Due to an emergency NASA has decided to land the space Shuttle in Montreal (at t

Question

Due to an emergency NASA has decided to land the space Shuttle in Montreal (at the Pierre Elliott Trudeau International Airport). The longest runway of this airport is 11,000 ft and 200 ft wide (3,353 m, 61 m). The Shuttle's total landing mass is 104,328 kg and its velocity is 350 km/hr (twice the landing speed of a commercial airliner). Upon touchdown, a chute (with D 16.5 m and CD 1.4) is deployed. The chute and orbiter's drag, along with the rolling resistance of the tires, slow the orbiter to a speed (say 60 km/hr) where the brakes can then be safely applied. Assess NASA's decision to land the Space Shuttle in Montreal. In your assessment consider only the deceleration of the vehicle due to the chute and neglect all the other effects. The parachute requires 5 seconds to be fully operational from the time of deployment. Assume further that the shuttle touches down 300 m from the beginning of the runway and upon brake application the vehicle needs approximately 200 m to come to a complete stop. (The nearby Mirabel International has a runway that is 12,000 ft long and 300 ft wide).

Explanation / Answer

According to the given problem,

Initial Distance Covered before deployment S1 = 300 m

Now, the vehicle is travelling at its landing velocity till the chute becomes fully operational. (as other effects are to be neglected)

Landing Velocity U = 350 km/hr = 97.22 m/s

Distance covered during the interval of deployment S2 = U*t = 97.22*5 = 486 m
(Neglecting the drag during partial opening of the chute and assuming there is no decelerating force)

After the chute opens, the vehicle decelerates to the terminal velocity with the help of the chute!!

Declerating force F = ma = D - T

Here, Thrust T = 0
Drag D = ½ Cd A V²

==> F = ma = ½ Cd A v²
==> a = (Cd A/ 2 m) v² = K v² , say

Now, a = dv/dt ==> dt = dv / a = dv / (K v²) ==> dv / v² = K dt ==> v = -1/(K t - A) = 1/(A- K t)

v = ds / dt ==> ds = - dt /(A+ K t) ====> s+B = - ln (A- K t) / K

Substituting the inital conditions [(i.e) at t= 0, s = 0 ; v = U], we get

s = - ln (1 - KU t) / K; v = U / (1 - KU t)

On further Simplification, we get S = [ln (U / V) ]/ K

It is quite evident that the end velocity using drag alone can not be zero.

Assuming the final velocity as given in the problem,
V = 60 km/hr = 16.67 m/s

Mass of the Shuttle m = 104328 kg
Acceleration due to gravity g = 9.81 m/s²
Coefficient of drag Cd = 1.4
Density of air = 1.22 kg/m³

Area of the chute A = D² / 4 = * 16.5² / 4 = 213.8 m²

==> K = Cd A/ 2 m
K = 1.4*1.22*213.8 / (2*104328) = 1.75 x 10³ m¹

Hence, S3 = [ln U / V) ]/ K = [ln (97.22 /16.67 ) ]/ (1.75 x 10³ ) = 1007 m

Distance moved after applying brake S4 = 200 m

Thus, Total Distance moved = S1 + S2 + S3 + S4 = 1993 m

Hence, NASA is not wrong .
Further working,
NASA cannot be wrong unless they decide to apply brakes after reducing the vehicle speed to 1.6 m/s (i.e) 6 km/s

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.