Question

A projectile is a distance r from the center of a heavenly body and is heading directly away. Classically, if the sum of its kinetic and potential energies is positive, it will escape the gravitational pull of the body, but if negative, it cannot escape. Now imagine that the projectile is a pulse of light energy E. Since light has no internal energy ,E is also the kinetic energy of the light pulse. Suppose that the gravitational potential energy of the light pulse is given by Newton’s classical formula U=-(GMm/r), where M is the mass of the heavenly body and m is an “effective mass” of the light pulse. Assume that this effective mass is given by $\mathbf{m}\mathbf{=}\mathbf{E}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}$.

Show that the critical radius for which light could not escape the gravitational pull of a heavenly body is within a factor of 2 of the Schwarzschild radius given in the chapter. (This kind of “semiclassical” approach to general relativity is sometimes useful but always vague. To be reliable, predictions must be based from beginning to end on the logical, but unfortunately complex, fundamental equations of general relativity.)

Short answer

The critical radius for which the light could not escape the gravitational pull of a heavenly body is within a factor of 2 of the Schwarzschild radius, is proved.

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The distance of the particle from the heavenly body is r.
• The energy of the projectile is E.
• The mass of the heavenly body is M.
• The effective mass of the light pulse is m.

Significance of the critical radius

The critical radius is mainly described as the lowest radius which an atom form. It is also described as the minimum size of a particle.

Determination of the critical radius

As the total energy of the particle is zero, then the equation of the total energy is expressed as:

KE + PE = 0

Here, KE is the kinetic and PE is the potential energy.

Substitute E for KE and$-\frac{GMm}{r}$for PE in the above equation.

$E-\frac{GMm}{r}=0$

Here, E is the energy of the particle, G is the gravitational constant, M is the mass of the heavenly body, m is the effective mass of the light pulse and r is the distance of the particle from the heavenly body.

Substitute $\frac{E}{c^2}$for in the above equation.

$$\begin{gathered} E-\frac{GME}{r{c}^2}=0 \\ E\left(1-\frac{GM}{r{c}^2}\right)=0 \end{gathered}$$

The next equation can be expressed as:

$$\begin{gathered} 1-\frac{GM}{r{c}^2}=0 \\ 1=\frac{GM}{r{c}^2} \\ r=\frac{GM}{c^2} \end{gathered}$$

The equation of the Schawzschild radius is expressed as:

$r_s=\frac{2GM}{c^2}$

Here, $r_s$is the Schawzschild radius.

Here, in the case of the Schawzschild radius, it can be identified that the critical radius is lesser than the factor of 2 than the Schawzschild radius.

Light can only escape the gravitational pull if the addition of the kinetic and the potential energy is greater than zero but, in this case, the addition is equal to zero.

Thus, the critical radius for which the light could not escape the gravitational pull of a heavenly body is within a factor of 2 of the Schwarzschild radius, is proved.