Calculate the ratio of the drag force on a jet flying at ${}^{\mathbf{1000}\mathbf{km}\mathbf{/}\mathbf{h}}$at an altitude of${}^{\mathbf{10}\mathbf{}\mathbf{km}}$to the drag force on a prop-driven transport flying at half that speed and altitude. The density of air is ${}^{\mathbf{0}\mathbf{.}\mathbf{38}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}}$at ${}^{\mathbf{10}\mathbf{}\mathbf{km}}$and ${}^{\mathbf{0}\mathbf{.}\mathbf{67}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}}$at ${}^{\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{km}}$.Assume that the airplanes have the same effective cross-sectional area and drag coefficient C.

Height of the jet is ${}^{10\mathrm{km}}$,

Velocity of the jet at 10km is ${}^{v_j=1000\mathrm{km}/\mathrm{h}}$,

Velocity of the jet at 5km is${}^{v_t=500\mathrm{km}/\mathrm{h}}$

The density of air at${}^{10\mathrm{km}=0.38\mathrm{kg}/{\mathrm{m}}^3}$,

The density of air at${}^{5\mathrm{km}=0.67\mathrm{kg}/{\mathrm{m}}^3}$.

This problem is based on the drag force which is a type of friction. This is the force acting opposite to the relative motion of an object moving with respect to the surrounding medium.

Formula:

The terminal speed is given by

$v_t=\sqrt{\frac{2F_g}{CpA}}$

Where C is the drag coefficient,${}^p$is the fluid density, A is the effective cross-sectional area, and${}^{F_g}$is the gravitational force.

For the passenger jet, the drag force will be,

$D_j=\frac{1}{2}C{p}_1{A}_j{v_j}^2$

and for the prop-driven transport, the drag force will be,

$D_t=\frac{1}{2}C{p}_2{A}_t{v_t}^2$

Where ${}^{p_1}$and ${}^{p_2}$represent the air density at 10 km and 5.0 km, respectively.

Thus, the ratio in question is,

$$\begin{gathered} \frac{D_j}{D_t}=\frac{p_1{V}^2}{p_2{V^2}_t} \\ =\frac{\left(0.38\mathrm{kg}/{\mathrm{m}}^3\right){\left(1000\mathrm{km}/\mathrm{h}\right)}^2}{\left(0.67\mathrm{kg}/{\mathrm{m}}^3\right){\left(500\mathrm{km}/\mathrm{h}\right)}^2} \\ =2.3 \end{gathered}$$

Therefore, the ratio of the drag force on a jet is ${}^{2.3}$