91Ó°ÊÓ

Consider a small, spherical particle of radius \(r\) located in space a distance \(R\) from the Sun. (a) Show that the ratio \(F_{\mathrm{rad}} / F_{\mathrm{grav}}\) is proportional to \(1 / r,\) where \(F_{\mathrm{rad}}\) is the force exerted by solar radiation and \(F_{\mathrm{grav}}\) is the force of gravitational attraction. (b) The result of part (a) means that, for a sufficiently small value of \(r,\) the force exerted on the particle by solar radiation exceeds the force of gravitational attraction. Calculate the value of \(r\) for which the particle is in equilibrium under the two forces. (Assume that the particle has a perfectly absorbing surface and a mass density of \(1.50 \mathrm{g} / \mathrm{cm}^{3} .\) Let the particle be located \(3.75 \times 10^{11} \mathrm{m}\) from the Sun, and use \(214 \mathrm{W} / \mathrm{m}^{2}\) as the value of the solar intensity at that point.)

Step by step solution

01

Interpret the given and required data

The spherical particle has a radius \( r \), and is located at a distance \( R \) from the sun. The goal is to (a): demonstrate the ratio \( F_{rad} / F_{grav} \) is proportional to \(1 / r\), and (b): determine the value of \( r \) where the particle experiences equilibrium between these forces. We also have a mass density (1.50 g/cm^3), the distance from the Sun (3.75 x 10^11 m), and the solar intensity (214 W/m^2) provided for the equilibrium calculation.

02

Derive the mathematical expression of forces

The force exerted by solar radiation, \( F_{rad} \), is given by \( F_{rad} = \frac{P}{c} \), where \( P \) is power and \( c \) is the speed of light. The power \( P \) can be denoted as \( P = IA \), where \( I \) is the solar intensity and \( A \) is the cross-sectional area of the sphere (\(\pi r^2\)). The gravitational force, \( F_{grav} \), is given by \( F_{grav} = \frac{GMm}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the Sun's mass, \( m \) is the particle's mass, and \( R \) is the distance from the Sun. The mass \( m \) of the particle can be noted as \( \rho V \), where \( \rho \) is the particle's mass density and \( V \) is the particle's volume (which is \( \frac{4}{3}\pi r^3 \)).

03

Show proportionality

Substitute the expressions of \( F_{rad} \) and \( F_{grav} \) into the ratio \( F_{rad}/F_{grav} \). Simplify the resultant equation and show that \(\frac{F_{rad}}{F_{grav}}\) is proportional to \(1/r\).

04

Calculate the equilibrium radius

Equilibrium between the two forces implies \( F_{rad} = F_{grav} \). By substituting and simplifying the equations as per the given values, you can solve for \( r \), the radius of the particle, which will give you the answer for the size of the particle at which the radiation pressure and the force of gravity will balance.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Solar Radiation

Solar radiation refers to the energy emitted by the Sun, which travels through space and reaches celestial bodies like Earth.
One of its components is light, which exerts pressure on any surface it encounters. This pressure is known as radiation pressure and it can have fascinating effects in space.
The force of solar radiation, often denoted as \(F_{\text{rad}}\), arises from the momentum transferred by absorbing or reflecting light. For a perfectly absorbing particle in space, this force is calculated as:

• \(F_{\text{rad}} = \frac{P}{c}\)

where \(P\) is the power of the radiation hitting the particle, and \(c\) is the speed of light. The power \(P\) depends on the solar intensity \(I\) and the cross-sectional area \(A\) of the particle:

• \(P = I \cdot A\)
• For a spherical particle, \(A = \pi r^2\)

Thus, solar radiation can exert a substantial force on particles, especially those with a large surface area compared to their mass.

Gravitational Attraction

Gravitational attraction is the force that pulls two masses toward each other, much like the force that keeps planets in orbit around the Sun. In space, this force plays a significant role in the motion of celestial objects.
For a small particle of mass \(m\) near the Sun, the gravitational force \(F_{\text{grav}}\) is described by Newton's law of universal gravitation:

• \(F_{\text{grav}} = \frac{GMm}{R^2}\)

where \(G\) is the gravitational constant, \(M\) is the mass of the Sun, and \(R\) is the distance between the Sun and the particle.
The particle's mass \(m\) can be expressed as a function of its density \(\rho\) and volume \(V\):

• \(m = \rho V = \rho \left(\frac{4}{3}\pi r^3\right)\)

This force naturally attracts the particle toward the Sun. The interaction between gravitational attraction and other forces, like radiation pressure, dictates the behavior of particles in space.

Equilibrium in Physics

Equilibrium in physics occurs when all the forces acting on an object are balanced, leading to a state where the object remains at rest or moves at a constant speed.
In the context of this problem, equilibrium is reached when the force of solar radiation \(F_{\text{rad}}\) balances the gravitational force \(F_{\text{grav}}\):

• \(F_{\text{rad}} = F_{\text{grav}}\)

By equating the expressions for \(F_{\text{rad}}\) and \(F_{\text{grav}}\) and solving for the particle's radius \(r\), we can determine the size of a particle that maintains equilibrium.
This balance is particularly crucial in understanding how small particles interact with solar radiation and gravity in space. It ultimately reveals the critical size \(r\) at which radiation pressure and gravitational pull are equal, preventing the particle from accelerating toward the Sun or being pushed away.