(a) Explanation Given information: The radius of the earth ( R ) is 6 , 370 km . The altitude of point C ( h C ) from the surface of the earth is 1,500 km. The range of trajectory ( s A B ) is 6,000 km. Calculation: Find the product of the constant of gravitation G and the mass M of the earth [ ( G M ) E ] using the equation: ( G M ) E = g R 2 Here, G is the gravitational constant. Substitute 9.81 m/s 2 for g and 6.370 × 10 6 m for R . ( G M ) E = 9.81 × ( 6.370 × 10 6 ) 2 = 398.06 × 10 12 m 3 /s 2 Find the radius of the point C using the equation: r C = R + h C Substitute 6 , 370 km for R and 1,500 km for h C . r C = 6 , 370 + 1 , 500 = 7 , 870 km Find the angle ∠ A O B ( ϕ ) using the equation: ϕ = s A B R Substitute 6,000 km for s A B and 6,370 km for R . ϕ = 6 , 000 6 , 370 = 0.94192 rad × 180 ° π rad = 53.968 ° Find the angle of elliptical trajectory ( θ A ) at A using the equation: θ A = 180 ° − ϕ 2 Substitute 53.968 ° for ϕ . θ A = 180 ° − 53.968 2 = 153.016 ° The radius of point A ( r A ) and point B ( r B ) is equal to the radius of earth. Write the equation of motion satellite and the planet at A of elliptical trajectory. 1 r A = G M h 2 ( 1 + ε cos θ A ) Here, h is the angular momentum per unit mass and ε is the eccentricity of trajectory. Substitute 153.016 ° for θ A . 1 r A = G M h 2 ( 1 + ε cos 153.016 ) (1) The angle of elliptical trajectory ( θ C ) at C is 180 ° . Write the equation of motion satellite and the planet at C of elliptical trajectory. 1 r C = G M h 2 ( 1 + ε cos θ C ) Here, h is the angular momentum per unit mass and ε is the eccentricity of trajectory (b) To determine Find the eccentricity of the trajectory.

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Chapter 12.3, Problem 12.115P

(a)

To determine

Find the velocity of the projectile at C.

(b)

To determine

Find the eccentricity of the trajectory.

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Chapter 12.3, Problem 12.114P

Chapter 12.3, Problem 12.116P

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Find the velocity of the projectile at C .

Solution Summary: The author calculates the velocity of the projectile at C using the equation: underset_4km/s.