Answer:
a) maximum mass of the Mars lander to ensure it can land safely is 200 kg
b) area of the parachute required is 480 m² which is larger than 400 m²
c) area of the parachute should be 12.68 m²
Explanation:
Given the data in the question;
V = 20 m/s
A = 200 m²
drag co-efficient CD = 1.855
g = 3.71 m/s²
density of the atmospheric pressure β = 0.01 kg/m³
a. Calculate the maximum mass of the Mars lander to ensure it can land safely?
Drag force FD = 1/2 × CD × β × A × V²
we substitute
FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²
FD = 742 N
we know that;
FD = Fg
Fg = gravity force
Fg = mg
so
FD = mg
m = FD/g
we substitute
m = 742 N / 3.71 m/s²
m = 200 kg
Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg
b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;
Given that;
M = 480 kg
Show that the parachute required would be larger than 400 m²
we know that;
FD = Fg = Mg = 480 kg × 3.71 m/s²
FD = 1780.8 N
Now, FD = 1/2 × CD × β × A × V², we solve for A
A = FD / 0.5 × CD × β × V²
we substitute
A = 1780.8 / 0.5 × 1.855 × 0.1 × (20)²
A = 1780.8 / 3.71
A = 480 m²
Therefore, area of the parachute required 480 m² which is larger than 400 m²
c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?
Given that;
g = 9.8 m/s²,
β" = 1 kg/m³
v" = 20 m/s
M" = 480 kg
we know that;
FD = Fg = M"g
FD = 480 kg × 9.8 m/s² = 4704 N
from the expression; FD = 1/2 × CD × β × A × V²
A = FD / 0.5 × CD × β" × V"²
we substitute
A = 4704 / 0.5 × 1.855 × 1 × (20)²
A = 4704 / 371
A = 12.68 m²
Therefore area of the parachute should be 12.68 m²
