Answer to Question #97036 in Astronomy | Astrophysics for Sushma

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 10 ki power 6 particles /m3

Answer 1

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

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

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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:

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

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:

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

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where n=106 particles/m3 ; mH = 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: $t = 2× 10^{15} seconds$

One year contains 3× 10⁷ seconds

So we get, t=67.000.000 years