Question

Calculate the ratio of the drag force on a jet flying at 950 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at two-fifths the speed and half the altitude of the jet. The density of air is 0.38 kg/m3 at 10 km and 0.67 kg/m3 at 5.0 km. Assume that the airplanes have the same effective cross-sectional area and drag coefficient C.

Answer 1

`(F_1)/(F_2)=3.55`

Step-by-step explanation:

F = Force

C = Drag coefficient equal for both aircrafts

ρ = Density of air

A = Surface area equal for both aircrafts

v = Velocity

`v_2=(2)/(5)v_1`

`F_1=(1)/(2)\rho_1 CAv_1^2`

`F_2=(1)/(2)\rho_2 CAv_2^2\\\Rightarrow F_2=(1)/(2)\rho_2 CA\left((2)/(5)v_1\right)^2`

Dividing the above two equations we get

`(F_1)/(F_2)=((1)/(2)\rho_1 CAv_1^2)/((1)/(2)\rho_2 CA\left((2)/(5)v_1\right)^2)\\\Rightarrow (F_1)/(F_2)=(\rho_1)/(\rho_2(4)/(25))\\\Rightarrow (F_1)/(F_2)=(0.38)/(0.67(4)/(25))\\\Rightarrow (F_1)/(F_2)=3.55`

The ratio of the drag forces is
`\mathbf{(F_1)/(F_2)}=\mathbf{3.55}`