A particle in the solar system is under the combined...

A particle in the solar system is 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.

A particle in the solar system is 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.

Key Concepts

Gravitational Attraction

This is the force that acts between any two masses according to Newton's law of universal gravitation. In the context of the solar system, the Sun’s gravity attracts particles, keeping them in orbit. The gravitational force depends inversely on the square of the distance from the Sun and directly on the mass of the objects involved.

Radiation Pressure

Radiation pressure arises from the momentum transferred by photons when they strike and are absorbed or scattered by a surface. For a particle that absorbs all incident light, this effect results in a force pushing the particle away from the source of the light (the Sun). It is proportional to the energy flux and the particle's cross-sectional area exposed to the radiation.

Radiation Force on Spherical Particles

When light is absorbed by a spherical particle, the radiation force acting on the particle is determined by the product of the radiation pressure and the particle's cross-sectional area. This force counteracts gravitational attraction, and its magnitude depends on both the intensity of the incident light and the size (area) of the particle.

Force Equilibrium and Critical Radius

The critical radius is derived by setting the outward radiation force equal to the inward gravitational force. This equilibrium condition marks the threshold where the net force on the particle changes sign. For particles with radii below this critical radius, the radiation force dominates, leading to the particle being blown out of the solar system.

Mass-to-Area Ratio

The mass-to-area ratio of a particle is crucial in determining how it responds to external forces. A lower mass-to-area ratio means that a given force (like radiation pressure) will cause a greater acceleration compared to gravity. This ratio, dependent on the particle's density and size, is critical for understanding whether radiation pressure or gravitational attraction will dominate its dynamics.

Transcript

00:01 In this question, we have a particle in a solar system.

00:04 It is under the influence of both the sun's gravitational attraction force and the radiation force, which comes as a result of the rays of the sun.

00:12 We're told that the particle is a sphere and it has a density of 1 times 10 to 3 kilogram per meters cubed and that all of the light that is incident is absorbed.

00:21 What we want to do in this question is show that there is a critical radius capital r under which the particle will be blown out of the solar system and calculate the value of this critical radius.

00:32 So what we first need to do is look at the expression for the volume and the mass.

00:37 We can use this to create an expression for the force due to gravity.

00:42 So what we can say is the force due to gravity is equal to the normal expression, so the gravitational constant multiplied by the two masses, divided by the separation between them squared, and what we can do is plug in this expression for m to create a further expression in terms of, with this small m substituted in.

01:06 So this gives us 4 pi g, gm, row r cubed, all over 3r squared.

01:28 Now we have done this, we can go and look at the power output of the sun.

01:32 We know from this expression that the radiation intensity is given by the power over the power, over the area, which is also equal to the power over 4 pi r squared.

01:44 And since the particle absorbs all of the energy, the radiation pressure is then equal to the intensity over the speed of light, which in terms of subbing in this expression for i, is equal to p p over 4 pi r squared c.

02:16 Now that we've done this, we know that all of the radiation that passes through a circle which has an area of pi r squared is perpendicular to the direction of the propagation, and as a result is absorbed by the particle...

Please give Ace some feedback