Question

A space vehicle returning from the Moon enters the Earth’s atmosphere at a speed of about 42,000 km/h. Molecules (assume nitrogen) striking the nose of the vehicle with this speed correspond to what temperature? (Because of this high temperature, the nose of a space vehicle must be made of special materials; indeed, part of it does vaporize, and this is seen as a bright blaze upon reentry.)

Short answer

The nitrogen molecules strike the space vehicle with aspeed of 42,000 km/h at atemperatureof \(1.53\; \times {10^5}\;{\rm{K}}\).

Step-by-step solution

Step 1:Understanding of root-mean square (RMS) speed

The square root of the mean of the square of the molecules’ speed in a gas at a certain temperature Tis termed as the root-mean square speed of that gas.

The RMS speed is given as

\({v_{{\rm{rms}}}} = \sqrt {\frac{{3kT}}{m}} \). …… (i)

Here, \(k\)is the Boltzmann’s constant,T is the temperature, and m is the mass of each molecule.

Given Data

The RMSspeed of nitrogen molecules is \({v_{{\rm{rms}}}} = 42,000\;{\rm{km/h}}\).

The molecular mass of each nitrogen moleculeis\(m = 28\;{\rm{u}}\).

Boltzmann constant is \(k = 1.38 \times {10^{ - 23}}\;{\rm{J/K}}\).

Determination of temperature of nitrogen molecules

From equation (i), the temperature of nitrogen gas is given as

\(T = \frac{{mv_{{\rm{rms}}}^2}}{{3k}}\).

Substitute the values into the above expression.

\(\begin{aligned}T &= \frac{{\left( {28\;{\rm{u}} \times \frac{{1.66 \times {{10}^{ - 27}}\;{\rm{kg}}}}{{1\;{\rm{u}}}}} \right){{\left( {42,000 \times \frac{5}{{18}}\;{\rm{m/s}}} \right)}^2}}}{{3\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)}}\\ &= \frac{{\left( {46.48 \times {{10}^{ - 27}}\;{\rm{kg}}} \right)\left( {1.36 \times {{10}^8}} \right)}}{{4.14 \times {{10}^{ - 23}}}}\\ &= 1.53\; \times {10^5}\;{\rm{K}}\end{aligned}\)

Thus, the nitrogen molecules strike the space vehicle with a speed of 42,000 km/h at a temperature of \(1.53\; \times {10^5}\;{\rm{K}}\).