標題:
Mass of asteroid
發問:
A space vehicle of mass 95 tonnes is intended to go into a circular orbit around asteroid at a path radius of 8.5 km. The asteroid is roughly spherical of diameter 11.09 km and have a mean density of 4100 kg m -3.Find mass of asteriod:A. 1.72 x 10 15 kgB. 1.98 x 10 15 kgC. 2.05 x 10 15 kgD. 2.93 x 10... 顯示更多 A space vehicle of mass 95 tonnes is intended to go into a circular orbit around asteroid at a path radius of 8.5 km. The asteroid is roughly spherical of diameter 11.09 km and have a mean density of 4100 kg m -3. Find mass of asteriod: A. 1.72 x 10 15 kg B. 1.98 x 10 15 kg C. 2.05 x 10 15 kg D. 2.93 x 10 15 kg E. 3.04 x 10 15 kg F. 3.92 x 10 15 kg 更新: Ysing Newton's gravitational constant G=66.7 x 10 -12 N m 2 kg -2, find gravitational force exerted on the vehicle by the asteroid: A. 257 N B. 265 N C. 309 N D. 420 N E. 489 N F. 503 N 更新 2: Find correct orbital speed: A. 3.01 ms-1 B. 4.79 ms-1 C. 5.21 ms-1 D. 6.35 ms-1 E. 7.28 ms-1 F. 8.88 ms-1 Find orbital period: A. 1.92 hrs B. 2.07 hrs C. 2.53 hrs D. 2.86 hrs E. 3.09 hrs F. 4.10 hrs
最佳解答:
1. Radius of theasteriod, r = 11.09 X 103 / 2 = 5.545 X 103 m By mass = density Xvolume = 4/3 ρπr3 = 4/3 (4100)π(5.545X 103)3 = 2.93 X 1015kg The answer is D. 2. Mass of the spacevehicle, m = 95 ton = 95 X 103 kg Gravitational force =GMm / R2 = (66.7 X 10-12)(2.93X 1015)(95 X 103) / (8.5 X 103)2 = 257 N The answer is A. 3. By mv2 / R= GMm / R2 (gravitational force accounts for the centripetal force) v = (GM / R)1/2 v = [(66.7 X 10-12)(2.93X 1015) / (8.5 X 103)]1/2 v = 4.79 ms-1 The answer is B. 4. By v = 2πR / T T = 2πR / v T = 2π(8.5X 103) / 4.79 T = 11 138 s T = 3.09 hrs The answer is E.
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