Answer to Question #87720 in Astronomy | Astrophysics for Shivam Nishad

Question

Write down the condition under which a large molecular cloud collapses to give rise
to new stars. Calculate the time of free fall of an HI cloud of density 106
particles/m

Answer 1

For collapse of a cloud with radius R to occur, need either

$$M_{cloud}>M_{Jeans}=\frac{3kTR}{2G \mu m_H}$$

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:

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

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:

$M_{cloud}>M_{Jeans}=\frac{3kTR}{2G \mu m_H}$

67.000.000 years