Explanation Given information: The altitude between the outer layer of earth to the point A ( A ) is 1500km . At the point A the speed ( v A ) is reduced to 40 percent. The radius of the earth ( R ) is 6370 km . The acceleration due to gravity ( g ) is 9.81 m / s 2 . Calculation: Calculate the distance from center of the earth to point A ( r ) using the relation: r = R + A Substitute 6370 km for R and 1500 km for A . r = 6370 + 1500 = 7870 km ( 1000 m km ) = 7.870 × 10 6 m Calculate the circular orbit velocity at C ( v C ) using the relation: v C 2 r = g R 2 r 2 Here, G is the universal gravitational constant and M is mass of the earth. Substitute 7.870 × 10 6 m for r , 9.81 m / s 2 for g and 6.370 × 10 6 m for R . v C 2 7.870 × 10 6 = 9.81 × ( 6.370 × 10 6 ) 2 ( 7.870 × 10 6 ) 2 v C 2 = 50579337.87 v C = 50579337.87 v C = 7111.9 m / s Calculate the velocity at the point A ( v A ) using the relation: Since the velocity is reduced to 40 percent, the velocity at A will be reduced to 60 percent. v A = v C × ( 60 100 ) Substitute 7111.9 m / s for v C . v A = 7111.9 × ( 60 100 ) = 4267.14 m / s Calculate the kinetic energy ( T A ) at the point A using the formula: T A = 1 2 m v A 2 Substitute 4267.14 m / s for v A . T A = 1 2 m ( 4267.14 m / s ) 2 = 9104241.89 m Calculate the potential energy ( V A ) in the point A using the formula: V A = − G M m r A Here, G is the universal gravitational constant and M is mass of the earth. Substitute g R 2 for GM and 7.870 × 10 6 m for r A

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Chapter 13.2, Problem 13.90P

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Find the speed of the spacecraft (vB) when it reaches the earth’s surface at point B.

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Chapter 13.2, Problem 13.89P

Chapter 13.2, Problem 13.91P

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Find the speed of the spacecraft ( v B ) when it reaches the earth’s surface at point B .

Solution Summary: The author calculates the distance from center of the earth to point A using the relation: r=R+A.

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Find the speed of the spacecraft (vB) when it reaches the earth’s surface at point B.

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