Explanation Show the free body diagram as in Figure (1). Express the semi major axis length a of the orbit using the relationship. v B 2 = 2 g R 2 [ 1 r − 1 2 a ] (I) Here, the radial distance at B is r , radius of earth is R , acceleration due to gravity is g and velocity of spacecraft is v B . Express the work energy equation. E = T B + V B = 1 2 m v B 2 + [ − m g R 2 r ] = 1 2 m v B 2 − − m g R 2 R (II) Here, kinetic energy of satellite is T B , potential energy of satellite is V B and mass of spacecraft is m . Express the angular momentum. h = r v θ = r ( v B cos θ ) (III) Here, tangential velocity is v θ . Express the eccentricity. e = 1 + 2 E h 2 m g 2 R 4 (IV) Express the radius of orbit at B . r = a ( 1 − e 2 ) 1 + e cos θ B (V) Here, angle made by satellite is θ B . Express the radius of orbit at C . R = a ( 1 − e 2 ) 1 + e cos θ C (VI) Here, angle made by satellite is θ C . Show the angle made by the semi major axis line as in Figure (2). Refer Figure (2) and calculate the angle β which locates the point C of impact with the earth. β = θ C − 2.73 ° (VII) Conclusion: Refer Figure (1), and express the radial distance at B . r = ( 5 R ) 2 + ( 2 R ) 2 = 29 R Refer Figure (1), and express the angle α α = tan − 1 [ 2 R 5 R ] = 21.8 ° Substitute 3200 m / s for v B , 9 .81 m / s 2 for g , 29 R for r and 6371 × 10 3 m for R in Equation (I). ( 3200 m / s ) 2 = 2 ( 9 .81 m / s 2 ) ( 6371 × 10 3 m ) 2 [ 1 29 × 6371 × 10 3 m − 1 2 a ] 1.2838 × 10 − 8 = 2.9147 × 10 − 8 − 1 2 a a = 3.066 × 10 7 m Substitute 3200 m / s for v B , 9 .81 m / s 2 for g , 29 R for r and 6371 × 10 3 m for R in Equation (II). E = 1 2 m ( 3200 m / s ) 2 − m ( 9

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ISBN: 9781118885840

Author: James L. Meriam, L. G. Kraige, J. N. Bolton

Publisher: WILEY

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Chapter 3.12, Problem 300P

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Determine the angle β which locates the point C of impact with the earth.

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Chapter 3.12, Problem 299P

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