Explanation The universal gravitational constant is 6.67 × 10 − 11 N ⋅ m 2 / kg 2 , mass of the earth is 5.97 × 10 24 kg , radius of earth is 6.37 × 10 6 m . The formula for gravitational potential energy is, U = − G m M r Here, U is the gravitational potential energy m is mass of module M is mass of moon r is the distance between moon and the module G is the universal gravitational constant The formula for kinetic energy is, K = 1 2 m v 2 Here, K is the kinetic energy v is the velocity The formula for total energy is, E = U + K Here, E is the total energy of the system. Substitute 1 2 m v 2 for K and − G m M r for U in the above expression. E = − G m M r + 1 2 m v 2 For projectile at a height above surface of earth: Substitute E 1 for E , r 1 for r and v c for v in the above expression. E 1 = − G m M r 1 + 1 2 m v c 2 (1) Here, v c is the escape speed of the earth The expression for value of r 1 is, r 1 = r e + h Here, r e is the radius of earth h is the height of projectile from the surface of the earth The formula for escape speed is v c = 2 G M r e (2) Here, G is the universal gravitational constant M is the mass of earth r e is the radius of earth For projectile on surface of earth: Substitute E 2 for E , r e for r and v c 3 for v in the above expression

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Chapter 12, Problem 53PCE

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The height above the earth at which speed of the projectile is one-third of the escape speed of the earth,

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Chapter 12, Problem 52PCE

Chapter 12, Problem 54PCE

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The height above the earth at which speed of the projectile is one-third of the escape speed of the earth,

Solution Summary: The author explains the universal gravitational constant, mass of the earth, distance between moon and module, and the formula for kinetic energy.

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The height above the earth at which speed of the projectile is one-third of the escape speed of the earth,

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