9. A 60.0-kg satellite is in orbit a distance 2000 km above the surface of the e

Question

9. A 60.0-kg satellite is in orbit a distance 2000 km above the surface of the earth. The mass of the earth is 5.976 x 10^24 kg and the radius of the earth is 6.378 × 10^6 m. a) What is the change in the gravitational potential energy if the satellite moves to a circular orbit 5000 km above the surface of the earth?

b) What is the change in kinetic energy if the satellite moves to a circular orbit 5000 km above the surface of the earth?

c) How much work must be done on the satellite to move it to a circular orbit 5000 km above the surface of the earth?

(a) 7.53 × 10^8 J (b) -3.76 x 10^8 J (c) 3.76 x 10^8 J

Explanation / Answer

Here,

mass of satellite, m = 60 Kg

height , h = 2000 km

mass of earth , M = 5.976 *10^24 Kg

radius of earth , r = 6.378 *10^6 m

a)

change in gravitational potential energy = G * M * m * (1/(h1 + r) - 1/(h2 + r))

change in gravitational potential energy = 6.673 *10^-11 * 5.976 *10^24 * 60 * (1/(2 *10^6 + 6.378 *10^6) - 1/(5 *10^6 + 6.378 *10^6))

change in gravitational potential energy = 7.53 *10^8 J

b)

as the change in kinetic energy = -change in potential energy/2

change in kinetic energy = -7.53 *10^8/2

change in kinetic energy = -3.765 *10^8 J

the change in kinetic energy is -3.765 *10^8 J

c)

work done on the satellite = change in gravitational potetnial energy + change in kinetic energy

work done on the satellite = 7.53 *10^8 -3.765 *10^8 J

work done on the satellite =3.76 x 10^8 J

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