Answer:
The value is [tex]A = 39315 \ m^2[/tex]
Explanation:
From the question we are told that
The velocity which the rover is suppose to land with is [tex]v = 1 \ m/s[/tex]
The mass of the rover and the parachute is [tex]m = 2270 \ kg[/tex]
The drag coefficient is [tex]C__{D}} = 0.5[/tex]
The atmospheric density of Earth is [tex]\rho = 1.2 \ kg/m^3[/tex]
The acceleration due to gravity in Mars is [tex]g_m = 3.689 \ m/s^2[/tex]
Generally the Mars atmosphere density is mathematically represented as
[tex]\rho_m = 0.71 * \rho[/tex]
=> [tex]\rho_m = 0.71 * 1.2[/tex]
=> [tex]\rho_m = 0.852 \ kg/m^3[/tex]
Generally the drag force on the rover and the parachute is mathematically represented as
[tex]F__{D}} = m * g_{m}[/tex]
=> [tex]F__{D}} = 2270 * 3.689[/tex]
=> [tex]F__{D}} = 8374 \ N[/tex]
Gnerally this drag force is mathematically represented as
[tex]F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}[/tex]
Here A is the frontal area
So
[tex]A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }[/tex]
=> [tex]A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }[/tex]
=> [tex]A = 39315 \ m^2[/tex]
