In microgravity astronaut training and equipment testing, NASA flies a KC135A ai
ID: 2046008 • Letter: I
Question
In microgravity astronaut training and equipment testing, NASA flies a KC135A aircraft along a parabolic flight path. As shown in the figure, the aircraft climbs from 24,000 ft to 28,210 ft, where it enters a parabola with a velocity of 139 m/s at 45.0° nose high and exits with velocity 139 m/s at 45.0° nose low. During this portion of the flight, the aircraft and objects inside its padded cabin are in free-fall; astronauts and equipment float freely as if there were no gravity.(a) What is the aircraft's speed at the top of the maneuver? m/s
(b) What is the aircraft's altitude at the top of the maneuver? ft
(c) What is the time spent in zero gravity? s
Explanation / Answer
Same one with different numericals During the ballistic flight, the aircraft moves under the sole influence of earth's gravity, just like a projectile. The components of its velocity when entering the zero-g flight are vx = 143 cos45 = 101 m/s vy = 143 sin 45 = 101 m/s and at the end of the parabolic flight path are v'x = 143 cos(-45) = 101 m/s v'y = 143 sin (-45) =-101 m/s In what follows I neglect the change of g in function of altitude and assume that it is always 9.8 m/s² (a) The velocity of the aircraft at the top of the maneuver is the same as its horzontal component vx, that is 101 m/s (b) The height it climbs: ?y = (0 - vy²)/2a = - (101²)/2×(-9.8) = 520 m The altitude of the top of the flight path y = y0 + ?y = 31,000 ft + 520 m = 9450 m + 520 m = 9,970 m (c) The time interval spent in zero-g ?t = ?(vy)/a = (v'y - vy)/-g = -202/-9.8 ?t = 20.6 s (d) I'm not sure about the upward acceleration of the motion in this phase. - If upward means vertical and upward, then the trajectory of the aircraft would be a parabola (not a circle) with positive concavity. At the bottom of that trajectory vy = 0 and vx remains the same as it was. The answer would be v = vx = 101 m/s - If upward doesnt necessary mean vertical, but a is a centripetal acceleration, then the trajectory is indeed a circle, and the aircraft speed at any position on the circle (not just the bottom) would be v = sqrt(aR) = sqrt(0.8×9.8×4,300) = 180 m/s²
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