Explanation Given: The satellite revolves the earth at elliptical path. Themass of the satellite is m = 60 kg . The radius of the satellite at position A is r A = 20 Mm . The radius of the satellite at position B is r B = 80 Mm . The mass of the earth M e = 5.976 × 10 24 kg . The gravitational constant is G = 66.73 × 10 − 12 m 3 / kg .s 2 The speed of the satellite at position A is v A = 40 Mm/h . Write the formula for kinetic energy ( T ) . T = 1 2 m v 2 (I) Here, m is the mass and v is the velocity. Write the formula for gravitational potential energy of the satellite. V g = − G M e m r (II) Write the formula for conservation of energy. T A + V A = T B + V B (III) Here, T A , T B are the initial and final kinetic energy and V A , V B are the initial and final potential energy. Conclusion: Calculate the velocity of the satellite at position A in terms of ( m/s ) . v A = 40 Mm h × 10 6 m × 1 h 1 Mm × 3600 s = 11111.11 m/s At position A : Calculate the kinetic energy. Substitute 60 kg for m and 11111.11 m/s for v in Equation (I). T A = 1 2 ( 60 kg ) ( 11111.11 m/s ) 2 = 370 , 37 , 02 , 962.963 J =3703 .702 × 10 6 J Calculate thepotential energy. Substitute 66.73 × 10 − 12 m 3 / kg .s 2 for G , 5.976 × 10 24 kg for M e , 60 kg for m and 20 × 10 6 m for r in Equation (II). V A = − 66.73 × 10 − 12 m 3 / kg .s 2 ( 5.976 × 10 24 kg ) ( 60 kg ) 20 × 10 6 m = 1196335440 J =1196 .335 × 10 6 J Here, the gravitational potential energy is negative because it above the datum (Earth). V A = − 11963 .35 × 10 6 J At position B : Calculate the kinetic energy

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.5, Problem 85P

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The speed of the satellite at the position B .

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Chapter 14.5, Problem 84P

Chapter 14.5, Problem 86P

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Chapter 14 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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The speed of the satellite at the position B .

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