Stars & galaxies, weekly problems

Problem 1.

M_{J} = (\frac{3 × 1.38 × 10^{−23} × 100}{2 × 1.67 × 10^{−27} × 6.67 × 10^{−11}} )^{3/2} (\frac{3}{4π × 1.67 × 10^{−27} × 10^{6}} )^{1/2}
× (n/cm^{3})^{−1/2}(T/100K)^{3/2}kg
= 1.5 × 10^{4}M_{⊙} (n/cm^{3})^{−1/2}(T/100K)^{3/2} .  (1)

After the shock, the mass of the cloud is twice the mass of the original cloud, M = 2×10^{4}M_{⊙}. The new Jeans mass is M_{J} = 1.5×10^{4}M_{⊙} (4)^{−1/2}(T′/100K)^{3/2}, where T′ is the temperature after the shock. Hence M = M_{J} for (T′/100K)^{3/2} = 8/3 or T′ = 192K.

For isothermal gas, T = 100K is below T_{max}, and so the cloud is Jeans unstable.

For adiabatic gas, the temperature after the shock is T′ = 4^{2/3}T = 252K, the cloud’s mass is below the Jeans mass and is hence is stable.