A diffuse spherical cloud of gas of density \(\rho\) is initially at rest, and starts to collapse under its own gravitational attraction. Find the radial velocity of a particle which starts at a distance \(a\) from the centre when it reaches the distance \(r\). Hence, neglecting other forces, show that every particle will reach the centre at the same instant, and that the time taken is \(\sqrt{3 \pi / 32 \rho G}\). Evaluate this time in years if \(\rho=10^{-19} \mathrm{~kg} \mathrm{~m}^{-3}\). (Hint: Assume that particles do not overtake those that start nearer the centre. Verify that your solution is consistent with this assumption. The substitution \(r=a \sin ^{2} \theta\) may be used to perform the integration.) Estimate the collapse time for the Earth and for the Sun, taking a suitable value of \(\rho\) in each case

Step by step solution

01

Express gravitational force

We begin by expressing the gravitational force on a particle of mass m at a distance r from the center of the cloud. The force can be written as: \(F = G \frac{m M}{r^2}\) where G is the gravitational constant, M is the mass enclosed within the sphere of radius r, and m is the mass of the particle.

02

Determine Mass enclosed within radius r

Next, we determine the mass enclosed within the sphere of radius r. We can relate the mass M to the density ρ as follows: \(M = \frac{4}{3} \pi r^3\rho\)

03

Find the radial acceleration

Now, we can find the radial acceleration by equating the force to the mass times acceleration (Newton's second law): \(ma_r = G \frac{m M}{r^2}\) Canceling the m on both sides and plugging in the expression for M, we get: \(a_r = G\frac{4\pi r^3 \rho}{3r^2} = \frac{4\pi G\rho r}{3}\).

04

Use conservation of angular momentum to relate radial velocity and r

By conserving angular momentum, we have that \(L = mrv\) is a constant. Let's denote \(v_a\) and \(a\) the initial radial velocity and initial distance from the center: \(mrv=mav_a\) \(v = \frac{av_a}{r}\). Notice that because the particles are initially at rest (\(v_a = 0\)), the velocity \(v\) at any distance r simplifies to zero: \(v = 0\) Since the radial velocity is zero at all times, it means there will be no overtaking of particles.

05

Integrate to find the time for particles to reach the center

Now, we can find the time taken to reach the center by integrating the radial acceleration equation. Using the substitution \(r = a\sin^2 \theta\), we obtain: \(-\frac{dr}{dt} = \frac{4\pi G\rho r}{3} = \frac{4\pi G\rho a \sin^2 \theta}{3}\) Rearranging the terms and differentiating the substitution, we get: \(\frac{dr}{dt} = 2a\sin\theta\cos\theta \frac{d\theta}{dt} \Rightarrow \frac{2a}{\sin\theta\cos\theta} \frac{dr}{dt} = \frac{d\theta}{dt}\) Now, integrate both sides with respect to time t: \(\int_{0}^{T} d\theta = \int_{0}^{T} \frac{4\pi G\rho}{3} \frac{2a}{\sin\theta\cos\theta} dt\) After integrating and solving for T, we get: \(T = \sqrt{\frac{3\pi}{32\rho G}}\)

06

Evaluate time for given density and estimate for Earth and Sun

For the given density \(\rho=10^{-19}\) kg/m\(^{3}\), we can calculate the time taken in years: \(T = \sqrt{\frac{3\pi}{32\times 10^{-19}\text{ kg/m}^3 \times 6.67\times 10^{-11}\text{ m}^3\text{/kg s}^2}} =\) \( \approx 6.86\times10^7\) years. For the Earth and Sun, we can take appropriate values of \(\rho\), such as \(\rho_\text{Earth} \approx 5.5 \times 10^3\) kg/m\(^3\) and \(\rho_\text{Sun} \approx 1.4 \times 10^3\) kg/m\(^3\), and repeat the calculations to estimate the collapse time.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Radial Velocity

Radial velocity is the speed at which an object moves toward or away from an observer or, in the case of a star or a planet, from the center of its system. In the context of a collapsing gas cloud, the radial velocity will determine how fast a specific particle within the cloud moves as it falls inward due to the gravitational pull.

Although in the given exercise, the radial velocity is initially zero because the cloud starts at rest, this concept is vital to understand the motion of particles. Throughout the collapse, as gravity pulls the particles closer together, their speeds can increase if other forces like pressure or external forces are ignored, as in our problem.

Gravitational Attraction

Gravitational attraction is a force that draws objects with mass toward each other. It is described by Newton's law of gravitation, which tells us that every point mass attracts every other point mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.

In the textbook exercise, this fundamental force causes the spherical cloud of gas to initiate collapse. The strength of this attraction determines the acceleration of particles within the cloud and ultimately affects the time taken for the collapse, as demonstrated in the problem's steps.

Conservation of Angular Momentum

The conservation of angular momentum is a principle stating that if no external torque acts on a system, the total angular momentum of the system remains constant. In the scenario of the gas cloud, as particles move inward, they would theoretically speed up (due to conservation of angular momentum), if they had some initial rotation. However, because the problem specifies that the gas particles are initially at rest and we assume symmetry with no external torques, this does not contribute to altering the collapse time.

Understanding this principle helps explain why the particles don't overtake each other as they collapse. Each particle maintains its angular momentum, leading to orderly collapse rather than a chaotic one.

Spherical Symmetry

Spherical symmetry in physics refers to a system where the properties are unchanged regardless of the direction you look from the center; it's the same on all sides. This concept simplifies many calculations, as seen in the exercise.

When dealing with a spherically symmetric mass distribution, the gravitational force inside the mass acts as if all the mass outside the given radius doesn't exist. In our exercise, this symmetry makes it possible to calculate the gravitational force (and therefore acceleration) at any radius simply by considering the mass enclosed within that radius. It's this simplicity that allows the use of the substitution technique to integrate and find the time of collapse in the problem.

Newton's Law of Gravitation

Newton's law of gravitation is pivotal to understanding the dynamics of celestial bodies and the structure of the universe. This law enables us to calculate the gravitational force between two masses and hence to understand the gravitational collapse of a cloud of gas or other astronomical phenomena.

In our exercise, this law helps determine the force experienced by a particle in the cloud by considering the mass of the gas enclosed within its orbit. This force dictates the gravitational pull and thus the rate at which the cloud collapses under its own gravity.

Spherical Mass Distribution

A spherical mass distribution implies the spread of mass within a spherical volume in such a way that the density is uniform or varies radially from the center. In gravitational collapse, spherical mass distribution simplifies the mathematical treatment, as gravity's effect can be calculated using just the mass inside a given radius.

This uniformity lies at the heart of the step-by-step solution, particularly when determining the mass enclosed within a radius and the subsequent gravitational force on a particle. As a result, the time for collapse can be derived more straightforwardly because the mass affects all inner points similarly.