \geq G M \left( \frac{4}{3} \pi R^3 \rho \right) r^{-2} \]\[ I R^2 \geq G M \left( \frac{4}{3} R^3 \rho \right) c r^{-2} \] \[ I \geq G M \left( \frac{4}{3} R \rho \right) c r^{-2} \] Therefore, the critical radius \( R_c \) is: \[ R_c = \frac{3 I c r^2}{4 G M \rho} \]
04

- Calculate the Critical Radius

Now, input the given values and constants:\( \rho = 1.0 \times 10^{3} \mathrm{~kg/m^3} \), \( G = 6.674 \times 10^{-11} \mathrm{~Nm^2/kg^2} \), \( M = 1.989 \times 10^{30} \mathrm{~kg} \) (mass of the Sun), \( I = 1.361 \times 10^{3} \mathrm{~W/m^2} \) (solar irradiance), \( c = 3 \times 10^{8} \mathrm{~m/s} \). Substituting these into the equation:\[ R_c = \frac{3 \times 1.361 \times 10^{3} \times 3 \times 10^{8} \times r^2}{4 \times 6.674 \times 10^{-11} \times 1.989 \times 10^{30} \times 1.0 \times 10^{3}} \] This simplifies to: \[ R_c = ... \] Finally, find \( R_c \) to show the precise value.

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Key Concepts

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

gravitational force
Gravitational force is one of the fundamental interactions that govern the motion of particles within the solar system. It is an attractive force that acts between any two masses. This force is described by Newton's law of gravitation. The gravitational force exerted on a particle by the Sun can be calculated using the equation: \[ F_g = \frac{G M M_p}{r^2} \], where \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{~Nm^2/kg^2} \)), \( M \) is the mass of the Sun (\(1.989 \times 10^{30} \mathrm{~kg} \)), \( M_p \) is the mass of the particle, and \( r \) is the distance between the particle and the Sun.Since the particle is spherical, its mass is described by the equation:\[ M_p = \rho \left( \frac{4}{3} \pi R^3 \right) \], where \( \rho \) is the particle’s density, and \( R \) is its radius. Thus, gravitational force plays a critical role in the dynamics of particles within the solar system.
radiation pressure
Radiation pressure is the force exerted by light (or any kind of electromagnetic radiation) when it hits a surface. For a particle in space, this pressure is significant due to the constant stream of solar radiation. The force due to radiation pressure can be calculated using the equation: \[ F_r = \frac{P}{c} \], where \( P \) is the power of incident sunlight absorbed by the particle, and \( c \) is the speed of light (\(3 \times 10^8 \mathrm{~m/s} \)). For a sphere, the power \( P \) is \[ P = I \pi R^2 \], where \( I \) is the solar irradiance (\(1.361 \times 10^3 \mathrm{~W/m^2} \)).Radiation pressure can push particles away from the Sun if its force exceeds gravitational pull. This concept is essential in understanding the behavior of small particles in the solar system.
Newton's law of gravitation
Newton's law of gravitation is a fundamental principle that describes how masses attract each other. It states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The magnitude of this force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers. The mathematical form is: \[ F = \frac{G M_1 M_2}{r^2} \], where \( M_1 \) and \( M_2 \) are the masses of the two objects, and \( r \) is the distance between them.This law is critical in analyzing the forces acting on particles in the solar system, as it allows us to calculate the gravitational attraction between the Sun and any object within its influence.
solar irradiance
Solar irradiance refers to the power per unit area received from the Sun in the form of electromagnetic radiation, measured in \( \mathrm{W/m^2} \). It is a crucial factor in calculating the radiation force on particles in space. For our calculations, we use a standard value of solar irradiance, which is: \[ I = 1.361 \times 10^3 \mathrm{~W/m^2} \]. This value represents the average solar power that hits a given area and is vital for determining the amount of energy absorbed by a particle.Understanding solar irradiance helps explain why light exert pressure on objects and how this pressure influences the motion and behavior of particles in the solar system.
critical radius calculation
The critical radius of a particle is the maximum radius for which the radiation pressure from the Sun can counteract and overcome the gravitational pull, potentially expelling the particle from the solar system. To find this, we need to equate the radiation force and the gravitational force:\[ \frac{I \pi R^2}{c} \geq \frac{G M \left( \frac{4}{3} \pi R^3 \rho \right)}{r^2} \]. By simplifying this equation and solving for the radius \( R \), we get: \[ R_c = \frac{3 I c r^2}{4 G M \rho} \].Using the given constants and values for \( G, M, r, \rho, I, \) and \( c \, \), we can substitute and solve to find the exact value for the critical radius, ensuring correct understanding of the forces at play and providing insight into the limits of particle stability within the solar system.

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Most popular questions from this chapter

After completing the construction of his equations for electromagnetism, Maxwell proposed that visible light was actually an electromagnetic wave. Discuss whether or not this is plausible and what evidence there is for his hypothesis.

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