Question

Calculate the time of free fall of an HI cloud of density 10^6
particles/m3
.

Answer 1

The acceleration g felt by a test particle for a spherically symmetric distribution of mass M and radius r.

From Newton's second law, the equation of motion for a test particle at the edge of the cloud is then

$$m \times g=G \times m \frac{M}{r^2} (1)$$

The mass M is equal to:

$$M= \frac{4 \pi}{3}\times r^3 \times ρ (2)$$

where ρ is the cloud density

We put (2) in (1):

$$g=\frac{4 \pi}{3}G \times r \times \rho (3)$$

If it starts initially at rest, then (if acceleration is constant) it will reach the center when

$$\frac{ g \times t^2}{2}=r (4)$$

We put (3) in (4) and solve for t:

$$t=\sqrt \frac {3}{2\times \pi \times G \times \rho} (5)$$

The cloud density ρ is equal to:

$$\rho=n \times mH (6)$$

where n=10⁶ particles/m³ ; m_H = 1.67 × 10^-27 kg

Using (6) we calculate the value of cloud density ρ: ρ= 1.67 × 10^-21 kg/m³

We put the value of cloud density ρ in (5) and get:

t=2× 10¹⁵ seconds

One year contains 3× 10⁷ seconds

In this case t=67.000.000 years

Answer: 2× 10¹⁵ seconds or 67.000.000 years