Make a plot of rS = 2GM as a function of R Here M is the mean total mass of dark matter enclosed in a sphere of R Solve the equation rS = R for R Compare it to cH0

To make a plot of rS as a function of R, we first need to understand the equation rS = 2GM. This equation represents the Schwarzschild radius, which is the distance from the center of a mass at which the escape velocity equals the speed of light. In other words, any object or particle that passes within this radius will be trapped by the gravitational pull and cannot escape.

In this case, M represents the mean total mass of dark matter enclosed in a sphere of radius R. We can assume that the dark matter distribution is spherically symmetric, so we can use the equation for the mass enclosed within a sphere:

M = (4/3)πR^3ρ

where ρ is the density of dark matter. We can then substitute this expression for M into the equation for rS:

rS = 2G(4/3)πR^3ρ

Simplifying this expression, we get:

rS = (8/3)πGρR^3

This is our equation for rS as a function of R. To plot this equation, we need to know the density of dark matter as a function of radius. This is not well-known, but we can assume that it follows a power-law distribution:

ρ = ρ0(R/R0)^-α

where ρ0 is the density at some reference radius R0 and α is the slope of the power law. We can then substitute this expression for ρ into our equation for rS:

rS = (8/3)πGρ0(R/R0)^(3-α)

This is the equation we can use to make our plot. We can see that rS increases as R increases, but the slope of the curve depends on the density profile of the dark matter halo.

To compare this to c/H0, we need to solve the equation rS = R for R. Substituting our expression for rS, we get:

(8/3)πGρ0(R/R0)^(3-α) = R

Simplifying this expression, we can rewrite it as:

R^(α+1) = (8/3)πGρ0R0^(α-2)

Taking the logarithm of both sides, we get:

(α+1)log(R) = log[(8/3)πGρ0] + (α-2)log(R0)

Solving for R, we get:

R = R0[(8/3)πGρ0 / R0^(α-2)]^(1/(α+1))

This is the equation we can use to compare to c/H0. We can see that R depends on the density profile of the dark matter halo and the value of α. If we assume a value of α = 1, which corresponds to a constant density profile, we get:

R = R0(3/4πGρ0)^(1/2)

This expression is independent of radius and only depends on the density of dark matter. We can compare this to c/H0, which represents the Hubble distance, the distance at which two objects are receding from each other at the speed of light due to the expansion of the universe. Using the current value of the Hubble constant, H0 = 67.4 km/s/Mpc, we get:

c/H0 = 4.55 Gpc

This is a much larger distance than the typical size of a dark matter halo, which is on the order of a few hundred kpc. Therefore, we can conclude that the Schwarzschild radius is much smaller than the Hubble distance and is not directly related to the expansion of the universe.