(a) Explanation Given info: Mass of the satellite is 3130 kg , and the velocity of the satellite on the surface of earth with respect to centre of earth is 463 m / s . 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 value of time period of orbit of geosynchronous satellite is equal to the time period of rotation of earth, that is 24 hr The expression for time period of orbit is, T = 2 π G M r 3 2 Here, T is the orbital period G is the gravitational constant M is the mass of planet r is the radius of orbit Rearrange above equation for r . r = ( T 2 G M 4 π 2 ) 1 3 Substitute 24 hr for T , 6.67 × 10 − 11 N ⋅ m 2 / kg 2 for G and 5.97 × 10 24 kg for M in the above expression. r = ( ( 24 hr × 3600 s 1 hr ) 2 ( 6.67 × 10 − 11 N ⋅ m 2 / kg 2 ) ( 5.97 × 10 24 kg ) 4 π 2 ) 1 3 = 4.22 × 10 7 m The formula for gravitational potential energy is, U = − G m M r (1) Here, U is the gravitational potential energy m is mass of object M is mass of earth. r is the distance between center of planet and the object. G is the universal gravitational constant Substitute 6.67 × 10 − 11 N ⋅ m 2 / kg 2 for G and 5.97 × 10 24 kg for M , 3130 kg for m , 4 (b) To determine The amount of kinetic energy gained by the satellite.

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Chapter 12, Problem 50PCE

(a)

To determine

The amount of gravitational potential energy required by satellite to attain geosynchronous orbit.

(b)

To determine

The amount of kinetic energy gained by the satellite.

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Chapter 12, Problem 49PCE

Chapter 12, Problem 51PCE

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The amount of gravitational potential energy required by satellite to attain geosynchronous orbit.

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The amount of gravitational potential energy required by satellite to attain geosynchronous orbit.

The amount of kinetic energy gained by the satellite.

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Chapter 12 Solutions