Question

In 1988, telescopes viewed Pluto as it crossed in front of a distant star. As the star emerged from behind the planet, light from the star was slightly dimmed as it went through Pluto’s atmosphere. The observations indicated that the atmospheric density at a height of 50 km above the surface of Pluto is about one-third the density at the surface. The mass of Pluto is known to be about 1.5×10²² kg and its radius is about 1200 km. Spectroscopic data indicate that the atmosphere is mostly nitrogen (N2). Estimate the temperature of Pluto’s atmosphere. State what approximations and/or simplifying assumptions you made.

Short answer

The temperature of Pluto’s atmosphere is, T = 105.6K .

Step-by-step solution

Identification of given data

• The height at which there is atmospheric density is,$h=50Km=5\times {10}^{4}m$ .
• The ratio of density is, $\frac{\rho }{{\rho }_0}=0333$.
• The mass of Pluto is, $M=1.5x{10}^{22}\mathrm{kg}$.
• The radius of Pluto is, $R=1200\times {10}^3=1.2\times {10}^{6}m$.

Concept of gravitational acceleration 

Free fall acceleration inside a vacuum is known as gravitational acceleration. Gravitational pull alone is responsible for this continuous increase in speed.

The gravitational acceleration is given by,

$g=\frac{GM}{R^2}$

Here G is the gravitational constant whose value is $6.67\times {10}^{-11}{m}^3.k{g}^{-1}.s^{-2}$ .

Determination of temperature of Pluto’s atmosphere

The gravitational acceleration of Pluto is determined using equation (i); we get,

$$\begin{gathered} g=\frac{6.67\times {10}^{-11}{m}^{3}k{g}^{-1}{s}^{-2}\times 1.5\times {10}^{22}kg}{1.44\times {10}^{12}{m}^2} \\ g=0.695m/s^2 \end{gathered}$$

The ratio of atmosphere densities gives Boltzmann factor for gravitational potential energy,

$\frac{p}{p_0}=exp\left(-\frac{mgh}{k_{B}T}\right)$

Here $k_B=1.38x{10}^{-23}JlK$is the Boltzmann constant.

Re-arranging the above equation, we get,

$$\begin{gathered} 0.33=exp\left(-\frac{mgh}{K_{B}T}\right) \\ \ln \left(0.33\right)=\frac{mgh}{K_{B}T} \\ T=\frac{mgh}{K_B.\ln (0.33)} \end{gathered}$$

Substitute values in the above equation, we get,

$$\begin{gathered} T=-\frac{4.65\times {10}^{-26}Kg\times 0.695m/s^2\times 5\times {10}^{4}m}{1.38\times {10}^{-23}J/K.\ln (0.33)} \\ T=105.6\left(\frac{1Kg.m^2/s^2}{1J/K}\times \frac{1J/K}{1Kg.m^2/s^2}\right) \\ T=105.6K \end{gathered}$$

Thus, the temperature of Pluto is 105.6K .