Problem A shuttle is to rendezvous with a space station which is in a circular orbit at an altitude of 250 mi above the surface of the earth. The shuttle has reached an altitude of 40 mi when its engine is turned off at point B. Knowing that at that time the velocity v o of the shuttle forms an angle  = 55 o with the vertical, determine the required magnitude of v o if the trajectory of the shuttle is to be tangent at A to the orbit of the space station. oo vovo R = 3960 mi 250 mi B A

Solving Problems on Your Own Problem A shuttle is to rendezvous with a space station which is in a circular orbit at an altitude of 250 mi above the surface of the earth. The shuttle has reached an altitude of 40 mi when its engine is turned off at point B. Knowing that at that time the velocity v o of the shuttle forms an angle  = 55 o with the vertical, determine the required magnitude of v o if the trajectory of the shuttle is to be tangent at A to the orbit of the space station. 1. Apply conservation of energy principle : When a particle moves under the action of a conservative force, the sum of the kinetic and potential energies of the particle remains constant. T 1 + V 1 = T 2 + V 2 where 1 and 2 are two positions of the particle. oo vovo R = 3960 mi 250 mi B A

Solving Problems on Your Own Problem A shuttle is to rendezvous with a space station which is in a circular orbit at an altitude of 250 mi above the surface of the earth. The shuttle has reached an altitude of 40 mi when its engine is turned off at point B. Knowing that at that time the velocity v o of the shuttle forms an angle  = 55 o with the vertical, determine the required magnitude of v o if the trajectory of the shuttle is to be tangent at A to the orbit of the space station. 1a. Kinetic energy: The kinetic energy at each point on the path is given by: T = m v oo vovo R = 3960 mi 250 mi B A

Solving Problems on Your Own Problem A shuttle is to rendezvous with a space station which is in a circular orbit at an altitude of 250 mi above the surface of the earth. The shuttle has reached an altitude of 40 mi when its engine is turned off at point B. Knowing that at that time the velocity v o of the shuttle forms an angle  = 55 o with the vertical, determine the required magnitude of v o if the trajectory of the shuttle is to be tangent at A to the orbit of the space station. 1b. Potential energy: The potential energy of a mass located at a distance r from the center of the earth is V g = - where G is the constant of gravitation and M the mass of the earth. G M mG M m r oo vovo R = 3960 mi 250 mi B A

Solving Problems on Your Own Problem A shuttle is to rendezvous with a space station which is in a circular orbit at an altitude of 250 mi above the surface of the earth. The shuttle has reached an altitude of 40 mi when its engine is turned off at point B. Knowing that at that time the velocity v o of the shuttle forms an angle  = 55 o with the vertical, determine the required magnitude of v o if the trajectory of the shuttle is to be tangent at A to the orbit of the space station. 2. Apply conservation of angular momentum principle: For a space vehicle of mass m moving under the earth’s gravitational force: r 1 m v 1 sin  1 = r 2 m v 2 sin  2 where 1 and 2 represent two points along the trajectory, r is the distance to the center of the earth, v is the vehicle’s speed, and  the angle between the velocity and the radius vectors. oo vovo R = 3960 mi 250 mi B A

Problem Solution Apply conservation of energy principle. T A + V A = T B + V B m v A 2 - = m v o 2 - G M mG M m rBrB G M mG M m rArA r B = ( 3960 mi + 40 mi )( 5280 ft/mi) = x 10 6 ft r A = ( 3960 mi mi )( 5280 ft/mi) = x 10 6 ft GM = g R 2 = (32.2 ft/s 2 )[(3960 mi)(5280 ft/mi)] 2 = x10 16 ft 3 s2s2 v A 2 - = v o 2 - (1.4077x10 16 ft 3 /s 2 ) ( x 10 6 ft) (1.4077x10 16 ft 3 /s 2 ) (21.12 x 10 6 ft) (1) oo vovo R = 3960 mi 250 mi B A

oo vovo R = 3960 mi 250 mi B A vAvA  A = 90 o Problem Solution Apply conservation of angular momentum principle. r B m v o sin  o = r A m v A sin  A (21.12 x 10 6 ft) v o sin 55 o = ( x 10 6 ft) v A sin 90 o (2) Solution of equations (1) and (2) gives: v o = 12,990 ft/s