Question

(a) A particle is dropped (from radius a with zero velocity) into the gravitational potential corresponding to a static homogeneous sphere of radius a and density ρ. Calculate how long the particle takes to reach the other side of the sphere. [Hint: the equation of motion is d²r/dt² = −GM(r)/r² .]

(b) Calculate the time required for a homogeneous sphere of radius a and density ρ with no internal pressure support to collapse to zero radius under its own gravity. [Apply the previous equation of motion to a particle on the surface.]

Answer 1

1. We have a static homogenous sphere of radius and uniform density . A particle is dropped from the surface of the sphere () with initial velocity, . We assume that the particle moves through the diameter of the sphere. Now, the problem has spherical symmetry as the potential depends only on the radial coordinate . The gravitational potential energy of the mass at a radial distance from the centre of the sphere is given by

Here, is the mass of the sphere distributed within an imaginary sphere of radius centred about the origin. It is given by

where is the volume of the imaginary sphere. Now, since the gravitational force is conservative, the force acting on the particle at a radial distance is given by

The negative sign shows that the force is always radially directed towards the origin (which means the force is attractive). Now, substituting the expression for from equation (2) into (3), we get the force as

From Newton's second law, the force is equal to mass times acceleration. Thus, the equation of motion is

or

The negative sign shows that the force is directed towards the origin while the radial vector is pointed away from the origin. You can see that the acceleration depends on the radial coordinate . Equation (5) becomes familiar in the following form:

with

Equation (6) represents the differential equation for simple harmonic motion with angular frequency

You can see that the particle executes simple harmonic motion in the region . It is interesting to note that the frequency of simple harmonic oscillation does not depend on the mass of the particle.

The time period (which is the time required for the particle to reach back to the starting point ) is given by

The time taken for the particle to reach the other side of the sphere is half the time taken for one complete oscillation (that is half the time period):

which is the required answer.

2. The gravitational force is always directed towards the origin. If there is no internal pressure there is nothing to balance the inward gravitational pull. This causes the sphere to collapse by its own gravitational field. The situation is no more static. However, we can analyze the condition as follows. Consider a particle on the surface of the sphere of raidus at time . Then, the sphere starts collapsing. That is the radius of the sphere decreases. However, the total mass is constant. This means that the density of the sphere increases as the sphere collapses. We need to set up the equation of motion of a particle on the surface of this collapsing sphere. At any instant (and at any radius r), the force acting on the sphere on the surface is

Newton's second law yields

Now, the total mass of the sphere at any point of its collapse is a constant. However, the density increases with decreasing as follows:

where is the volume of the collapsing sphere at any instant.

We can rewrite equation (13) in terms of the initial density and the initial radius as follows:

Now, the mass of the sphere can be written as

Using (15) in (12), we get

Now, the velocity of the particle on the surface of the sphere (which is same as the velocity of collapse of the sphere) is given by

Now, using the chain rule, we can write

Substituting (18) in (16), we get

The differential equation (19) is separable. Thus, we integrate (19) to solve in terms of :

Now after integration, we get

To find the constant of integration, we set the initial condition that for . Thus,

Using (22) in (21), we get

Thus, the velocity at any instant is given by

Now, from (17), the time taken for collapse can be found as follows:

Now, changing the variable of integration from to , we get

Now, we can make a trigonometric substitution and runs from to :

which is the required answer. You can see that the time for the collapse depends only on the initial density.

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