problem 13.194 a shuttle is to rendezvous with a space station which is in a circular orbit at an...
TRANSCRIPT
Problem 13.194
A shuttle is to rendezvous with aspace station which is in a circularorbit at an altitude of 250 mi abovethe surface of the earth. The shuttlehas reached an altitude of 40 miwhen its engine is turned off at pointB. Knowing that at that time thevelocity vo of the shuttle forms anangle = 55o with the vertical,determine the required magnitude ofvo if the trajectory of the shuttle is tobe tangent at A to the orbit of thespace station.
o
vo
R = 3960 mi
250 mi
B
A
Solving Problems on Your OwnProblem 13.194
A shuttle is to rendezvous with a spacestation which is in a circular orbit at analtitude of 250 mi above the surface ofthe earth. The shuttle has reached analtitude of 40 mi when its engine isturned off at point B. Knowing that atthat time the velocity vo of the shuttle
forms an angle = 55o with the vertical, determine the requiredmagnitude of vo if the trajectory of the shuttle is to be tangent atA to the orbit of the space station.
1. Apply conservation of energy principle : When a particlemoves under the action of a conservative force, the sum of thekinetic and potential energies of the particle remains constant.
T1 + V1 = T2 + V2
where 1 and 2 are two positions of the particle.
o
vo
R = 3960 mi
250 mi
B
A
Solving Problems on Your OwnProblem 13.194
A shuttle is to rendezvous with a spacestation which is in a circular orbit at analtitude of 250 mi above the surface ofthe earth. The shuttle has reached analtitude of 40 mi when its engine isturned off at point B. Knowing that atthat time the velocity vo of the shuttle
forms an angle = 55o with the vertical, determine the requiredmagnitude of vo if the trajectory of the shuttle is to be tangent atA to the orbit of the space station.
1a. Kinetic energy: The kinetic energy at each point on the pathis given by:
T = m v212
o
vo
R = 3960 mi
250 mi
B
A
Solving Problems on Your OwnProblem 13.194
A shuttle is to rendezvous with a spacestation which is in a circular orbit at analtitude of 250 mi above the surface ofthe earth. The shuttle has reached analtitude of 40 mi when its engine isturned off at point B. Knowing that atthat time the velocity vo of the shuttle
forms an angle = 55o with the vertical, determine the requiredmagnitude of vo if the trajectory of the shuttle is to be tangent atA to the orbit of the space station.
1b. Potential energy: The potential energy of a mass located ata distance r from the center of the earth is
Vg = -
where G is the constant of gravitation and M the mass of the earth.
G M mr
o
vo
R = 3960 mi
250 mi
B
A
Solving Problems on Your OwnProblem 13.194
A shuttle is to rendezvous with a spacestation which is in a circular orbit at analtitude of 250 mi above the surface ofthe earth. The shuttle has reached analtitude of 40 mi when its engine isturned off at point B. Knowing that atthat time the velocity vo of the shuttle
forms an angle = 55o with the vertical, determine the requiredmagnitude of vo if the trajectory of the shuttle is to be tangent atA to the orbit of the space station.
2. Apply conservation of angular momentum principle: For a spacevehicle of mass m moving under the earth’s gravitational force:
r1 m v1 sin1 = r2 m v2 sin 2
where 1 and 2 represent two points along the trajectory, r is thedistance to the center of the earth, v is the vehicle’s speed, and the angle between the velocity and the radius vectors.
o
vo
R = 3960 mi
250 mi
B
A
Problem 13.194 SolutionApply conservation ofenergy principle.
TA + VA = TB + VB
m vA2 - = m vo
2 - G M mrB
G M mrA
12
12
rB = ( 3960 mi + 40 mi )( 5280 ft/mi) = 21.12x106 ft
rA = ( 3960 mi + 250 mi )( 5280 ft/mi) = 22.2288x106 ft
GM = g R2 = (32.2 ft/s2)[(3960 mi)(5280 ft/mi)]2 = 1.4077x1016ft3
s2
vA2 - = vo
2 -(1.4077x1016 ft3/s2)(22.2288x106 ft)
12
12
(1.4077x1016 ft3/s2)(21.12x106 ft)
(1)
o
vo
R = 3960 mi
250 mi
B
A
o
vo
R = 3960 mi
250 mi
B
A
vA A = 90o
Problem 13.194 Solution
Apply conservation of angularmomentum principle.
rB m vo sino = rA m vA sin A
(21.12x106 ft) vo sin 55o = (22.2288x106 ft) vA sin 90o (2)
Solution of equations (1) and (2) gives:
vo = 12,990 ft/s