Space Is Full Of Hot Gas

Classical Mechanics Level 5

A hot, spherically symmetric cloud of an ideal gas ( H 2 H_2 for example) is in intergalactic space. The gas is in equilibrium and the density ρ ( r ) \rho(r) of the gas is given by ρ ( r ) = ρ 0 r − 2 \rho(r)=\rho_0r^{-2} . What is the ratio of the temperature at a distance r 0 r_0 to the temperature at a distance 2 r 0 2r_0 ?


The answer is 1.00.

Roger Kepstein by Roger Kepstein

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3 solutions

Jatin Yadav
Mar 18, 2014

Assumptions :

  • The cloud is very large, i.e. it extends to ∞ \infty

  • At the end of cloud, the pressure is 0 0 , i.e. at r = ∞ r = \infty , P = 0 P=0 .

    Solution :

The gravitational field at a distance r r is given by: g ( r ) = G ∫ 0 r ρ 0 x − 2 × 4 π x 2 d x r 2 = 4 π G ρ 0 r \displaystyle g(r) = \frac{G \int_{0}^{r} \rho_{0} x^{-2} \times 4 \pi x^2 dx}{r^2} = \frac{4 \pi G \rho_{0}}{r}

Now, d P = − ρ g d x dP = -\rho g dx

⇒ d P = − ρ 0 x − 2 × 4 π G ρ 0 x d x \Rightarrow dP = - \rho_{0} x^{-2} \times \frac{4 \pi G \rho_{0}}{x} dx

⇒ ∫ 0 P ( r ) d P = − 4 π G ρ 0 2 ∫ ∞ r x − 3 d x \Rightarrow \displaystyle \int_{0}^{P(r)} dP = -4 \pi G \rho_{0}^2 \int_{\infty}^{r} x^{-3} dx

⇒ P ( r ) = 2 π G ρ 0 2 r − 2 \Rightarrow P(r) = 2 \pi G {\rho_{0}}^{2} r^{-2}

Now, T = P M ρ R T = \frac{P M}{\rho R}

= 2 π G ρ 0 2 r − 2 M ρ 0 r − 2 R \displaystyle \frac{2 \pi G {\rho_{0}}^{2} r^{-2} M}{\rho_{0} r^{-2} R}

⇒ T = 2 π G ρ 0 M R \Rightarrow \large \displaystyle \boxed{T = \frac{2 \pi G \rho_{0} M }{R}}

Thus, T T is constant everywhere. Hence, our answer is 1 1 .

10 Helpful 0 Interesting 1 Brilliant 0 Confused

Why not use the ideal gas equation?

siddharth shah - 7 years, 2 months ago

Yea I did this method too, but before arriving at this solution, I tried another method and reached a dead end.

Basically, my approach was to calculate the Force which a thin spherical shell has to support, and then dividing by the area of the shell to find the pressure.

For a spherical shell at radial distance r r , I took g ( r ) g(r) and integrated to find the force from the mass of the cloud which the shell has to support. I define σ ( r ) = 4 π ρ 0 \sigma(r) = 4\pi \rho_0 as the mass of a thin spherical shell, and m ( r ) = 4 π ρ 0 r m(r) = 4 \pi \rho_0 r as the mass of the sphere up to radius r r . Making use if shell's theorem, my integral went:

F = ∫ r ∞ G σ ( r ) m ( r ) r d r = ∫ r ∞ G ( 4 π ρ 0 ) 2 r d r F = \int^{\infty}_{r} \frac{G \sigma(r) m(r)}{r} dr = \int^{\infty}_{r} G\frac{(4\pi \rho_0)^2}{r} dr

which evaluates to ∞ \infty

And since the area at a radial distance r r is finite, then wouldn't the pressure be infinite too?

Jiahai Feng - 7 years, 2 months ago

Hats off to u man really nice soln.

aryan goyat - 4 years, 6 months ago
Damodar Prabhu
Apr 11, 2014

It is already stated in the problem that the gas is in equilibrium!

So I think we can conclude that there is no heat interaction inside the gas sphere or in other words the temperature inside the gas is the same!(simple Thermodynamics)

Correct me if I'm wrong,guys.

2 Helpful 0 Interesting 0 Brilliant 1 Confused

Brilliant!

Daniel Moura - 6 years, 10 months ago
Avraam Aneleitos
Apr 20, 2014

At every distance r r from the center there are gas atoms at different pressure which is given by P = F S = F 4 π r 2 P=\frac{F}{S}=\frac{F}{4πr^2} .Since the gas is in equilibrium F F must be constant so P P is inversely proportional to the square of r r just like ρ ρ .So from P = ρ R T M P=\frac{ρRT}{M} we get that T = c o n s t . T=const. (I'm not too sure about that).

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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