Question

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing vehicle and parachute is 2270 kg, the drag coefficient is effectively 0.5, the atmosphere density is 0.71 that of Earth (take Earth atmosphere density as 1.2 kg/m3), and the Martian gravitational acceleration is 3.689 m/s2, find the required total frontal area (in m2) of the lander plus a parachute to land at the given velocity. Assume the landing vehicle has achieved terminal velocity as it falls through the Martian atmosphere.

Answer 1

Answer:

The value is      $A = 39315 \ m^2$

Explanation:

From the question we are told that

    The velocity which the rover is suppose to land with is  $v = 1 \ m/s$

    The  mass of the rover and the parachute is  $m = 2270 \ kg$

     The  drag coefficient is  $C__{D}} = 0.5$

      The atmospheric density of Earth  is  $\rho = 1.2 \ kg/m^3$

     The acceleration due to gravity in Mars is  $g_m = 3.689 \ m/s^2$

     

Generally the Mars  atmosphere density is mathematically represented as

          $\rho_m = 0.71 * \rho$

=>        $\rho_m = 0.71 * 1.2$

=>        $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute  is mathematically represented as

          $F__{D}} = m * g_{m}$

=>       $F__{D}} = 2270 * 3.689$  

=>       $F__{D}} = 8374 \ N$  

Gnerally this drag force is mathematically represented as

         $F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

So  

         $A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=>       $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=>       $A = 39315 \ m^2$