Karman line is a theoretical construct that separates the earth's atmosphere from outer space. It is defined to be the height at which the lift on an aircraft flying at the speed of a polar satellite $(8 \,km / s )$ is equal to its weight. Taking a fighter aircraft of wing area $30 \,m ^2$, and mass $7500 \,kg$, the height of the Karman line above the ground will be in the range .............. $km$ (assume the density of air at height $h$ above ground to be $\rho( h )=1.2 e ^{\frac{ h }{10}} \,kg / m ^3$ where $h$ is in $km$ and the lift force to be $\frac{1}{2} \rho v^2 A$, where $v$ is the speed of the aircraft and $A$ its wing area).
KVPY 2021, Advanced
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(B)
For equilibrium :
$mg ^{\prime}=\frac{1}{2} \rho v ^2 A$
$mg \left(1-\frac{2 h }{ R }\right)=\frac{1}{2} \times 1.2 e ^{-\frac{ h }{10}} v ^2 A$
$\Rightarrow 7500 \times 9.8\left(1-\frac{2 h }{ R }\right)=0.6 e ^{-\frac{ h }{10}}\left(8 \times 10^3\right)^2 \times 30$
$\downarrow$ Neglect this term for a while and we get $h \simeq 95 \,km$
So, with $\left(1-\frac{2 h}{R}\right)$, we will get little lesser value or $h$.
$\therefore(B)$ is the answer.
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